Algebra 1 Full Course

Hello and welcome to algebra one lesson one in this lesson we're going to learn about variables and expressions so now that we've completed our pre-algebra course and we have a great command over arithmetic we're going to move on to some tougher material now generally when you finish a pre-algebra course or when you first Start an algebra 1 course you begin learning about something known as a variable okay so what a variable is it's a symbol generally a letter and it's going to be a lowercase letter used to represent an unknown quantity okay another way you

can kind of think about this it's going to be a placeholder so i put here that x y and z are the most commonly used in algebra but you Can use any lowercase letter that you want just for right now these are the ones that you're pretty much going to work with x y and z all right so here's some examples so mark works at a car wash and earns 63 dollars per day plus tips so if i wanted to kind of model this mathematically and say how much does he earn every day well i

know that he earned sixty three dollars at minimum right sixty three Dollars per day so let me start by putting sixty three dollars down now we have an additional amount we have this keyword here that says plus and then we have tips we don't know what the amount of tips is going to be right just like a waiter or anybody who works in the service industry that gets tips you don't know what you're going to get in tips that day you might get a customer that overly Tips or a customer that undertips so that amount's

going to vary so we can represent this unknown with a variable so i'd put 63 dollars plus and let's use the common variable of x so plus x so this represents the amount the amount of tips now we can use this to model what this person mark is going to Receive every day he gets 63 plus x where x equals the amount of tips that he's going to receive so let's say the day is over and he knows that he got seven dollars in tips so x in this case would equal seven if i want

to know how much he made for that day all i would do is i would take this right here this seven and i would plug It into the x right here because x equals seven he got seven dollars in tips x was the amount of unknown tips he was going to get and so we plug in a 7 for x and we'd end up with 63 dollars plus seven and i'll put dollars there just to be consistent and this will equal 70 dollars right that's how much he received for that day but the next day

it's going to be different so let's say The next day x is going to equal 9. well then i'd take 63 and i'd add 9 to that right i'd make 2 more dollars or i'd have 72. so the following day you would have 63 dollars plus and let me do this in a different color so we don't get confused nine dollars right so in that case x is equal to nine so we see that as we change the value for x the value Is going to change in this case x is equal to 7 i

plug in a 7 for x and i get 70 right 63 plus 7 is 70. and the other case x equals 9. so i plug a 9 in for x 63 plus 9 is going to give me 72 so it earns 72 in this case let's take a look at another one in case you're a little confused so a gallon of milk costs an unknown amount Jason wishes to purchase two gallons well again we can use a variable to model this situation so i know that he's gonna buy two gallons so that's two and it

costs an unknown amount so i can use a variable let's use y in the scenario to represent that so i need to multiply the cost per gallon times the number of gallons the number of gallons is 2 the cost per gallon is unknown so i would Have 2 times y which we're going to write as just 2y okay when a number is next to a variable it implies multiplication so 2y like that is telling me that i have 2 times y or 2 times y you remember back in pre-algebra when i said we stopped using

this symbol to imply multiplication and the reason is because it often gets Confused with the variable x so we'll either use this symbol or we'll use parentheses or replace the number next to the variable and this is the most common you'll just have 2 times y like this just 2y that implies multiplication now if i give you a cost per gallon of milk i can tell you how much jason's going to spend on his milk purchase right if i say that y is equal to One dollar and i'm not going to put the dollar symbol

in this time i'm just going to put one well then really all i'm going to do is plug a 1 in for y right this is the unknown the unknown cost per gallon of milk if i tell you that it equals one dollar i could just plug a 1 in there and i Could say okay well 2 times 1 is 2. so 2 times 1 is 2. so he spends 2 on 2 gallons of milk if that amount changes right if it changes if i say well the price of milk just went up so let's

say that now y is going to be equal to 3 okay we had a big price increase well now i'm going to plug in a 3 for y so instead of two times one i Would have two times three and that's going to give me six so now he's going to spend six dollars on two gallons of milk right and that makes sense two times three is six so if jason buys two gallons of milk at three dollars per gallon he spends six dollars in the other scenario if he buys two gallons of milk at

one dollar a gallon he's going to spend two dollars all right so let's get into some more kind Of definitions a term is a single number variable or a number times one or many variables so for example you might see something like 4x okay that's a term or you might see something like 9 y that's a term or you might see something like 24 x y z Or you might see something like negative 3 a these are all examples of a term right a single term again it's a number by itself so if i just

had let's say the number seven that's a term or it's a number times one or many variables so in this case you have one variable one variable one variable in this case you have many variables and these variables can be raised to powers So for example i could have had an example of a term where i have 24 x squared y to the fourth z to the eighth let's say n to the 12th that's a term right a single term now here's another kind of definition thing when a number is next to a variable or

variables it is known as a coefficient so in all these scenarios where i'm saying okay 2 times y right 2y is how we would say it but 2 times y really because it's multiplying 2 is the number that's multiplying the variable y it's known as a coefficient 2 is the coefficient of y if i had let's say 4 x squared z 4 is the coefficient right it's the number that's multiplying the variable Or again variables if you have more than one so this is again a coefficient and these are kind of very basic definitions that

even if you don't get them in this video you can write them down and put them on flashcards if you want we're going to talk about them over and over and over again by the time you get to the 10th lesson you will know all of these terms right it won't be an issue for you all right so a number that is hanging out on Its own is known as a constant so you're going to get to a point in this lesson where we start talking about algebraic expressions and you might see something that looks

like 5x plus 7. so this number this 7 that's hanging out by itself is a constant this is a constant its value is not going to change it's what we've worked with in pre-algebra right or we're just Doing basic operations if i do 7 times 3 that value is always going to be 21 right it's not going to change because we're working with constants if i have 5x plus 7 well i don't know what the value of x is again it's allowed to vary it's allowed to change and so the value for this is going

to change based on what x is if i change x i change the value of this whole algebraic expression So that's kind of the difference here that's why we say this is a constant because the number seven is not going to change if i have something like three x squared minus two two is hanging out by itself two's value is constant it's not going to change so that's why we refer to it again as a constant alright so here's a pretty important definition an algebraic expression is one or more Terms separated by plus or minus

symbols so again your terms in an algebraic expression are separated by the plus and minus symbols very important so let's look at an example let's say we have something like 5x plus 2y minus 4z so if you want to look at that algebraic Expression and say where are the individual terms well every time you see a plus or minus sign you've cut off a term so this is a term this 5x because there's a plus here right that cuts off this term from this term then this 2y is a term right you have a minus

sign here and you have a plus sign over here so it kind of it kind of separates everything let me just kind of draw a line here in A different color this kind of separates everything and then here where you have 4z again that's a term so you have 5x 2y and 4z that make up the terms here and it's called an algebraic expression because there are variables involved if we just kind of have numbers involved we just refer to that as an expression right so if you had something like you know 2 times 3

that's an expression right we work with That in pre-algebra so 2 times 3 is 6. all right so the main thing about an algebraic expression is that the value will change as its value given for the variables involved are changed so let me give you some examples let's say we had something simple like two x plus three and i say i wanna let x let x Equal four well what i'm going to do is everywhere i see an x in the algebraic expression i'm going to plug in a 4. so i'm going to take this

value and plug it in here everywhere i see an x so that means that i'm going to have 2 times 4 plus 3. and again i can put parentheses around 4 And put it next to 2 and it implies multiplication remember 2x is the same as 2 times x so when i replace x with 4 i'm still going to do the multiplication operation but now since it's not a variable it's a number that i replace the variable with i can just calculate right 2 times 4 is 8. so this would be 8 plus 3 and

then 8 plus 3 is 11. and again i did the multiplication first because of the please excuse my dear aunt sally right Remember your order of operations this is where you're really going to use them all right so let's say we change the value of x let's say let's say we want to let x now equal i don't know let's say 5. so instead of plugging in a 4 for x now i'm going to say well 2 times 5 plus 3 is going to equal what again i'm plugging in for x whatever value i let

x equal so That's why there's a 5 there and so we just calculate 2 times 5 is 10 then 10 plus 3 is 13. and i did that in one step versus 2 up here hopefully you can follow right 2 times 5 is 10 then 10 plus 3 again is 13. so as i change the value of x the value of the algebraic expression will change let's do one more let's say again two x plus three let's let x equal Negative 2. so i'm going to plug in a negative 2 for x right very very

simple so 2 times negative 2 plus 3 equals what two times negative two remember positive times negative is negative two times two is four so that's gonna give me negative four then plus three and at this point we know how to add integers right negative four plus three Is negative one all right so again in each case we got a different value for the algebraic expression because we chose a different value for our variable x so in some cases we can simplify an algebraic expression and we're going to do this using the distributive property so

we learned this back in pre-algebra and to give you a basic reminder of this Let's say we had something like three times this quantity 6 minus 3. again because we have parentheses here i'm basically saying what is 3 times the difference of 6 and 3 right so 6 minus 3 is what that's 3 so 3 times 3 is 9. but we can use the distributive property to do this as well remember because multiplication is distributive over addition or subtraction so what i can do is i can distribute the 3 to each Term inside of the

parentheses remember if i have a single number that's a term right a single number that's a term so 3 times 6 would give me 18 but let's write it out like this and then minus you'd have three times three three times three i'll get the same answer as i did before right i'm going to get 9 either way 3 times 6 is 18 and we're subtracting away 3 times 3 which is 9 and 18 minus 9 is not all Right so either way you do it you get the same answer so we're going to use

that property here to simplify we have 3 times this quantity x minus 4. so what i'm going to do is i'm going to distribute this 3 to each term inside the parentheses so 3 times x is 3x then minus 3 times 4 which i can write 3 times 4 i can just say 12. right we know that's 12 at this point so i've simplified 3 times this Quantity x minus 4 as 3x minus 12. now i don't have a single number here as an answer because i still don't know what the value of x is

but i've simplified it a little bit or some people in some cases are going to want to write it this way and we'll learn about that later on in algebra 1 where we're going to reverse what we just did that's a process called factoring and again we're not going to talk about that Now that'll be in a much further lesson all right let's take a look at another one we have 4 times the quantity x minus 7 plus 2. so what i'm going to do is i'm going to multiply 4 times x and i'm going

to distribute so that'll be 4x and then we'll have 4 times 7 that's 28 and then plus 2. so at this point i don't know what 4x is right i don't have a value of x but i do know what negative 28 plus 2 is Right i can do that and again if i have minus 28 that's the same thing as plus negative 28 so that's why i said it that way so negative 28 plus 2 is going to give me negative 26. so we would have 4x minus 26. all right now let's talk about

like terms so like terms are terms with the exact same variable or Variables if there's more than one that have the exact same exponents all the variable parts have to be the same i can't stress that enough so for example 5x and 3x are like terms are like terms it does not matter what the coefficients are right remember that's the number that's multiplying the variable or again variables it only matters what the variable or Variables are in this case we have x in this case we have x it's the exact same variable and so we

have like terms let's look at 5x and 3y i have x and then i have y those are not the same not the same right they're different letters so they're not the same and so we do not have like terms these are not like terms what about 2x and then 3x cubed Well they both have the variable of x but they're not raised to the same power that's got to be the same everything has to be the same so this stops it from being like terms so these are not like terms not like terms what

about here we have 7x to the fourth power 4y to the fourth power well we have the same exponent notice how we have Each variable raised to the fourth power but the variable itself is different one is x one is y you can't have anything be different again the coefficients can be different 7 or 4 or whatever the number parts are that can be whatever the variable part has to be the same and in this case it's not this is x this is y they're different and so they're not like terms Here we have three

z squared x and then two x z squared so this one might look like they're not like terms but in fact they are the variables are presented in a different order but i want you to remember back from pre-algebra that multiplication can occur in any order right that's called the commutative property of multiplication so if i have you know 3 times 2 that's the same as 2 Times 3. in each case it's 6. so i can multiply variables in any order i can have z squared times x or x times z squared it doesn't matter

right so i have the exact same variables here raised to the same power i have z squared and z squared so put a check mark there and then i have x and x put a check mark there and again the coefficients can be different so because everything is the same we have like terms Like terms okay so let's talk a little bit about what we're going to do with these like terms all right so when we combine like terms the variable part stays the same so if i have x squared and x squared that doesn't

change we're just going to be adding and subtracting the coefficients or again the number parts All right so let's take a look at 2x plus 3x nice and easy problem one should imagine if you had two oranges and you added to that three oranges well everybody knows they'd have five oranges the easy way to do it is just to leave the variable the same just say okay i know i'm going to have x and just add the coefficients so 2 plus 3 is 5 that's 5x if you were to visualize what's going on 2x is

x plus x So x plus x and then 3x is x plus x plus x if you kind of view it this way it makes sense okay i have 1 2 3 4 5 of these guys 5x but you don't want to do that every time it's going to slow you down right so you just want to say okay the variable part is going to stay the same so that x is going to stay the same so i just write that first Then i'm just going to add the coefficient parts right so 2 plus 3

is what that's 5 so i end up with 5x and again if i have 2 of something plus 3 of the same thing i'm going to have 5 of something so 2 apples plus three apples gives me five apples two honda civics plus three honda civics gives me five honda civics right so on and so forth another way to kind of think about it using the distributive property and it Involves factoring which we haven't gotten to yet but if i were to write this for you like this let's say that i pulled the x out

which is common to both and i wrote this as x times the quantity 2 plus 3. well i could do 2 plus 3 first and get 5 multiply 5 times x ended up with 5x or i could again distribute the x x times 2 is 2x plus x times 3 which is 3x and again you you see where We start with this 2x plus 3x so kind of any way you want to use to visualize what's going on here it's 2 of something plus 3 of the same thing is giving you 5 of that thing

two apples plus three apples five apples two oranges plus three oranges five oranges you know so on and so forth here we have two x plus three y now we don't have like terms here so we can't combine them so let me try to make sense of this for you If i have two let's say oranges and i try to add that to three apples well i still have two oranges and three apples right i can't combine them and say okay well i have five orange apples right i don't in the case where we had

the same variable right we had like terms yeah i'm adding the same thing so if i have again two oranges plus three oranges i have five oranges if i have two apples plus three apples i have five apples but If i have two oranges plus three apples i just have two oranges and three apples so there's no way to kind of you know combine these they're it's still going to just be 2x plus 3y so if you don't have like terms you really can't simplify right in this scenario right here we have 4x squared plus

3x squared plus z now look for your like terms i have x squared and x squared So these two are like terms these are like terms and so we can combine them and i missed an r there the z is not common to anything right so we're not going to have like terms with just the z so what you do in this scenario is you just combine what you can so 4x squared plus 3x squared again leave the variable part the same so Leave that x squared part the same and then just add your coefficients

so what is 4 plus 3 that's 7. so i have 7x squared and then just copy the rest so plus c so when we simplify this algebraic expression we end up with 7x squared plus c right again you combine what you can right in this case we could combine the like terms 4x squared and 3x squared all right here's one that's a little bit More challenging we have negative 2 times the quantity 4x minus 7 plus 4 times the quantity x minus 3. so the first thing i'm going to use is the distributive property to

kind of simplify this a bit so i'm going to multiply negative 2 times 4x and that's going to give me negative 8 x and then i'm subtracting away negative 2 times 7 which is going to be negative 14. now i'd end up with minus a negative 14 which is the same as plus 14. now plus i'm going to do the same thing over here we have 4 times x that's 4x and then minus 4 times 3 that's 12. and so now we're going to look for like terms so i have negative 8x and 4x those

are like terms right the variable is x right there's no power you can think of the power as being one right so in each case you have the same thing and then you have 14 and negative 12. Those are just numbers or constants those can be combined so let's start out with just negative 8x plus 4x well the variable is going to stay the same so that's just going to be x and then think about the coefficients so what is negative 8 plus 4 well negative 8 plus 4 is going to give me negative 4.

so this is going to be negative 4x and then i'm just going to combine the Number parts so i have 14 minus 12 that's going to give me 2. so plus 2. so the answer here would be negative 4x plus 2. and i can't simplify this any further because i don't know what the value of x is so i can't combine this with 2 right i don't i can't really simplify this unless you give me a value for x if you said hey let x equal this then i could give you a value for the

algebraic expression but without that i'm kind of stuck here hello and welcome To algebra one lesson two in this video we're going to learn about equations so in our last lesson we learned a lot of definitions that were going to help us out throughout our study of algebra two of those definitions that we're going to review here one would be that of a term right a term is a single number or a number that is multiplied by one or many variables and those variables can Be raised to powers now we also have something known as

an algebraic expression and an algebraic expression is a single term right so it could just be one term or a collection of terms separated by plus or minus signs all right now let's move on and talk a little bit about equations so an equation is a statement that two algebraic expressions are equal all right so this next point is going to be very very important An equation always includes the equality symbol so always includes the equality symbol so when you've looked at the algebraic expressions we only put an equal symbol in once we get a

value for it right so as an example let's say i had something like 3x minus 4. this is an algebraic expression this is an algebraic expression [Music] now i can't really do anything with this when i have an algebraic expression i can simplify it or i can plug in for the variable if you give me a value so if i said let x equal 2 i could plug a 2 in there and i could evaluate right and in that particular case you might see an equal sign because i'm saying hey if x equals 2 this

algebraic expression is equal to this value right but it didn't start out With an equal sign right there's nothing there other than 3x minus 4 at this point so that's an algebraic expression an equation is going to look like this we have 2x is equal to 8. so that's the main difference if you're just starting out in algebra and you get a test and you see 2x equals 8 well okay it has an equal sign to start so this is an equation This is an equation that's your number one way to tell right it's got

an equal sign it's an equation you see something like 7x squared minus 1 okay no equal sign this is an algebraic expression so let's talk a little bit about equations now and we're not going to get to solving equations until the next lesson we're just going to learn some of the basics of equations now so that we Can just kind of lightly introduce the concept to ourselves all right so we start out with this 2x equals 8. so for an equation to be true the left and the right side must be the same value so

if i look at the right side the value is eight it's just a number the number eight if i look at the left side i have a two times x i have two times x so what you have to Ask yourself is what value what value can i substitute to get a value of 8. and why do i want a value of 8 well for an equation to be true again the left and the right side need to be equal the right side is 8. the left side is 2 times x so what i'm saying

is 2 times some number is equal to 8. Now can you think of a number that when you multiply by 2 you get 8 and almost all of you know that that number is 4. so in other words if i replace x with 4 i get a true statement meaning the left and the right side would both be 8 right they'd both be the same value so this is true if x equals 4. and the way you check it is you just Plug in just like we did in the last lesson when we're talking about

algebraic expressions now we're kind of carrying that over we're going to plug in a 4 for x and we're going to see if the left and the right side are equal so 2 times 4 should be equal to 8 and of course it is 8 equals 8. and when you do this type of replacement you're just looking for the same value on both sides so 8 and 8 that checks out If i had replaced x with anything else i would have gotten a false statement and the reason for that is there's no other number that

when you multiply by 2 you get 8. right it doesn't exist so if i was to pick let's say 3 for example let's say i said that x was equal to 3 well 2 times 3 is not equal to 8 right i'd get 6 equals 8. that doesn't make any sense right 6 And 8 are not the same they're not equal so this is wrong that's wrong okay so you're looking for the value for that variable that makes the equation true meaning the left and the right side will be the same in this case that's

going to be x equals 4. all right so now let's look at 3 plus x equals 5. so again i want a value for x that when i substitute it for x i get a True statement so 3 plus what 3 plus question mark gives me 5 gives me 5. that's what i'm looking for 3 plus what gives me 5. well we all know that that answer is 2. so if x is equal to 2 i get a true statement right if x equals 2 this equation is true and again just plug in for the

x so 3 plus 2 is equal To 5. the left side will become 5 3 plus 2 is 5 and that equals 5. i'm looking for the same value on the left and the right side of the equation okay if i would choose any other number here i would get a false statement let's say for example i said that x was equal to i don't know negative 1. so if i plugged in a negative 1 for x i put 3 plus negative 1 equals 5. 3 plus negative 1 is 2 so i'd Have 2 equals

5. so if you solve an equation and the left and the right side are not equal you did something wrong right you need the left and the right side to be the same this is wrong wrong the only value for x that's going to work is 2 right because 3 plus 2 is going to give us 5. all right let's take a look at a little practice exercise we want to determine if each is a Solution to the given equation so we're going to start out with x over 6 is equal to 2 and then

here are the values that we're going to try out as solutions so let me kind of write this down here we have x over 6 is equal to 2. let me scroll down and get a little room going so i'm going to start out with the value of 5. so i'm going to say is x equals 5 a solution Well to figure that out i'm just going to plug a 5 in for x so 5 plugged in for x over 6. does that equal 2 no it doesn't right it doesn't 5 6 is less than

1 so it can't be equal to 2 right so this doesn't work as a solution now the next value they want us to try is 1. so i would plug a 1 in for x i'm going to plug a 1 in there And i'm going to see if i get a true statement so 1 over 6 is equal to 2. is that true well no it's not 1 6 is not equal to 2. so that doesn't work as a solution either let's try the last one the last one is 12. so i'm going to plug

a 12 in for x now and see if that works out so we would have 12 over 6 equals 2. well 12 over 6 that's 12 divided by 6 That is 2. so the left side is going to become 2 and the right side is 2. so the left and the right side are equal they are the same value and so yes 12 is a solution here okay 12 is a solution here so we can say that x equals 12 is our answer all right let's take a look at another one we have 3 times

the quantity 4k plus 4 and this is equal to negative 84. so again what i'm going to do is take each Value here and plug it in for our variable in this case it's k and see if we get a true statement see if the left and the right side are equal if we do then we know that that value is a solution for the equation right so i'm going to start out with negative 8 and i'm just going to plug it in for k and so we'll have 3 times this quantity 4 times negative

8 then plus 4 And this equals negative 84. so i'm going to start out inside the parentheses with 4 times negative 8 that's negative 32 and then when i'm done with that i would continue inside the parentheses i would add 4. negative 32 plus 4 is negative 28 so this would be 3 times negative 28 and this equals negative 84. all right so i know that 3 times Negative 28 would be negative and then 3 times 20 is 60 3 times 8 is 24 so 60 plus 24 is 84 so this would be negative 84

over here and over here it's negative 84. so that checks out so negative 8 is a solution for this equation okay this is a solution you could say k is equal to negative 8. now we'll try the other ones but they're Going to fail so let's go ahead and erase this real quick put a check mark right here let's go with 7. let's see if that works out for us so we're going to put 3 times the quantity 4 times instead of k i'm plugging in 7 plus 4 equals negative 84. 4 times 7 is

28 then 28 plus 4 is 32 so i'd have 3 times 32 And that's going to equal negative 84. now without doing the multiplication i already know this isn't going to work because this is negative over here and this is two positives being multiplied together so even if the number part worked out which it's not going to we'd end up with a positive on one side and a negative on the other so we know it's going to fail we can go ahead and say well what is 3 Times 32 3 times 30 is 90 3

times 2 is 6. so this is 96 and that's not equal to negative 84. that's that doesn't work right this is wrong so 7 is not a solution for the equation we can go back up here and i'm going to mark that out that didn't work now the last value we're going to try is 0. so we're going to plug in a 0 for k 3 times this quantity 4 times 0 plus 4 equals negative 84. all right so 4 times 0 is 0 then 0 plus 4 is 4. so i'd have 3 times 4

is equal to negative 84 and obviously that's not going to work 3 times 4 is 12 and that does not equal negative 84. so the left and the right side are not the same value right i have negative 84 on the right i have 12 on the left not the same so this doesn't work this is wrong And so zero is not a solution so we saw that only k equals negative eight is what works in our equation right and it works because when we plug in a negative 8 for k we get the same

value on the left and on the right side again that value is negative 84. all right let's take a look at another one we have n squared equals 8n so here's our proposed solutions we have Negative 4 we have 0 and we have 8. so let's start out with negative 4. so i'm going to plug that in everywhere i see an n you have to be careful because in a lot of cases we're just plugging in once here we have two ends that we need to plug in for we have to plug in there and

also there i'm going to start out by putting parentheses around negative 4 and then i'm going to square that and this equals 8 times negative 4. Notice how i use parentheses when i'm substituting in very important here because i have a negative number that is raised to a power and i don't want to make a sign mistake right whatever value i have for n in this case that's negative 4 i want to square it so i need to make sure i use parentheses okay so negative 4 squared is going to be positive 16. and we're

saying this is equal to 8 times negative 4 which is Negative 32 this obviously does not work out and so this is not a solution right negative 4 doesn't work let's try zero so i would plug in a 0 for n so i'd have 0 squared equals 8 times 0. and that is in fact going to work as a solution 0 squared is 0 and 8 times 0 is also 0. so the same value occurs on the left and the right side that value is zero and so this is a solution right this is a

solution All right let's try eight now so we have 8 that we're plugging in for n that's going to be squared and then this is equal to 8 times 8. so 8 squared is 8 times 8 that's 64. and this is equal to 8 times 8 that's 64. so that works also we get 64 equals 64. so 8 is also a solution to the equation all right let's take a look at one last problem so we have x minus 6 over 4x Minus 5 over 4x equals 1 over x so we have a lot of

places we need to plug into so i have two proposed solutions i have negative 1 and then i have 15. so for negative 1 we're going to plug that in here here here and here so everywhere there's an x so i would have negative 1 Minus 6 over 4 times negative 1 then minus 5 over 4 times negative 1 and this should be equal to 1 over negative 1. so let's crank this out real quick so negative one minus six is negative seven so that's negative seven over four times negative one that's negative four then

we're subtracting away We have five over four times negative 1 is negative 4 and this should be equal to 1 over negative 1 which is negative 1. so if we look we have a common denominator right now and so we can just say what is negative 7 minus 5 i just work with the numerators well negative 7 minus 5 is negative 12. and so if i had negative 12 over negative 4 negative 12 over Negative 4 that's positive 3. that's positive 3. positive 3 is not equal to negative 1. so this does not work all

right let's try 15 now and again everywhere i see an x i'm going to plug in a 15. so i'd have 15 minus 6 over 4 times 15 then minus 5 over 4 times 15 this should be equal to 1 over 15. so let's crank this out real quick so 15 minus 6 in the numerator is 9 Over 4 times 15 in the denominator that's 60 then minus 5 over 5 over 4 times 15 again 4 times 15 is 60 and this is equal to 1 over 15. so we have the same denominator here what

is 9 minus 5 that's 4. so we get 4 over 60 is equal to 1 15. so at this point it doesn't seem like we have the same value on each side but if we were to remember that we simplify Our fractions remember that 60 is divisible by 4. 60 divided by 4 is 15. so this i could cancel and make 15 cancel and make 1. so really what we end up with if we simplify that is 1 15. so we have the same value on the left and the right hand side and so 15

is a solution for this equation so we can go back up here and we can just put a check mark here yes x equals 15 is a solution here hello and welcome to Algebra one lesson three in this video we're going to learn about the addition property of equality so in our first two lessons of our algebra one series we learned some important vocabulary that's going to help us throughout our course the first thing we kind of talked about was that a variable is a symbol and generally it's a letter a lowercase letter like x

or y or z or or any one that you Want to use that represents some unknown value then we kind of moved on and we talked about something else known as a term now a term is a single number or it could also be a single variable or it could be a number times one or many variables where your variables are possibly raised to powers so as an example a term could be again just a number so let's say 7 Or it could just be a variable let's say y or it could be a number

times one or many variables so i could have something like 7x or i could have 2 x y where i can have 3 x squared y cubed z something like that each is an example of a term now kind of the next thing we talked about was an algebraic expression so this is a single term okay so a single term by itself Or it's a collection of terms separated by plus or minus symbols so for example let's say i had 7x minus 2. so i have a term here and a term here and it's separated

by a minus symbol so this is an algebraic expression or i could just have a single term i could just have the number three right that's also an algebraic expression or i could have something much more complex let's say i had 4x squared minus 2 y Plus 17. again each term okay each term is separated by the plus and minus signs so this is a term 4x squared 2y is a term and 17 is a term so the next we kind of talked about equations so i want you to recall that an equation is a

statement that two algebraic expressions are equal now we fully understand what an algebraic expression is at this point let's say i had something like 2x plus 3 2x plus 3. again we know that this is an algebraic expression this is an algebraic expression right i have a term 2x that's added to another term that's 3 right that's following the definition of an algebraic expression and let's say this is equal to another algebraic expression just the number 11. and again this can also be an algebraic Expression because an algebraic expression again is a single term or

it's a collection of terms separated by plus or minus symbols so in each case we have an algebraic expression and now we've kind of thrown into the mix this equals sign so what this tells me is that the left side is the same as or is equal to The right side you can kind of think of some definitions of things being equal in your life if we have equal pay that means we make the same amount and so our ultimate goal when we're solving an equation is to find the value or in some cases the

values that when substituted in for the variable or in some cases you'd have variables that makes the equation true it means making the equation to where the left Side and the right side are the exact same value and again let me kind of further clarify this i'm telling you that you can plug something in for x and simplify over here and the left side would also be 11 right at this point 2x plus 3 doesn't look like it's the same thing as 11. but if i plug in the number 4 for x let's see what

happens on the left i would have two times four Again all i did was i replaced the variable x with four plus three equals eleven so two times four will be done first because of our order of operations 2 times 4 is 8 so we'd have 8 plus 3 equals 11 and then on the left side 8 plus 3 is 11. so 11 equals 11. so when i replace my variable x with the number 4 what happens is through simplification i End up with the same exact value on the left as the right and so

our equation we'd say is true now because the same value occurs on the left and the right we can say we have a true statement now 4 is the only value that happens to give me a true statement if i replace x with anything else it would be false so let me show you an example let's say that i said okay i think x is Equal to 7 and i plug in a 7 for x what's going to happen here well you probably can guess that i'm not going to have the same value on the

left and the right 2 times seven would be done now plus three equals eleven two times seven is fourteen fourteen plus three equals eleven fourteen plus three is seventeen seventeen is not equal to eleven Seventeen is not the same value as eleven right they're not equal if i made 17 and you made 11 we don't have equal pay so at this point we can say that this is false right this is a false statement because 17 does not equal 11. and right here i'm saying that they are equal so it's kind of a contradiction right

17 is not equal to 11 so we say it's a false statement So again only x equals 4 is a solution to this equation so again officially if you want to write this down in your notes a solution to an equation is any value that makes the equation true when it replaces the variable so in the last example we saw that if we replace the variable with 4 we got a true statement the left and the right side were both 11. if we replace the variable with anything Else we won't get the same number on

each side and so it's not a solution all right so so far up to this point i've been giving you solutions to our equations that we're working with and we've kind of just plugged in and we've seen if it works or if it doesn't work but how do you actually go about getting the solution well what we're going to focus on right now we're going to focus on Very very simple equations so the first chapter in your textbook on solving equations is going to be dedicated to solving linear equations and it's going to be linear

equations in one variable so one variable is involved and they look like this so you might see this in your textbook ax plus b equals c and x is our variable but really you can Use anything i could write this as a y plus b equals c or i could write a z plus b equals c you can use anything as your variable it doesn't matter but typically we use x so that's why this is presented like this in your textbook and i want to talk a little bit about kind of some of these letters

that we're using here a b and c the first thing is that they can be any real numbers and we haven't really Defined what a real number is but just kind of anything you can think about at this point fractions decimals negatives you know integers anything you've worked with can be plugged in for a b or c with one restriction we cannot have a equal to 0. and the reason for that is if i plug a 0 in for a i would have 0 times x and 0 times anything is 0. so if i plugged

in a 0 for a 0 times x is 0 That's gone and so i don't have a variable anymore right i just have b equals c and that's only true if b and c are the same number right so that doesn't work for us so a b and c can be any real number only a can't be zero now b could be zero c could be zero okay that restriction is just on a let's talk a little bit about what a is called so a is the number that is multiplying your variable when a number

is Multiplying a variable it's called a coefficient a coefficient and that's very important to write that down or memorize it in some way because that's going to come up a lot so if i had something like negative 7x we say that negative 7 the number that's multiplying x is the coefficient of x or if i had let's say 32z 32 is multiplying the variable z so 32 is the coefficient of z now then we have plus b equals c now there's no variable involved in b or c so when that happens we call these constants

so this is a constant and this is called a constant okay so this is kind of a number hanging out by itself so for example if i had something like negative 3 x Minus 7 equals 2 this negative 7 and this 2 those are both constants and then this negative 3 here the number that's multiplying x that's a coefficient right that's the coefficient of x all right so today we're going to kind of start off let me kind of erase this i'm going to show you what we're going to be working through and then we're

going to jump right in So we're going to start off with a very simple type of linear equation one variable it's where a equals 1 so a is 1 so you would have 1x and 1 times something is just itself so if you see 1 times x or 1 is the coefficient of x you can just write that as x okay you don't need to put the 1 next to it and majority of the time you're not going to so then plus b equals c so what i'm saying is you're going to see x Plus

some number equals c so let's say x plus 3 equals 7 or x minus 4 equals 2. remember if i have minus that's the same thing as plus negative right i could write that as x plus negative 4 equals 2 if i wanted to all right so without further delay let's talk a little bit about how we solve an equation so in order to solve an equation of the form that i just showed you we're going to use three properties so the first one is very very simple It's the fact that a number plus its

opposite is always 0. so for example something like 3 plus negative 3 is 0 right the opposite of 3 is negative 3 or negative 141 plus its opposite which is positive 141 is also 0 or 10 360 plus the opposite of that which is negative 10 360 is zero very very simple property but you need to remember that Now the other one is that if you add zero to a number it remains unchanged it's called the additive identity right so zero when added to a number leaves the number unchanged so for example five plus zero

equals five six plus zero equals 6. x plus 0 equals no different because we have a variable x z squared plus 0 equals Z squared okay so very important that you understand those two properties now the most important property that you're going to learn and you definitely want to write this down is called the addition property of equality and in fact that's the title of this video so this tells us that we can add the same number to both sides of an equation without changing the solution so here's the key Without changing the solution so

i'm going to change what the equation looks like but the solution meaning the number that i plug in for the variable that makes it true is not going to change so my original equation that i have has the same solution as the modified one that's very important because i want to know that the answer that i get in the end works in the original equation so let's start off with a basic example We have x plus 4 equals 10. and in the world of solving equations it doesn't get any easier so i want to start

out with just telling you that your ultimate goal your number one goal is to isolate the variable so i'm always going to start out by just highlighting this or circling it so that i can mentally prepare myself to say hey i want to isolate x so what i mean by isolating x my goal at The end is to have x by itself on one side of the equation it could be on the left it could be on the right but nothing else can be over there with x i want x to be equal to some

value to some number this is going to end up being my solution x equals some number that number is the solution the number that i can plug in for x and it should result in a true statement Provided i did my work correctly so how do we go about isolating x how do i get x by itself on one side well i want you to look at what's being done to x right now we have 4 that's being added right so x plus 4. and think back a moment ago when i told you that if

you add 4 if i have 4 plus negative 4 i get 0 right i get 0. so what happens if i add negative 4 to The left side of the equation well let's do that real quick so we would have x plus 4 plus negative 4 equals and then over here we have 10. now on the left side of the equation if i simplify 4 plus negative 4 is 0. so i'm going to rewrite this as x plus 0 equals 10. now x plus 0 as i just told you if you Add 0 to a

number it remains unchanged same thing with a variable so x plus 0 is just x right if i was to simplify so now x is by itself i've met my goal to isolate but there's a problem i can't just go around adding stuff to one side of the equation to make it legal to maintain my solution i've got to do it to both sides otherwise it will not work so i added negative 4 over here to the left side And i'm going to do the same thing on the right side so plus negative 4. so

now that i've done that i'm going to maintain my solution meaning the original solution for x plus 4 equals 10 which you can probably guess is 6 at this point will still be true when i get down here throughout the steps your equation is going to look different each time but it's going to maintain that same solution which is what we need So if i add negative 4 and 10 in this step here instead of writing 10 i'm going to have 6 and then down here i would write 6 again because x plus 0 equals

6 then we have x equals 6. so once i get to this point where x equals some number i have my solution i have my value that when it replaces the variable it gives me a true statement so x equals 6. so we already know how to check and see if we got the Right answer i take a 6 i plug it back in for x and i see if i get the same value on the left and the right side so let me erase this and i would just plug in a 6 for x

so i would have 6 plus 4 equals 10. we all know that 6 plus 4 is 10. so the value on the left is 10 the value on the right is 10 and so we get a true statement right so x equals 6 is the correct solution here right because again when i plug a 6 in For x i end up with the same value on the left as on the right now let's say for example you didn't get the right answer let's say you goofed up and you understood the principle here but you missed

the step let's say you did x plus 4 and you subtracted 4 away or you added negative 4 again those are the same thing so sometimes you'll see me put minus 4 sometimes you'll see me put plus negative 4. Same exact thing then equals let's say you just put 10. and i forget to add negative 4 on this side so this cancels and i have x and now i'm saying x equals 10. well x doesn't equal 10. but when you go back and check it you're going to figure out hey i made a mistake 10

plus 4 is 14. 14 is not equal to 10. so that's how you catch your simple mistakes especially in the beginning you take your value plug it back in for the Variable and see if the left and the right side are equal very important to check your work at this stage of the game all right let's take a look at another one all right let's say we have x minus 7 equals negative 12. so again our goal is to isolate the variable we want to isolate x and let me highlight it so what's being done

to x right now i always think about okay what's being Done well i'm subtracting 7 away i want x by itself so remember i can use that trick where i have x plus 0. well if i have x minus 7 and it might make you more comfortable to write that as x plus negative 7 what can i add over here to get x plus 0 right because we know that simplifies to x and i'd have x isolated on one side remember if i had the opposite of negative seven which is Positive seven negative seven and

positive seven would give me zero so plus seven okay then equals negative twelve but remember i can't just go around adding whatever i want to one side of the equation if i'm going to do that i've got to do it to the other side so that i maintain my solution so plus 7 over here as well so now we've got it on this side and This side so we are good to go all right let's scroll down a little bit and we'll have x plus negative seven plus seven so we know that negative seven plus

seven is zero so this would be x plus zero and this equals negative 12 plus 7 which is negative 5. i can simplify x plus 0 to just x and this equals negative 5. and in the future equations i'm going to stop writing x plus 0. we know that x Plus 0 is just x so i'm just going to line out this and then we're going to just say okay that's x now all right so we get our solution which is x equals negative 5 and we can check it by plugging in a negative 5

for x so if i plug a negative 5 in there i'd have negative 5 minus 7 equals negative 12. negative 5 minus 7 is negative 12. we get negative 12 equals negative 12. And so yes my solution here of x equals negative 5 is correct all right what about x plus 5 equals 2 well again i want to isolate my variable x so i want to isolate isolate and so let me highlight this guy right here and in order to do that again if i'm adding 5 to x what can i do to make that

5 go away Well if i have 5 and i add negative 5 or i subtract away 5 again that's the same thing that's going to go away right that 5 will disappear i'll just have x so if i have x plus 5 plus negative 5 or i can write minus 5 but again if i'm doing that over here if i'm subtracting 5 away i've got to do the same thing over here i've got a minus 5 over here as well so now i can go through and simplify so 5 minus 5 is 0. so i'm

not going to Write x x plus 0. you know that's just going to simplify to x so i'm just going to cross this out and just put that this is x and this equals 2 minus 5 is negative 3. so we get x equals negative 3. again you can plug negative 3 back in for x and show that we get a true statement so you'd have negative 3 there plus five equals two so negative three plus five is the same thing as five minus three and we know that's two so we get two Equals two

and so again we can say our solution here x equals negative three is correct all right let's take a look at the final problem here so we have z minus 9 equals negative 10. so now my variable is z and i always want to isolate it and i know a lot of students get confused when they see something other than x at first but again your variable can be whatever In this case it's z that's our unknown now to isolate z we look at what's being done to z so right now i'm subtracting 9 away

you can think about that as adding negative 9 if that's more comfortable for you so we'll have z let's go ahead and put plus negative nine and if i think about this side only forget about this for a second how can i make this negative nine go away well again If i have negative 9 and i add the opposite of that which is positive 9 negative 9 plus 9 would be 0. so over here this would go away and i would just have z but to make it legal i've got to also do that same

action which is to add 9 to the right side so in other words i added it over here i've got to add it over here otherwise i'm going to end up messing up my solution we need our solution to stay The same so we get the right answer so over here now i would have z is equal to negative 10 plus 9 is negative 1. so that's my solution z equals negative 1. now again we plug back in for z in the original equation let me erase this and we'll say instead of z i'll have

negative 1 minus 9 equals negative 10 and negative 1 minus 9 is negative 10 negative 10 is equal to negative 10 same value on each Side of the equation and so we can say that our solution z equals negative one is correct hello and welcome to algebra one lesson four in this video we're going to learn about the multiplication property of equality so in the last lesson we finally started talking about how to solve an equation and we started talking about the first type of equations that we look at and those are linear equations in

one Variable so let me kind of redefine that for you so a linear equation one variable looks like this so it's ax plus b equals c again your variable doesn't have to be x but again that's the most commonly used in algebra so we kind of show that when we show a generic formula now i also told you that a b and c could be any real number that you could think of i had one exception to that Your value for a which is the coefficient of x cannot be zero so a cannot be zero

because again if zero multiplies x x is going to disappear right zero times anything is always zero so we can't have x dropping out because this is an equation and we need to have a variable so other than that again a b and c can be anything that you want it to be now specifically in the last lesson we Looked at the case where a was one right we had one x plus b equals c 1 times x anything times 1 is just itself so really we saw x plus b equals c so x plus

some number equals some other number so as an example here we have x minus nine equals negative two And i can rewrite this as x plus negative nine equals negative two so x plus some number equals some number so we learned how to solve an equation of this form basically we do it in one step using something known as the addition property of equality the addition property of equality tells us that we can add or also subtract the same number 2 or in the case of Subtraction from both sides of an equation without changing the

solution that's the key as i go through the steps to solve an equation i need my solution to stay the same because if i get a solution at the end and it doesn't work in the original equation what use is it so the key is i can add the same number again to both sides of the equation and it doesn't change the solution now the way we use that our goal when solving an Equation is to isolate the variable in this case that would be x now if i look on the left side of the

equation i have this plus negative 9. i've got to think of a way to get rid of that because i just want x by itself i want x equals something so i can have a solution remember that if i add negative 9 and positive 9 i get 0 and if i had x plus 0 that would simplify to just x So what i'm going to do is on the left side i'm going to add the opposite of negative 9 which again is positive 9 and because i do that to the left side i must also

do it to the right side so i added 9 here and i added 9 here and now on the left side negative 9 plus 9 is 0. so i would have x plus 0 x plus 0 which simplifies to just x On the right side i'd have negative two plus nine which is seven and so i'd have x equals seven as my solution and if you want to you can take a seven and plug that in for x in the original equation and say 7 minus 9 equals negative 2 and that's true the left side

would simplify to negative 2 and the right side is negative 2. so the same value is on both sides of the equation and so we know that the solution x Equals 7 is correct now suppose i give you something like 3x equals 9. how do we isolate x here again this is our goal to isolate x well now you'll notice that i don't have anything that's being added to x now i have something that's multiplying x so how do we isolate x in this case well we need to think about a few different properties first

and then we're going to revisit this problem so the first thing Is that a number times its reciprocal is always 1. so we think about reciprocals when we think about fractions right if i have something like two-fifths and i multiply by the reciprocal of two-fifths remember i take the denominator and i put in the numerator and i take the numerator and put in the denominator so a number like two-fifths times its reciprocal which is Five-halves will always be one and why is that the case well if i cross-cancel this five will cancel with this five

and give me one this two will cancel with this 2 and give me 1. so i basically have 1 times 1 which equals 1. now this is always true a number of times this reciprocal is always 1. and you can see a kind of more advanced example of that let's say i have the number 4. what is the reciprocal of 4 remember you could write 4 as 4 over 1 and then the reciprocal would be take 1 and put it in the numerator take 4 and put in the denominator and again everything's going to cancel

this 4 is going to cancel with this 4. you really think of this 1 canceling with this 1 i mean it's just going to be 1. so you have 1 times 1 which equals 1. so again a number times its reciprocal is always 1. all right another fact that You might want to think about is that a non-zero number divided by itself is also always one so for example 10 divided by 10 equals one 1037 divided by 1037 is 1. any non-zero number divided by itself is 1. you have to exclude 0 in that because

you can't say 0 divided by 0 equals 1 because you can't divide by 0. right 0 can never be in the denominator so That's why we say any non-zero number divided by itself is one so there's one more thing i want you to note and that's that if you multiply a number by one it remains unchanged so it doesn't matter what it is so 17 times one equals 17. negative 15 times 1 equals negative 15. so x times 1 equals x or z squared times 1 equals z squared or if I wrote it a different

way and put 1y that's just equal to y right that's just equal to y so a lot of times we'll see 1 is the coefficient of a variable and we just write that variable without the 1 right because it's the same thing all right so now let's talk about the multiplication property of equality so we can multiply or divide both sides of an equation by the same non-zero number and that's important 0 is not involved here And maintain the same solution if you use 0 to multiply remember if you multiply 0 by something it becomes

0. so 0 is not going to work for our multiplication property of equality has to be the same non-zero number all right so let's revisit our 3x equals 9. so again our goal is to isolate the variable x so we want to isolate this and what can we do to get x by itself well I have 3 multiplied by x equals 9. i want you to think back to what i just said a minute ago a number divided by itself is one so if i divided if i divided the left side here by the coefficient

of x which is three if i divided this by three and to make it legal divided this by 3. what would happen over here is i would have 3 divided by 3 which is 1 And then 1 times x would just be x right this would be 1x equals 9 divided by 3 is 3. 1x equals 3 or x equals 3. so when you have a number as we call it a coefficient that that is multiplying a variable in this format all you need to do to isolate the variable is divide both sides of the

equation by that number right by the coefficient of your Variable another way you can think about this is i could think of three as three over one let's say i had three over one times x equals nine so if i have three over one again a number times its reciprocal is always going to be 1. so i could have also said okay i'm going to multiply this side by the reciprocal of 3 over 1. the reciprocal of 3 over 1 is 1 over 3 or 1 3 and i'm going to multiply this by 1 3.

And you can see that multiplying by a third and dividing by three that's the exact same thing if i multiply by a fourth or i divide by four that's the same if i multiply by a fifth or i divide by five that's the same so what we get here is this 3 cancels with this 3 and i'll basically have 1 times x or just x equals this 9 will cancel with this 3 and give me a 3 3 times 1 is 3. so x Equals 3. and again we always want to check our solution we

might have made a mistake somewhere so take the 3 and plug it back in for x you'd have 3 times 3 is equal to 9 and then over here on the left 3 times 3 is 9. so you get 9 equals 9. same value on the left as the right so yeah we can say that our solution here x equals 3 is correct all right what about 2 3 x equals 10. So the coefficient of x what's multiplying x here is two thirds so how do i isolate x in that scenario well again if i

want to get x by itself one thing i can do is i can multiply both sides by three halves that's the reciprocal of two thirds remember if i have two thirds and i multiply this by three halves that's going to become one now the only other thing over here is x so times x So this is going to cancel with this and give me one this is going to cancel with this and give me 1. so i'll have 1 times x or just x or you can write 1x if you want and then simplify that

just to x doesn't really matter but over here i've got to make sure that because i multiplied by 3 halves on the left side i do it to the right side otherwise you won't get the correct answer so 10 times 3 halves this 10 would cancel with the 2 and give me a 5. 5 times 3 is 15. so 1x equals 15 and again this will simplify to x equals 15. so let's check this x equals 15 real quick so i'm plugging a 15 in there and i'd have two thirds times 15 equals 10. so

this 15 would cancel with this 3 and gave me a 5 5 times 2 is 10. so i'd get 10 equals 10. again the same value on the left as the right so x equals 15 is correct so one thing i just want to Bring your attention to when you're in your textbook on this section about the multiplication property of equality you basically have two different scenarios you have a fractional coefficient for x so like this one two thirds times x equals ten and i'm just writing this out like this just for us to be

able to visually see what's going on when you have a fractional coefficient it's going to be faster for you to Multiply both sides by the reciprocal of that fraction if i wanted to i could divide both sides by two-thirds i can divide both sides by the coefficient of x i can always do that so if i divide two-thirds by two-thirds that's going to be what two-thirds divided by two-thirds the first fraction stays the same the second fraction we take the reciprocal that's three halves and we multiply remember we we did that Before we multiplied by

three halves so i'm kind of saving time because this cancels this cancels right this becomes 1. so we see that this would become 1 right so you just have x on this side now over here 10 divided by 2 thirds so 10 divided by 2 thirds and remember the way we do that is we write 10 is 10 over 1 and we multiply by the reciprocal of two-thirds which is three halves again that's what i did so This cancels with this and becomes 5 5 times 3 is 15. so this simplifies to 15 over here

and we got that by just multiplying both sides by the reciprocal so it was kind of like a faster way than doing the division right because you got to go through and kind of set things up so if you see a fractional coefficient for x you want to multiply both sides by the reciprocal of that fractional coefficient If you don't have that if you have for example an integer coefficient let's say it's negative 2. so let's say i have negative 2x equals 10 for example well in this case yeah i can go through and multiply

by the reciprocal the reciprocal of negative 2 is negative one-half i can do that but really it's kind of just faster just To divide in this scenario right so in this scenario i wouldn't i wouldn't do that i would just say okay the coefficient of x is negative two so to get x by itself i would divide both sides of the equation by negative two because negative two divided by negative two is one so i'd have one times x or just x then 10 divided by negative 2 is negative 5. so in this case the

solution would be x Equals negative 5. so again if you have a fractional coefficient for your variable you want to multiply both sides of the equation by the reciprocal of that fraction if you have an integer coefficient you just want to divide both sides of the equation by that integer all right let's look at the next one so we have negative 10x equals negative 60. so here's an example where we have an integer coefficient right we have Negative 10 that's multiplying x so to isolate my variable x to isolate x to isolate x i would

divide both sides of the equation by negative 10. so if i divide this side by negative 10 and this side by negative 10 this will cancel with this and give me one one times x is just x over here negative 60 divided by negative 10 is just 6. and that's how quick i get my solution Right x equals 6. and if you check this just plug back in for x plug a 6 in there what have negative 10 times 6 equals negative 60. of course that's true negative times positive is negative 10 times 6 is

60. so i get negative 60 equals negative 60 and so we can say this solution here x equals 6 is correct all right what about negative 4x equals 20. so again i have an integer coefficient For x so if negative 4 is multiplying x and i want x by itself again i want to isolate i want to isolate my variable x well i'm just going to divide both sides of the equation by negative 4 right the coefficient of x really that's simple because negative 4 divided by negative 4 is going to give me 1 right

1 times x is just x and i got to make sure i do it to this Side so then i have x is equal to 20 divided by negative 4 is negative 5. plug negative 5 back in for x in the original equation so you will have negative 4 times negative 5 equals 20. negative times negative is positive 4 times 5 is 20. so 20 equals 20. so yeah this is the correct solution x equals negative 5. all right what about something like z over 8 equals negative 6. So the first thing is you got

to realize if i have z over 8 i can think about it as z divided by 8 or i could really think about this as 1 8 times z equals negative 6. so if i think about it as 1 8 times z well yeah this is a fractional coefficient for z remember we talked about the fact that if you had a fractional coefficient for your variable when you go to think about isolating the Variable i would just multiply both sides by the reciprocal of that fractional coefficient so the reciprocal of 1 8 is 8 over

1 or just 8. so times 8 over 1 times 8 over 1. so this will cancel and you'll just have z on the left side on the right negative 6 times 8 over 1 is the same as negative 6 times 8 that's negative 48. now another way to kind of think about This and again there's multiple ways to look at it i have z divided by 8. so kind of to get z by itself since it's divided by 8 i need to multiply by 8 kind of to undo the division so you can think about

as multiplication undoing division division undoing multiplication right you always want to do kind of the opposite operation so since i have z and i'm dividing by eight to undo that i Multiply by eight and because i do it to this side i must also do it to this side so then this will cancel with this right eight over eight is just one one times z is just z and negative 6 times 8 again is negative 48. so kind of either way you think about that you get z equals negative 48 as your solution all right

so let's check our solution of z equals negative 48. so again i'm just going to take Negative 48 and i'm going to plug it in for z so i would have negative 48 divided by 8 equals negative 6 and of course we can already see that this is true a negative divided by a positive is a negative 48 divided by 8 is 6. so i'd have negative 6 is equal to negative 6. all right let's take a look at one final problem so we have x over 6 is equal to negative 13 over 6. so

again i can think about this in multiple ways I can just say okay i have x divided by six so to undo the division i can use multiplication so x over six equals negative 13 over six since i'm dividing x by six i can multiply by 6 right 6 over 6 would be 1 right those would cancel and i'd have 1 times x or just x on the left but again to make this legal i've got to also multiply by 6 on the right side and then 6 would cancel with 6 here and give me

1. 1 times negative 13 is negative 13. so x equals negative 13 is the solution now another way you could have thought about this if i see a variable over a number remember i can break that up and say okay well this is the same as if i had 1 6 times x Equals negative 13 over 6. for some students that'll make it mentally easier for you to see okay well i have this times this so i know that i can multiply both sides by the reciprocal if i have a fractional coefficient for x so

i can multiply both sides by 6 over 1. and multiplying by 6 over 1 is the same thing as what we did earlier when we multiply by 6. right it's the same thing It's just this might be mentally easier for you to remember so then this will cancel with this and give me a 1. so 1 times x is just x and then over here this will cancel with this and give me 1. 1 times negative 13 is negative 13. so again either way you do it you're going to get that x equals negative 13

and when we check this problem if i plug a negative 13 up here Well look how easy it is to see i would have negative 13 over 6 right again i just plug this in for x equals negative 13 over six yeah same value on the left and the right so our solution here x equals negative 13 is of course correct hello and welcome to algebra one lesson five in this video we're going to learn about solving linear equations in one variable okay so in the last two lessons we've Learned how to solve very simple

versions of a linear equation in one variable so we kind of started out by learning something known as the addition property of equality and then we moved on to talk about the multiplication property of equality but in this lesson we're going to combine the two and learn how to solve any linear equation in one variable so let me first define what a linear equation of one Variable is it looks like this and you're probably going to see this in your textbook ax plus b equals c now in the previous two lessons you probably heard me

say that a b and c are real numbers and we really haven't talked about what a real number is yet but for right now just think of it as any number that you can think of so now let's kind of move further in this definition and we're going to put a Restriction on a so that's this right here so you see here where i have a does not equal 0. and why can't a be 0. well if i plug a 0 in for a i would have 0 times x and 0 times x is 0

so then i just have b equals c and that's not going to work unless b and c are the same number and in almost every case they're not going to be so again let me reiterate ax plus b equals C is kind of our general form for a linear equation in one variable where a b and c are real numbers so any numbers that you can think about and a can never be zero so this can never never be zero all right let's talk a little bit about the addition property of equality again this tells

us that we can add or subtract Any number to or in the case of subtraction from both sides of an equation and remember we use it to solve these simple equations that look like this x plus 4 equals 7 right so in this case we have 4 that's being added to x we want to isolate x so what we do is we subtract we subtract 4 from both sides of the equation right that's how we isolate x because we're adding 4 to x we subtract 4 so x can be by itself right 4 plus Negative

4 or 4 minus 4 is 0 and x plus 0 is just x so i cancel this out here this becomes a zero and then over here seven minus four is three so let me scroll down a little bit and we have we have x equals three and i taught you how to check things we plug the 3 in for x in the original equation you'd have 3 plus 4 which is 7 you'd have 7 equals 7. so that checks out all right so the Other property that we've already talked about is the multiplication property

of equality now this tells us that we can multiply or divide both sides of an equation by the same non-zero non-zero number okay so you can't do it with zero but anything else so that's another example of a restriction so we use this to solve stuff like 6x equals 36 right so in this case i'm multiplying x by 6. so to kind of undo that and to get x by itself i'm Going to divide right just think about the opposite operation if i'm adding something to x i want to subtract if i'm subtracting away from

x i want to add if i'm multiplying something by x i want to divide if i'm dividing x by something i want to multiply so in this case i am multiplying x by 6 so i want to divide by 6 right i have to do that to both sides of the equation to preserve the solution so this cancels with this and Becomes 1 1x is just x 36 divided by 6 is 6. and of course if you want to check this if i plugged a 6 in for x there 6 times 6 is 36 you'd

have 36 equals 36 so that is correct okay so now that we kind of have the hang of the basics let's talk about a general procedure for solving a linear equation in one variable and this is for any linear equation that you run across we're going to look in the next lesson At what to do when you have fractions or decimals but even if you didn't use that what we're going to cover in the next lesson you can still solve it using this procedure all right so the first thing you want to do is simplify

each side completely some of the things you're going to have to do you are going to have to combine any like terms that you might have on each side and you're also going to want to remove any parentheses that you have Using the distributive property all right so the next thing you have to do is isolate the term that includes the variable on one side of the equation so you're going to do this using the addition property of equality so kind of think about it as okay i learned the addition property of equality first that's

always what i'm going to use first all right so the next thing we're going to do is isolate the variable and you're going to do this using the Multiplication property of equality and remember we learned this second so this is kind of the next step that you do so you do the addition property of equality first then the next step you do is the multiplication property of equality now this is going to give you your solution and then you just want to check the answer all right so let's take a look at negative 4 minus

we have 4 times the quantity 5 plus 5x and this is equal to negative 64. now again the first thing we want to do is simplify each side separately so over here on the left i have negative 4 and then for the sake of simplicity let's write this as plus negative 4 times this quantity 5 plus 5 x because when i'm doing my multiplication here i want to include the negative when i'm multiplying and you don't have to Write plus negative four you can just have a minus out in front of the four and just

kind of include that in the sign right at this point in the game you want to do things to kind of speed up your arithmetic all right then this equals negative 64. nothing i can really do with negative 4 at this point i'm just going to rewrite that and then over here i'm going to use my distributive property so i'll multiply Negative 4 times 5 that's negative 20. so we'll have plus negative 20 or of course you can write minus 20 does not matter and then i'm going to multiply by 5x now so negative 4

times 5x would give me negative 20x so i'm going to write minus 20 x or of course again i could write plus negative 20x again it doesn't matter so then this equals negative 64. Now i'm looking to combine any like terms that i have i can't do anything with this because i don't have any other variable term involved in this equation right i just have that one but over here i have negative 4 plus negative 20. i can combine those two so negative 4 plus negative 20 is negative 24. so this would be negative 24

minus 20x Equals negative 64. okay so now i'm at the stage where i want to isolate the term that includes the variable so that term is this right here this minus 20x or negative 20x however you want to think about that so how do i isolate that well we already know that we use the addition property of equality and essentially if i have this negative 24 over here if i add 24 to both sides i'm going to get rid of it Right so if i have negative 24 i add 24. if i was to come

across positive 24 i would subtract 24 right you always want to do the opposite so i'm just going to add 24 over here and then to make that legal i'm going to add 24 to this side of the equation now this is going to cancel right negative 24 plus 24 is 0. so the left side of this equation is negative 20x on the right side i have negative 64 Plus 24. so what's that going to be well mentally we can do this we have a 4 and a 4 here so really i'm doing 64 minus

24 which is 40 and i'm slopping a negative on it so this is going to be negative 40. okay so now we're at our third step and remember that's to isolate the variable we do this using the multiplication property of equality now if i have negative 20 times x to get x by itself i need to divide right Remember this is multiplication the opposite of that would be division i need to divide by negative 20. again just look at what's being done to x if negative 20 is multiplied by x the opposite of multiplication is

division okay so if i'm multiplying negative 20 by x i divide by negative 20 so that these two will cancel out and become one and i just have x by itself now because i do it to this side i must Also do it to this side and negative 40 divided by negative 20 would give me 2. so what i get is my solution is that x is equal to 2. all right so we definitely want to check this and make sure we got the right answer so let me erase everything and we'll go back up

to the top all right so we have x equals 2. we have negative 4 we have negative 4 Minus 4 times the quantity 5 plus 5 where i see the x i'm going to plug in a 2 and then close the parentheses and then equals negative 64. okay so let's crank this out we're going to start out inside the parentheses here and we're going to start out with 5 times 2. 5 times 2 is 10 and then if i add 10 to 5 i'd have 15. so i'll have negative 4 minus 4 times 15 and

this should equal Negative 64. so i'm going to do multiplication first so you can think about this as negative 4 times positive 15 that would give me negative 60. so negative 4 minus 60 equals negative 64. and of course it does right negative 4 minus 60 is negative 64. so we get negative 64 on the left we have negative 64 on the right and so yes x equals 2 is the correct solution all right let's take a look at another One we have negative 3 times the quantity 5p minus 5 plus 2p equals 67. so

again the first thing i want to do is simplify each side separately so to do that over here i'm going to use my distributive property i'm going to multiply negative 3 times 5p and that's going to give me negative 15p and then next i'm going to multiply negative 3 times negative 5. so negative 3 times negative 5 is positive 15 and then plus 2p equals 67. all right now i want to combine any like terms that i have so over here i have negative 15p and i have 2p so negative 15 plus 2 is negative

13 the variable comes along for the right so this is negative 13 p plus 15 equals 67. there's nothing else to Simplify on each side and so now i move on to the next step where i'm going to isolate the variable term on one side of the equation now if i look at my variable term it's negative 13p what's being done to it well i'm adding this 15 right here so in order to isolate negative 13p i need to do the opposite operation i need to subtract 15 away or you think about it as adding

negative 15 to both Sides right so i need to subtract 15 over here and then over here to make it legal i need to also subtract 15. so this is going to cancel and become zero and i have isolated my variable term on one side so i have negative 13 p here and then over here 67 minus 15 that's going to give me 52 right 7 minus 5 is 2 6 minus 1 is 5. all right so now what's left we want to isolate The variable right so we want to isolate p and how do

we do that well we always want to look at what's being done to that variable right so in this case we have negative 13 that's multiplying our variable p so to undo multiplication we do division right so we're going to divide both sides of the equation by the coefficient of p which is negative 13. very very simple So this cancels with this and i'm just going to have p and then over here what is 52 divided by negative 13 well it's going to give me negative 4 and that's my solution p is going to equal

negative 4. all right let me erase everything and then we're going to go back and check all right so let's see if we got the right answer we're going to have negative 3 times Inside the parentheses we have 5 plug in a negative 4 for p so times negative 4 minus 5 close the parentheses plus 2 times negative 4 equals 67. so we're going to start out inside the parentheses here we always multiply first 5 times negative 4 is negative 20. negative 20 minus 5 is negative 25. so this would be negative 3 negative 3

times negative 25 Plus we have 2 times negative 4 equals 67. so we're going to multiply before we add negative 3 times negative 25 is 75 plus we have 2 times negative 4 that's negative 8. and this equals 67 all right so what is 75 plus negative 8 or what is 75 minus 8 well that's going to be 67 so we have 67 equals 67 and so p equals negative 4 is the correct solution all right let's take a look at another One we have 30 equals negative 10 times the quantity 9x plus 10 plus

8 times the quantity negative 5 minus 10 x so again i want to simplify each side separately so i have 30 on the left nothing i can really do with that on the right i'm going to use my distributive property and i'm going to multiply negative 10 times 9x so that's going to be negative 90x Plus i'm going to have negative 10 times positive 10 which is negative 100 so i can write plus negative 100 i could just write minus 100. so then plus now we have 8 times negative 5 that's going to be negative

40. so i'm just going to write minus 40. and then we have 8 times negative 10x and that's going to be negative 80 x All right so now we want to combine like terms and basically you can see that we have negative 90x and we have negative 80x those can be combined and we also have negative 100 and we have negative 40 those can be combined as well so we have 30 on the left that's not changing on the right negative 90x minus 80x or negative 90x plus negative 80x however you want to think about

that That's going to give me negative 170 x then i'm going to have negative 100 minus 40 that's minus 140. so now that we've simplified each side completely meaning i can't do anything else to either side to make it more simple i'm going to isolate the variable term so in this case that's going to be this right here this Negative 170x now again again again what can we do here to isolate that we have to look at what's being done to the variable term i'm subtracting away 140. so if i'm subtracting away 140 i can

always look to add the opposite so the opposite of negative 140 is positive 140 and i got to do it to both sides to make this legal so that cancels becomes zero 30 plus 140 is 170. so this equals over here just negative 170 x and to get x by itself to isolate the variable which is our next step we look at what's being done to x and in this case we have multiplication so again you always look to do the opposite if you're multiplying you divide if you're dividing you multiply if you're adding you

subtract so it's it's just that simple so in this case we Have multiplication so we look to division so divide this by the coefficient of x which is negative 170. divide this by negative 170. so this and this are going to cancel and become 1. 1 times x is just x and then over here 170 divided by negative 170 is negative 1. so you have negative 1 equals x or you can flip that A lot of people like to have the variable on the left myself included so if i see something like this i just

kind of flip it and say okay well x equals negative 1. it doesn't really matter these two mean the exact same thing right it does not matter whatsoever all right so let's erase everything and check our answer all right so we have 30 equals negative 10 times this quantity 9 times negative 1 plus 10 Then plus 8 times negative 5 minus 10 times negative 1 and then close parentheses okay so we have 30 over here on the left and then equals inside the parentheses here 9 times negative 1 is negative 9. negative 9 plus 10

is 1. so i would have negative 10 times 1 which is just negative 10 right we don't even need to kind of write that then plus we have 8 times and then we have inside of here Negative 5 minus 10 times negative 1. so i can just think about this as negative 10 times negative 1 that's going to be positive 10 so negative 5 plus 10 would be 5 so i'd have 8 times 5 here now in this next step i would do my multiplication 8 times 5 is 40. and so now negative 10 plus

40 would be 30. so i'd have 30 equals 30 and so my solution x equals negative 1 is correct okay let's take a look at one final problem so we have negative 10 times the Quantity negative 4 minus 6x this is equal to 3 times the quantity negative 3x minus 8 then plus 5 x all right so we're going to start by again simplifying each side separately so on the left negative 10 times negative 4 is 40. then negative 10 times think about this is negative 6x that's going to be plus 60x and this equals

over here 3 times Negative 3x is negative 9x and then 3 times negative 8 is negative 24 so minus 24 plus 5 x okay so now i want to combine like terms nothing i can combine on the left on the right i can combine the negative 9x and the 5x that's going to give me negative 4x so i'm going to write this as 40 plus 60x equals again this will be Negative 4x when i add negative 9x and 5x minus 24. all right so i want all the variable terms on one side of the equation

that's how i kind of isolate the variable term on one side and i have just a number on the other so in the previous examples it's been kind of easy to do that here what you're going to notice is that you have a variable term here and here Now what you want to do is use the addition property of equality to get both on one side remember i can move anything so if i have negative 4x over here i can just add 4x and make it disappear on this side but i have to do the

same thing over here to make it legal so in other words if i add 4x here it's going to disappear on this side negative 4x plus 4x that's 0. so it's gone But it reappears over here because i have to also do it on this side to make it legal so now i'll have 40 plus 60x plus 4x is 64x and this equals negative 24. this is still not what i want because i want a variable term on one side of the equation and i want just a number on the other so in order to

do this again if i'm Adding 40 to the variable term 64x to make this 40 go away i'm just going to subtract it away right but i have to do it to both sides that's the key so minus 40 over here and minus 40 over here so now this is gone and i just have 64x 64x on the left side so i've isolated the variable term on the right side negative 24 minus 40 is negative 64. now i just want my variable by itself so I just want x so how am i going to get

that again look at what's being done to x if we're multiplying like we are here you have 64 that's multiplying x so you want to use division right to undo the multiplication so if i divide both sides of the equation by the coefficient of x which is 64 i'm going to get x by itself this cancels with this remember 64 divided by 64 is 1 1 times x is just x over here negative 64 divided by 64 this gives me negative 1. so negative 1 x equals negative 1 is my solution all right so let's

erase everything and check and so we'd have negative 10 times the quantity negative 4 minus 6 times negative 1. this equals 3 times this quantity negative 3 times negative 1 minus 8 then plus 5 times negative 1. and again all i'm doing is i'm plugging In a negative 1 ever everywhere that x exists so that means i plugged in a negative 1 here here and here and i'm just going to make sure the left and the right side are going to be the same value right make sure they're equal so on the left i start

inside the parentheses and i'm going to start with negative 6 times negative 1 that's 6 and So i'd have negative 4 plus 6 that's going to give me 2. so this left side would be negative 10 times 2. on the right side inside the parentheses negative 3 times negative 1 is 3 3 minus 8 is negative 5. so i'd have 3 times negative 5 and then plus 5 times negative 1. let's do all our multiplication negative 10 times 2 is negative 20. 3 times negative 5 is negative 15 And plus 5 times negative 1 is

negative 5. you can kind of eyeball this and see it works out on the left you have negative 20 on the right negative 15 plus negative 5 is negative 20. and so negative 20 equals negative 20. we have the same value on each side of the equation so we know that x equals negative one is the correct solution hello and welcome to algebra one lesson six In this video we're going to learn about solving linear equations with fractions or decimals so up to this point we really haven't worked a lot with equations that have fractions

or decimals i kind of did that to keep it very simple in the beginning but as you kind of progress through your chapter on solving linear equations in one variable you're going to see a lot of problems with fractions and decimals So when they occur we can use a special trick to get rid of them so let's first talk about how to solve a linear equation with fractions now a lot of people hate fractions so when they occur inside of an equation and you're just starting to learn equations it makes it that much worse right

you're already struggling to figure out how to solve an equation and now i'm throwing fractions in there and you're like oh what am i Going to do when you first start out the problem you're going to clear your fractions by multiplying both sides of the equation by the lcd remember that's called the least common denominator of all fractions in the equation so for example we have 7 4 x plus 1 half x equals 63 over 20. what you'll do is look at all your fractions so here's a fraction here's a fraction here's a fraction okay

I want you to look at the denominators of each fraction so you have a 4 you have a 2 and you have a 20. so 4 2 and 20. those are my denominators and we want the lcd remember the lcd is equal to the lcm the least common multiple of the denominators so what is the lcm of 4 2 and 20. [Music] well remember from pre-algebra the way we figure this out we would factor each number so 4 would be 4 would be 2 times 2. 2 is a prime number that's not going to factor

and 20 is going to be 5 times 4 4 is 2 times 2. very very easy to do and to build the lcm We take each prime number from each prime factorization and we throw it in there the only thing is that you have to remember if a prime factor is repeated so in other words in four we have two factors of two and two we just have that number and in 20 we have two factors of two i put the largest number of repeats between all of the prime factorizations So with the number two

it occurs twice at most right it occurs twice here it occurs twice here so i'm going to put two in when i build my lcm and if that doesn't make any sense to you go back to the pre-algebra lesson on finding the least common multiple and it will make sense to you then all right now the next thing i want to put in here i have a 2 and a 2 i have a 2 here a 2 and a 2 here i only have a 5 left to put in so throw in a 5 and

2 Times 2 is 4 4 times 5 is 20. so the lcm for 4 2 and 20 is 20. so that means my least common denominator here is 20. now once i have figured that out all i do is multiply both sides of the equation by that and the fractions will go away so i'll have 20 times 7 4 x plus 1 half x equals 63 over 20 times 20. Now remember this is legal because of the multiplication property of equality i can multiply both sides of an equation by the same non-zero number and preserve

the solution so on the right it's pretty easy this just cancels with this and i just get the number 63. [Music] on the left i've got to use my distributive property and i've got to distribute 20 to each term here so i'm Going to end up with 20 times 7 4 x so 20 times 7 4 x plus 20 times one half x now you can look and see okay between 20 and 7 4 this is going to cancel with this and give me 5. 5 times 7 is 35 so i'm going to have 35

x plus and then over here this is going to cancel with this and give me a 10. 10 times 1 is just still 10 then times x so I'm going to have 10x there this equals 63 and we'll scroll down a little bit more so now i have an equation that's clear of fractions and the solution is going to be the same as if i had just worked with the fractions so i would start off by just combining like terms 35x plus 10x is 45x and this equals 63. now to get x by itself i

divide both sides of the equation by the coefficient of x which Is 45 this will cancel with this and just give me an x and 63 over 45 that's not going to be a whole number so i just have to reduce the fraction so 63 is 9 times 7 [Music] 45 is 9 times 5. so the greatest common factor there is going to be 9. i can cancel this with this And i'll have seven fifths right so x equals seven fifths you're going to get a lot of fractional solutions moving forward i kind of stayed

away from it in the beginning because i wanted to make everything nice and easy for you to understand before we start really ramping it up and dealing with fractions and decimals all right so i want to do two things i want to check This so let's erase everything here and we're going to go back up to the top so remember x equals seven fifths so i'm gonna plug a seven fifths in everywhere i see an x so seven fourths times seven fifths plus one half times seven fifths equals sixty three over 20. all right so

over here nothing i can cross cancel so 7 times 7 Is 49 and 4 times 5 is 20 plus over here nothing i can cross cancel 1 times 7 is 7 over 2 times 5 that's 10. now in order to add i've got to have a common denominator so i'm going to go ahead and just multiply this by 2 over 2 and this will be 14 over 20. 14 over 20 and this equals 63 over 20. and of course the left and the right Side are going to be equal 49 plus 14 is 63. this

is over 20 and this is equal to 63 over 20. so our solution x equals seven-fifths is correct one thing i want to show you if we would have just left the fractions as is and just work with them you're going to get the same answer so clearing the fraction is just more of a trick that you can use if you just don't like working with fractions sometimes it just makes it a lot easier right nice and Simple so let's say i have 7 4 7 4 x plus 1 half x equals 63 over 20.

if i want to i can combine like terms here to simplify on the left so move this kind of x over to the side and get a common denominator multiply this by 2 over 2. so then i'll have 7 4 let me scroll down here i'll have 7 4 x plus 1 times 2 is 2. so 2 over 2 times 2 is 4. so 2 4 x equals 63 over 20. So to combine like terms here i'm just going to add the numerator 7 plus 2 is 9. so i would have 9. the common denominator

is 4 so 9 4 then times x equals 63 over 20. now how do i isolate the variable here we talked about in one of our beginning lessons if i have a fractional coefficient for x and i want x by itself remember if i multiply a number by its Reciprocal the result is 1. so if i multiply 9 4 by 4 9. remember the reciprocal is found by taking the denominator and putting it into the numerator taking the numerator and putting it into the denominator i'm going to get one right if i multiply these two

so this this this this that all cancels so this is 1x now because i do it over here i've got to do it over here so this cancels with this and gives me Seven this cancels with this and gives me five so what am i gonna have i'm gonna have x equals seven fifths and that's exactly what we got by clearing the fractions now in this particular case because i'm very good at working with fractions right i've been doing a long time it didn't take me any longer to work this problem with the fractions involved

for yourself you might not be that good With fractions so you might want to just take the time and clear the fraction so that you don't have the added stress of trying to solve an equation and trying to work with the fractions together now when we get to decimals because most of you work with calculators you might not want it cleared it might not save you any time it just depends all right let's take a look at another one so we have 3 halves x plus 1 plus 4 3 x Equals negative 31 eighths plus

two-thirds x all right so what i'm going to do is again clear the fractions from the equation so let's look at our denominators we have a 2 a 3 an 8 and a 3. now i'm looking for the lcd the least common denominator and the way i find that i want the least common multiple of the denominators so 2 3 and 8. and i'm not going to put 3 in Twice right it's the same number so 2 doesn't factor so i'm just going to throw that in there 3 doesn't factor i'm just going to throw

that in there and 8 is 3 factors of 2. so how many factors of 2 am i going to put in when i'm building my lcm only three right so i'm gonna put one and then twos because i already have one over here from when i did this remember when you build your lcm you want the largest number of repeats Between any of the prime factorizations so between eight you have one two three and two you have one the largest number of repeats is three so three goes in when you build this okay so we

have two times three which is six times two which is 12 times 2 again which is 24. so that's going to be my lcm for those numbers and so my least common denominator is 24. All right so lcd is equal to 24 and so i multiply 24 by both sides of the equation so i'm going to put some brackets here very important that you use brackets or parentheses because you need to distribute okay you need to distribute this value of 24 to each term i see a lot of students will just write 24 next to

this and then they multiply by that and then They're done that's you need to multiply 24 by every term to make this legal okay so over here negative 31 eighths plus two-thirds x and we'll enclose this inside some brackets as well and then we'll use our distributive property so what i'm going to do i'm going to multiply 24 by 3 halves x so 24 times 3 halves x and as i do it i'm just going to do the Calculation kind of save some time so 24 would cancel with 2 and give me 12 12 times

3 is 36 so this would be 36x this is 36 x okay plus next i'd have 24 times 1 that's very simple that's just 24. then plus next i'd have 24 times four-thirds x so 24 times four-thirds x the 24 would cancel with the 3 and give me an 8. 8 times 4 is 32. so this would Be 32x 32 x and then this equals i'm going to do the same thing over here so 24 times negative 31 8. so negative 31 8 times 24 this cancels with this and gives me a three three times

negative 31 is negative 93. so negative 93 and then i have one more 24 times two-thirds x so that's going to be plus two-thirds x times 24 And this 24 would cancel with this three and give me eight eight times two x is 16x so this would be 16x okay so now we're free of fractions now i just go about my usual business which is to simplify the left and the right side so this over here 36x and 32x we can combine like terms that's going to give us 68x so 68x plus 24 equals over

here i really can't do anything i just have negative 93 Plus 16x now as we learned in previous lessons we've got to get the variable terms on one side of the equation i have 16x over here and i have 68x over here so i need to move one of them i always move the one on the right to the left i like my variable to end up on the left so i'm going to move this over here and the way i'm going to do that i'm Just going to subtract 16x from both sides of the

equation that'll go away and then minus 16x over here 68 minus 16 is 52. so this would be 52x plus 24 equals negative 93. all right so now i want to get the variable term by itself and to get this by itself i need to subtract away 24 right because i have plus 24 Over here that's next to the 52x so minus 24 minus 24. so that's going to go away the left side is now just 52x and on the right side if i have negative 93 minus another 24 i'm going to end up with

negative 117. all right the final step here is to isolate the variable completely and if i look at it i have what i have 52 times x so to get x by itself i divide by 52 i divide by that Coefficient of x all right so divide both sides of the equation by 52 this will cancel with this and give me 1. 1 times x is just x so that's going to be equal to negative 117 divided by 52 so this won't result in an integer or whole number so we're just going to reduce the

fraction so negative 117 i'm going to write negative 1 times 13 times 3 times 3 and for 52 i'm going to write 13 times 2 Times 2. so what can i cancel what can i cancel i can cancel this 13 with this 13 and i'm really left with what i'm left with negative 1 times 3 times 3 or negative 9 over 2 times 2 or 4. so my answer here is going to be that x equals negative 9 4. x equals negative 9 fourths so i'm going to plug in a negative nine fourths everywhere i

see an x so i'm Gonna have three halves times negative nine fourths plus one plus four thirds times negative nine fourths is equal to negative 31 8 plus two thirds times negative nine fourths all right so if we have three halves times negative nine fourths what can we cancel Well nothing so we just have to do the multiplication 3 times negative 9 is negative 27 2 times 4 is 8. so then plus 1 plus over here this 4 would cancel with this 4 and i'd have negative nine thirds negative nine thirds and equals we have

negative 31 eighths plus over here i can cancel before i multiply negative nine divided by three would be Negative three so this cancels with this and gives me a 3. and this will cancel with this and give me a 2 down here so this completely canceled out that's just a 1 now so we'd have basically just negative 3 halves so plus negative 3 halves all right so on the left i got to think about getting a common denominator so 8 is going to factor into 3 factors of 2. 3 is not going to factor so

the best i can do there for my lcd is going to be 24 right 8 times 3. so negative 27 over 8 times three over three plus for one i'm just going to write it as 24 over 24. it's the same thing as one and then plus we'll have negative nine over three times eight over eight put this equal to negative 31 8 plus negative 3 halves all right so what is negative 27 times 3 well That's negative 81 so this would be negative 81 over 24 plus 24 over 24 plus negative 9 times 8

is negative 72 over 24. let's go ahead and figure this out real quick so if i have negative 81 plus negative 72. so 81 plus 72 kind of mentally 1 plus 2 is 3 8 plus 7 is 15. so that'd be 153 attach the negative so that's Negative 153 plus 24 and we can kind of do that in our head also because really think about 153 minus 24 that's going to give you 129 and then just slap the negative on there so we would have negative negative 129 over 24. so this is equal to and

then over here really i can easily see that the lcd is Going to be 8. so i just need to multiply this side by 4 over 4 and i'd have negative 31 8 plus negative 3 times 4 is negative 12 over 2 times 4 is 8. so negative 31 plus negative 12 is going to be negative 43. so we'll have equals negative 43 8. so a lot of you at this point will stop and go okay i got the wrong answer what happened always make sure that when you get Fractions they're reduced to their lowest

terms or they're simplified these fractions are the same the one on the left is just not simplified if i think about negative 129 i can write this as negative 1 times 43 times 3. i can write 24 as 8 times 3 and we can see that we can cancel this 3 with this 3 and i'm left with negative 1 times 43 or negative 43 over 8 that's what i have over here negative 43 over 8. and you can't simplify that any further right on either side it doesn't matter because you've got the same number on

either side but if you want to think okay well can i simplify this further just for the sake of completeness you can't right because 8 is 2 times 2 times 2 and negative 43 is not divisible by 2. so you'll end up with negative 43 8 equals negative 43 eighths and so you know your solution which we Found was x equals negative nine fourths is correct all right let's take a look at another type of problem you're going to come across so we have 37 18 plus two-thirds x equals negative one plus two-thirds times this

quantity negative four-thirds x plus nine-fourths if i begin my problem the normal way if i say okay what's the lcd of all the fractions involved and i go to multiply it it's not going to clear all the fractions Right away what you need to do is use your distributive property first so if you come across a problem where you have fractions that are inside of a set of parentheses or a set of brackets use the distributive property first and simplify so what i want to do so i have 37 18 over here plus two thirds

x equals negative one plus i'm going to remove the brackets Two-thirds times negative four-thirds x i would have two-thirds times negative four-thirds x so two times negative four is negative eight three times three is nine so this would be negative eight ninths so i'm gonna put minus minus eight ninths times x and then again two thirds times nine fourths so two thirds times nine fourths And what i can go ahead and do is cancel this with this that'll give me a three cancel this with this and that'll give me a two so this is canceled

completely it's just a one and so i'll have three halves there so this will be plus plus three halves okay so once you've done that go ahead and think about okay what's the lcd now so i Have 18 i have 3 i have 9 and i have 2. and at this point in your math career you should be able to kind of look at that and say okay i know the lcd is 18. right so i know most of you can do that if you can't again take the steps so lcm of 2 3 9

and 18. so i know that 2 doesn't factor I know that 3 doesn't factor 9 is 3 times 3. so i'm only going to put one more copy of 3 in there because 9 is 3 times 3 i only want the largest number of repeats to go in 18 is 2 which i already have in there times 3 which is already in there times 3 again you could think about it as 2 times 9 or you know 3 times 6 however you want to do it it's going to break down into 2 times 3 times

3 and we already have that there So if we multiply we get 18. so that's the lcm of those numbers and so for the lcd it's 18. all right so now we're just going to multiply both sides of the equation by 18. again you want to use brackets or parentheses make sure that you multiply each term by 18. [Music] all right so over here on the left if i Do 18 times 37 18 we know that 18 would cancel with 18. right so this would just be 37 then plus we have 18 times 2 3x

18 would cancel with 3 giving me a 6 6 times 2 is 12 so this would be 12x and this equals over here 18 times negative 1 is negative 18. over here 18 times negative 8 9. think about 18 canceling with 9 and Giving me a 2. 2 times negative 8 is negative 16. so this would be negative 16 times that variable x then plus lastly i have 18 times 3 halves 18 will cancel with 2 and give me a 9 9 times 3 is 27 okay now we just simplify so i can't do anything

to the left i'm just going to copy that down so 12x plus 37 i'm just going to reverse the order On the right i can simplify because i can add negative 18 and 27 that's going to give me positive 9. so i'd have negative 16x plus 9. i want all my variable terms on one side of the equation so because i have one there and one here i always like my variable on the left i'm going to move that guy over there so i'm going to add 16x to both sides of the equation so that's

going to go away 12x plus 16x Is 28x so plus 37 equals 9. i'm going to get my variable term by itself on the left so that means i'm going to be subtracting away the 37 from both sides of the equation so that's gone so i'll have 28x equals what is 9 minus 37 that's going to be negative 28. and we divide both sides of the equation by 28 the coefficient of x in order to isolate the variable Completely and i'm going to have x is equal to negative 1. okay x is equal to negative

1. all right so we could erase all this and check so x again equals negative 1. so we'd have 37 18 plus 2 times negative 1 which is just negative 2 so i'm just going to write minus two thirds and this is equal to negative one plus two thirds times this quantity we'll have negative Four times negative one so that's positive four over three plus nine 4. so on the left side i just multiply this by 6 over 6 and i'll have a common denominator right so i'd have 37 minus 2 times 6 is 12

over the common denominator of 18. 37 minus 12 is 25 so this would be 25 over 18. 25 over 18 And 25 is 5 times 5 18 we know is 2 times 3 times 3 so really we can't make that any more simple over here if i think about four thirds plus nine fourths i'd multiply this by four over four and multiply this by three over three and so negative one plus two thirds times 4 times 4 is 16 over 12 plus 9 times 3 is 27 over 12 and 16 plus 27 would be 43

so i'd have 43 12. so let me just erase this and write 43 twelfths so if i'm multiplying here this is going to cancel with this and give me a six and so what i'm going to have is negative one plus i'll have one times 43 or 43 over 3 times 6 or 18. so to get a common denominator here i'm going to write this negative 1 as negative 18 over 18. So we just need to think about what is 43 minus 18 well that's 25 right so this becomes 25 18 just like on the

left side of the equation and so we know that our solution x equals negative 1 is correct so let's talk a little bit about when your equation has decimals in it in most cases you're using calculators on your exams or your homework and working with decimals when you have a calculator is not any more difficult so This is just something i'm teaching as a matter of completeness you do need to know how to do it because you might be asked for it on a test but i don't think it would actually save you any time

especially if you're working with a calculator but if you're faced with decimals i just want you to think back to when we learned that if you multiply by 10 or 10 raised to a whole number that is one or larger we're gonna move our decimal Point one place to the right for every zero in our power of ten so for example let's say i have the number six and i multiply by one 100 well i have two zeros in this power of 10 and so i take 6 i write a decimal point after 6 because

you can always write 6 as 6.0 and then i can move the decimal point 2 places to the right to get my answer because i have two Zeros and that power of 10. so this is really 600 right we know that 6 times 100 is 600 we also could write 100 as 10 to the second power and kind of the shortcut is the look at the exponent there and say okay i have two places that i'm going to move my decimal point to the right because if you have a 10 raised to the power of

2 it's going to be a 1 followed by 2 zeros and once you Get good at this you can be very very quick with these calculations so let's say i have 3 times 10 to the fourth power i see that there's a 4 here as my exponent that means that i would have a 1 followed by 4 zeros right that's what 10 to the fourth is so this would be a 3. put my decimal point down and i'm going to move that decimal point four places to the right because there's a four there so one

two Three four and i'm going to end up with thirty thousand right because ten to the fourth power ten to the fourth power again is a one followed by four zero so one two three four or ten thousand and you can kind of mentally see that three times ten thousand is thirty thousand it's pretty easy to do so if we do this and we involve you know kind of some Decimals we can kind of use the same principle to clear the decimals so if we have 6.32 and i want to make this into a whole

number meaning i don't want to see this decimal point anymore i have one two decimal places here all i need to do is multiply it by a number that has a one followed by two zeros right so i'd multiply it by a hundred or ten squared so times a hundred so this would be Equal to six 632 right because this would go one two places to the right so essentially what's going to end up happening is we're going to look at our equation and we're going to see the largest number of decimal places between everything

involved so we're then going to multiply both sides of the equation by a power of 10 that will clear that decimal so you want to clear the largest number of decimal places so if the largest number of Decimal places you have is 3 then you're going to multiply by a thousand right it's a power of 10 that has a 1 followed by 3 zeros and if you clear the one that has the largest number of decimal places you're going to clear all of them all right so let's take a look at 20.01 equals negative 6.7

minus 10.7 n minus 8.6 i'm gonna look through this equation i have one two decimal places here i have one decimal place here one here and one here so the largest number of Decimal places between any of these numbers involved would be 2. now in order to clear the equation of decimals i need to multiply by a power of 10 that will allow me to move all the decimal points in the equation two places to the right so to do that you're gonna multiply by a hundred okay just think about okay i need two zeros

in the power of ten so that's ten squared or a hundred So again i'm going to use parentheses or brackets to put 100 outside 20.01 equals and i have 100 outside of here too negative 6.7 minus 10.7 n minus 8.6 very very easy now we're just going to use our distributive property so 100 times 20.01 again move your decimal point two places to the right so this would be 2001 this is equal to Over here 100 times negative 6.7 you'll have negative you'll have 6. this is going to go one so i'll have a 7

and then one more place to the right and so a zero so negative 670 and then minus here we have 10.7 n so i'm going to have 10 this is going to go one place to the right that's going to be a 7. one more place to the right that's going to be a zero and then my n so that's minus 1070 n and then minus here we have 8.6 so That's going to be 860 right this is going to go one two places to the right so it's going to go right here and then

right there so we have a transformed equation now that doesn't have any decimals involved and again you just go through after you have this and do your regular procedure so i would combine like terms here on the right negative 670 minus 860 is negative 1530 so this will be negative 1070 n And then minus 1530 and this is equal to over here on the left we have 2001. so i still want to isolate my variable term and to do that i'm going to add 1530 to both sides of the equation and so 2001 plus 1530

is 3531 and this equals negative 1070 n all right now to get n by itself i just divide both sides of the equation by the coefficient so divide by negative 1070 divide by negative 1070 and this will cancel with this and then 3531 divided by negative 1070 is going to give me negative 3.3 so i get negative 3.3 equals n now notice how i wrote it like that versus as a fraction you can do either because we're working with decimals i chose to write this as a decimal you can do either one it does Not

matter so i'm going to rewrite this as n equals negative 3.3 and we're going to go back up to the top and check it okay so we're going to have n equals negative 3.3 here so we're going to have 20.01 is equal to negative 6.7 minus 10.7 times again negative 3.3 minus 8.6 okay so Over here on the left 20.01 is not going to change don't need to do anything there on the right i need to do my multiplication first what is negative 10.7 times negative 3.3 it's gonna give me 35.31 35.31 okay so now

i have negative 6.7 plus 35.31 minus 8.6 so i'm going to add my numbers with like signs first what is negative 6.7 minus 8.6 that's going to be negative 15.3 so i'm going to write this as 20.01 equals negative 15.3 plus 35.31 and now we just need to add over here what is negative 15.3 plus 35.31 that's going to give me 20.01 so 20.01 equals 20.01 same number on the left and the right so n equals negative 3.3 is the correct solution all right let's take a look at another one we have 17.8321 plus 2.589y

equals 0.8 times the quantity y plus 4.37 plus 4.6y now just like we saw with fractions when you're trying to clear decimals you've got to use your distributive property first if you have decimals inside the parentheses so 17.8321 plus 2.589 y equals distribute this inside here so 0.8 times Y is 0.8y and then distribute this inside here so plus so 0.8 times 4.37 is 3.496 so 3.496 plus 4.6 y so now i'm going to look to see what number has the highest number of decimal places i have one two three four here i have one

two three here i have one here one two three here and i have one here so the highest number of decimal places Is four so that means i need to multiply by a number that has a one followed by four zeros so i'm going to multiply both sides of this equation by a one followed by one two three four zeros or the number ten thousand so i'm gonna have ten thousand times seventeen point eight three two one plus two point five eight nine y again make sure to put this inside of Parentheses you're really multiplying

ten thousand by each of these numbers you can't make the mistake of just multiplying it by 17.8321 it's also got to be multiplied by 2.589 y as well okay so over here i'm going to have 10 000 also multiplied by and i'm going to put this off the screen for a minute so we'll have 0.8 y plus 3.496 Plus 4.6 y again inside of parentheses okay so starting over here on the left if i multiply 10 000 times 17.8321 it's going to be 178321 right this will go four places to the right and so i

end up with 178 321 so then plus now if i multiply 10 000 by 2.589 y i'm basically going to have 2 5 8 9 and then a 0 at the end again this goes four Places to the right and so it would pass the nine and then go one more right so that's why i have the zero here we go one two three four places to the right so that ends up giving me 25 890 and then times the variable y and then over here on the right i have ten thousand times point eight

y so if i have point eight and this goes four places to the right it's going to go 1 2 three four so that's gonna be eight thousand then again times the variable y then plus over here i have three point four nine six this is going to go four places to the right so it's gonna pass the six by one more place so i'm gonna need to write three four nine six zero right this is gonna go one two three four places to the right and then plus we have four point six y so

ten thousand times four point six we Think about 46 here and i'm gonna need to put three more zeros one two three this is going to go one two three four places to the right and then don't forget the y and then we have our transformed equation unfortunately it's so big it doesn't fit on the screen but we go through the same process on the left i can't simplify any more so i'm just going to write 178 thousand 321 Plus 25 890 y this is equal to over here on the right i can combine like

terms and i need to put a comma in there i have 8000 y plus 46 thousand y that's going to give me fifty four thousand y then plus and i put a comment here thirty four thousand nine hundred sixty so again i want all my variable terms on One side of the equation and so i like my variable term to be on the left i like my number to be on the right so i'm going to go ahead and subtract 54 000 y from both sides of the equation that's going to go away and just

to save a little time i'm going to simultaneously subtract 178 321 from both sides of the equation so why am i doing that well what's going to end up happening is i'm going to have a variable term on This side this is gone and i'm just going to have a number on the right side all right so what is 25 890 minus 54 000 that's going to be negative 28 110 and then y and this is equal to what is 34 960 minus 178 321 that's going to give me negative 143 361. And now as

a final step we just get y by itself we divide both sides of the equation by negative 28 ten right the coefficient of y so negative twenty eight thousand one hundred ten so negative one hundred forty three thousand three hundred sixty one divided by negative twenty eight thousand one hundred ten is five point one so i'm going to have y equals 5.1 now let's erase everything and we're Going to check it so again y equals 5.1 so i'm going to have 17.8321 plus 2.589 times five point one [Music] equals point eight times the quantity for

why i'm putting in five point one plus four point three seven plus four point six times five point one so starting on the left 5.1 times 2.589 is 13.2039 so i'd have 17.8321 plus 13.2039 now if i add these two 17.8321 plus 13.2039 is going to give me 31.036 now over here i have 5.1 plus 4.37 to be done first that's going to give me 9.47 and if i multiply that by 0.8 i'm going to get 7.576 now plus 4.6 times 5.1 that gives me 23.46 and if i combine these two so 7.576 Plus 23.46

i'm going to end up with 31.03 so the same number on both sides of the equation so i know my solution y equals 5.1 is correct hello and welcome to algebra one lesson seven in this video we're going to learn about linear equations with no solution or an infinite number of solutions all right so so far we have only worked with one type of equation this equation is known as a conditional Equation let me highlight that a conditional equation so it is true only under certain conditions so let's see an example so here's a very

simple example of a conditional equation everything we've worked with so far has been a conditional equation so negative 2x equals 4. so we all know how to solve this at this point right we want to isolate x so we divide both sides of the Equation by negative 2 right the coefficient of x this cancels with this and becomes 1. so i have x equals 4 divided by negative 2 is negative 2. so that's my solution now the reason it's called a conditional equation is because this equation is true under this specific condition that x is

equal to negative 2. if i plug in a negative 2 for i would have negative 2 times negative 2 Equals 4 and then this left side simplifies to 4 so 4 equals 4. so that all checks out now if i choose anything else for x this is no longer a true statement if i was to say if i was to say x is equal to 7 for example well negative 2 times 7 is not equal to 4. negative 2 times 7 is negative 14. that's not equal to 4 this is wrong Wrong okay only if

i use x equals negative 2 will this equation be true all right so the next type of equation is known as an identity okay an identity an identity is always true okay no matter what we replace our variable with so we can replace it with a billion negative 2 4 27 negative 1 4 it does not matter It will always be true the key thing is that an identity has an infinite number of solutions so here's an example of an identity so we have 3x minus 12 equals 3 times the quantity x minus 4. and

before we go through and show why this is an identity just in your head pick any number you want doesn't matter what it is and plug it in for x and i bet it works i'm going to pick the number i don't know negative 8. so i'm Going to plug in a negative 8 here and a negative 8 here and let's see if we get a true statement so 3 times negative 8 is negative 24 minus 12 equals 3 times the quantity negative 8 minus 4 so negative 8 minus 4 and negative 8 minus 4

is negative 12. so i would have 3 times negative 12 on the right so this is negative 36 over here negative 24 minus 12 is negative 36 also And you might say okay well you solved that already you knew that x was equal to negative 8. pick anything else let's pick 4 for example let's pick 4 and see what happens so i'll plug in a 4 here and i'll plug in a 4 here so 3 times 4 is 12 over here minus 12. over here i'll have 3 times the quantity 4 minus 4 which is

0. so 3 times 0 is 0 over here 12 minus 12 Is 0. 0 equals 0. so that's a true statement let's try one more so now let's try i don't know negative 3 negative 3. so 3 times negative 3 minus 12 is equal to 3 times the quantity negative 3 minus 4. so over here 3 times negative 3 is negative 9 negative 9 minus 12 is negative 21. over here negative 3 minus 4 is negative 7 3 times negative 7 is negative 21. So again again again i get the left and the right side

as the same value so how can that be the case well the trick to this identity is if you haven't noticed it already the left and the right side are the exact same thing they're just in different forms this over here is just a simplified version of this over here if i use my distributive property on the right side so the left would Still be 3x minus 12. 3 times x is 3x and then minus 3 times 4 that's 12. so i have the exact same thing on both sides of the equation so of course

whatever i plug in here and here i'm going to get the same value because i'm doing the exact same thing if i plug in a 2 here and i plug in a 2 here in each case i'm multiplying 3 by 2 and then i'm subtracting away 12 on both sides so it doesn't matter what number i pick i do The exact same thing to it and so the left and the right side will always be equal and so there's an infinite number of solutions because i can choose any number that i want right it's literally

unlimited as far as how many solutions you would have so if you were to come across this in your textbook or your teacher gives it to you for homework or test just write that the solution is all real Numbers and as we move throughout algebra we're going to see some different notation that we're going to use to say hey all real numbers for right now we're just going to write it out okay we're not ready for the notation yet so we're just going to write all real numbers that's your solution all right so lastly we

see equations with no solution these equations are known as contradictions so contradictions So no matter what you choose for n it's never going to work and to see why let's let's simplify the right hand side of the equation so negative 2 equals we have 2 we have negative 6 times n that's negative 6n then we have negative 6 times negative n that's positive 6n so what you're going to see is that if i simplify the right hand side here the variable n is going to go away completely Negative 6n plus 6n those are opposites those

are gone so what i'm left with is just nonsensical statement that negative 2 equals 2. well it doesn't this is wrong negative 2 doesn't equal 2. so nothing is ever going to work here right no matter what you do because if i put something in for n i'm just going to subtract it away so the result of this would be 0 right i plug something in for n i subtract that same number away i get 0. I multiply 6 times 0 i get 0. so i'm just left with negative 2 equals 2 right once i

simplify this is what i'm going to have no matter what i choose for n negative 2 equals 2. so there's no value of n that can ever be chosen that will make this a true statement and so this is an example of a contradiction and so when this occurs you can put no solution and again we'll have some fancy notation For that moving forward but for right now just put no solution hey there's there's nothing i can do here so let's look at some examples and let's see if we can figure out through simplification what

type of equation they are so this one i have 63 minus 55n equals negative 11 times this quantity 5n minus 11. so on the left 63 minus 55n on the right i'm going to use my distributive property to remove the Parentheses so negative 11 times 5n that's negative 55n and then i'll have negative 11 times negative 11. that's positive 121. i told you this in one of the videos i made in a previous lesson or practice that i can't remember which one if you have the exact same number on two sides of the equation you

can get rid of it in other words i have negative 55 n here i have negative 55 n here i Can just get rid of it why is that the case well through the addition property of equality i can legally make it disappear if i add 55 n here and i add 55 in here i've added the same number to both sides of the equation in both cases it's gone it sums to zero and i'm left with 63 equals 121 which again is nonsense this is nonsense okay in case you don't know So there's no

solution here no solution this is another example of a contradiction right if you end up with your variable disappearing and you get you know one side doesn't equal the other you have a contradiction there's not a solution right so this is nonsense all right let's take a look at this one we have negative six times the quantity negative seven plus x is equal to negative six times the Quantity x minus seven so again let's simplify negative six times negative seven is forty-two negative six times x is negative six x so minus six x over here

negative 6 times x is negative 6x over here negative 6 times negative 7 is plus 42. now if i look closely i can see that these are the same right this is an identity and so upon First glance and i'm looking at this equation up here it looks like a regular equation but once you simplify and kind of really pay attention i can reorder this to negative 6x plus 42 and it's the exact same thing as over here negative 6x plus 42 so whatever value i choose for x doesn't matter what it is i plug

it in i multiply by negative 6 and i add 42 to it the same thing is going to happen on Both sides of the equation and so i can pick whatever i want for x and it'll be a solution so this would be all real numbers again this is an identity all right let's take a look at another one we have negative 4 minus 9 times the quantity x minus 6 this is equal to negative 3 times the quantity 3x plus 5 and then plus 2. again i'm going to simplify each side of the equation

i'm Going to look to see what type of equation i have so negative 4 i have negative 9 times x that's negative 9x or minus 9x negative 9 times negative 6 is plus 54. this is equal to here i have negative 3 times 3x that's negative 9x then i have negative 3 times 5 that's minus 15 and then plus 2. now again again again I've told you that if you have the same number on both sides of the equation you can get rid of it so i have negative 9x here i have negative 9x here

i can cancel them out again if i added 9x to both sides of the equation this is gone so what am i left with on the left negative 4 plus 54 is 50. on the right negative 15 plus 2 is negative 13. again 50 equals negative 13 that's nonsense Nonsense okay does not work so this has no solution another example of a contradiction and again if you look at this equation to start it looks like a normal equation something we would have gotten in previous lessons right something you would just crank out you're going to

get these on a test usually at the end of the section on linear equations in one variable your teacher is going to throw this at you and expect you to have Forgotten about these type of equations and then you end up trying to solve it and you just why is the variable gone well 50 equals negative 13 that doesn't make any sense so just remember if the variable is gone and one side doesn't equal the other there's no solution right it's a contradiction all right let's take a look at another example so we have 9

times the quantity x minus 6 equals 9x minus 54. So if i use my distributive property 9 times x is 9x minus 9 times 6 that's 54. so i have the same thing on the left and the right 9x minus 54. this is an identity right i can pick anything for x and it works so all real numbers all real numbers would be my solution here and again this type of equation is known as an identity hello and welcome to algebra 1 lesson 8. in this video we're going to talk about Applications of linear equations

this is going to be the first section of that and a lot of you will know this as solving word problems so before we kind of get started i just want to tell you that once you've mastered the material that we've presented in the first seven sections of this course you're ready to move on and start solving word problems now this is where a lot of students really get frustrated When you start solving word problems it's unlike solving the problems you've gotten before now in addition to solving an equation you also have to figure out

how to get the equation so that's what makes word problems so difficult you have to read through and figure out your own equation before you even solve it so the first thing i want to do here is i want to go through a brief procedure that we can use when we're solving a word problem now your textbook might Have something that's different if you want to follow that do so if you want to follow mine do so if you want to follow your teachers do so the main goal here is to get a procedure that

you can follow each and every time until you feel completely comfortable with word problems alright so solving a word problem so the first thing and probably the most important thing is to read the problem carefully and determine what you are Asked to find i can't tell you how many students i've tutored and you know we do a sample problem and they don't even know what they're looking for so this is probably the most important thing just to read the problem carefully again and determine what you are asked to find i always make my tutoring students

write out hey this is what i'm looking for so that they know what the overall goal is All right now the next thing is to assign a variable to represent the unknown sometimes there's going to be more than one unknown and as we get through algebra you're going to have a lot more complicated scenarios than you're going to face when you're just working with linear equations in one variable for the problems we're going to look at here if we have two unknowns we're going to be able to represent one of the Unknowns in terms of

the other all right so the next thing we're going to do is we're going to write out an equation and the way we do that we read back through our problem and we're going to see that something is equal to something else they're always going to give you a way to set up an equation so that you can get an answer now we get to the easy part we just solve the equation so this is the part that we already know how to do Then part five we state the answer in terms of the problem

so in regards to a word problem that's generally going to be a nice little sentence describing your answer then lastly we want to check the result and make sure it's reasonable so for example if you're asked how many people are on a train and you get an answer of 24.65 you're probably wrong right because there's not 0.65 of a person so you're looking for a nice whole Number or let's say they ask how many people are on a train and you get negative seven well there's that doesn't make any sense so when you're solving word

problems you need to make sure that your answer is reasonable if it's not you probably made a mistake alright so let's start off by looking at some different types of problems that you will encounter the first type is called sums of quantities so you're going to see these Right away when you get to your algebra one section on applications of linear equations and these problems involve finding two individual amounts sometimes it's more than two when the sum is known so that's the key so you're finding individual amounts when the sum is known so let's take

a look at an example so the phantoms won 8 more games than the aztecs last year their combined wins were 32. Find the number of games won by each team last year so kind of our first step is to read the problem and determine what you're asked to find so we've read the problem i recommend reading it more than once and writing down specifically what your goal is that way you're going to have a laser focus so again the phantoms won eight more games than the aztecs last year their combined wins were 32. Find the

number of games won by each team last year so this is my goal i want to find the number of games won by each team last year so at the end of the day i'm going to be writing a nice little sentence saying that the aztecs won this many games and the phantoms won this many games based on the information given in the problem all right so now we're going to move on to the next step and this step is a little bit Challenging for students so i want you to pay close attention so this

step is assign a variable so how do we assign a variable what do we do well remember we're assigning a variable to represent the unknown in this case we have two unknowns and our two unknowns would be how many games so how many games did the aztecs win And then also how many games did the phantoms win these are our two unknowns so when we get to the stage of assigning a variable it really doesn't matter which one i assign a variable to so in other words we could say let x equal the number of

games the aztecs one now you might be saying okay well if x equals the number of games that the Aztecs won what are you going to do about the phantoms well you're going to express the number of games that the phantoms won in terms of this variable x so let me show you how to do that let's scroll up to the top so if i let x equal the number of games that the aztecs won well the phantoms won 8 more games than the aztecs last year again let me kind of repeat this x is

the number of games the aztecs won The phantoms won 8 more than that so i can just say that then x plus 8 is going to be equal to the number of games the phantoms one very very simple and i know at first it's a little confusing but again if you read back through the problem it makes perfect sense so let's go back up there so the Phantoms won eight more games than the aztecs last year okay eight more their combined wins were 32. find the number of games won by each team last year so

again if i'm letting x be the number of games that the aztecs won last year well then x that number again plus eight will be the number of games that the phantoms won so now we have represented Both of our unknowns in terms of this one variable x okay so again we let x equal the number of games the aztecs won and so then x plus eight is the number of games the phantoms won now the next step is to write out an equation and in order to do that you have to make use of

the information given to you in the problem so again we're going to read it one more time so the phantoms won eight more games in the aztecs last year their combined wins and This is key right here their combined winds or 32. so if their combined winds were 32 that means that if i take the number of games that the aztecs won which is x and i add to that the number of games that the phantoms won which is x plus 8 that amount should be 32 so that's how you go through and set up

your equation again this is the number of games that the aztecs won okay Number of games number of games the aztecs won and then this is the number of games the phantoms won number of games the phantoms won if i sum those two amounts together they're combined wins i get 32. right this is the combined so i can go through and solve the equation find the value of x i'll know the number of games the aztecs won i can add eight to that value and find the Number of games the phantoms won let's go ahead

and crank this out so if we have x plus this quantity x plus 8 so we'd have x plus x that's 2x plus 8 equals 32 subtract 8 away from each side of the equation this will cancel and we'll have 2x equals 32 minus 8 is 24. divide both sides of the equation by 2 And i'll have x is equal to 12. so again what does that mean x equals 12 is not the solution don't circle that on your paper and go i'm done you're not because with a word problem you have additional work you

have to go through and make sense of your answer so that's why after we finish step number four solving the equation for step five you're going to state the answer and in order to be able to state the answer you've got to be Able to go back up and make sense of it so x is the number of games the aztecs won so that means the aztecs won 12 games so the aztecs won 12 games while the phantoms won remember they won eight more games so 12 Plus eight is 20. right remember you had x

plus eight to represent the number of games that the phantoms won 12 is x so we now know the value for for x which is 12. plug that in 12 plus 8 would be 20. so while the phantoms won 20 games so the aztecs won 12 games while the phantoms won 20 games so now you've answered your word problem you just want to read back through the problem and Make sure that your answer makes sense so the phantoms won eight more games than the aztecs last year well yeah the phantoms won 20 games the aztecs

won 112. 12 plus 8 is 20. so that checks out their combined wins were 32. 12 plus 20 is 32. so that checks out also so it looks like we have the correct answer again the phantoms won 20 games and the aztecs 112. so one question you might have is what if i started out by saying okay well the Number of games that the phantoms won would be represented with x so let's try it that way let x equal the number of games the phantoms 1. so then x minus 8 is now the number of

games the aztecs won So you can do it either way it doesn't really matter you're going to get the same answer so we let x equal the number of games the phantoms won and so then x minus 8 is the number of games the aztecs won and again their combined wins were 32 so i can do the same thing as far as setting up an equation so x plus this quantity x minus 8 equals 32 x plus x is 2x Minus 8 equals 32 add 8 to both sides of the equation that's gone i'll have

2x is equal to 40. divide both sides of the equation by 2. and what i'm going to have now is that x equals 20. and you might say well you got a different answer before you set x equals 12 remember we changed what x was representing so of course the answer is going to change but we still got the Correct answer and the same answer as before x equals 20 here because again we let x equal the number of games the phantoms won okay so the phantoms won 20 games and then x minus 8 is

the number of games the aztecs won so 20 minus 8 is 12. so again you get the same answer the phantoms won 20 games and the aztecs won 20-8 or 12 games alright let's take a look at another one So there are three girls in a family we have jessica jennifer and ashley ashley is three years older than jessica jennifer is five years younger than jessica their combined age is 34. find the age of each girl so when you read this problem for the first time again your goal is to figure out what you're being

asked to find we're being asked to find the age of each girl this is another problem Where we're trying to find the individual amounts given that we have the sum of the individual amounts so their combined age is given to us it's 34. so if there are three girls jessica jennifer and ashley they represent our three unknowns we want to find the age of each one so when we move on to our next step which is to assign a variable to represent the unknown well you have Three unknowns here you don't know the age of

ashley you don't know the age of jessica and you don't know the age of jennifer now i'm going to set a variable equal to one of their ages and then i'm going to express the other girls ages in terms of that one variable now here's a little trick when you read through your problem if you have one of the individuals that is involved in both comparisons so In this case you have ashley is three years older than jessica and then jennifer is five years younger than jessica so jessica is involved in the comparison with ashley

and also with jennifer so i'm going to go ahead and let x i'm going to let x equal jessica's age now i can express the other girls ages in terms of this variable x so i can say then For ashley we're told that ashley is three years older than jessica so if she's three years older than jessica and jessica's age is x well then x again jessica's age plus 3 is going to equal ashley's age and then what about jennifer well it's also telling me that jennifer is five years younger than jessica so that tells

Me i can put then x which again is jessica's age minus five okay minus five because jennifer is 5 years younger that's going to be jennifer's age so we have x as jessica's age then we have x plus 3 as ashley's age right because ashley is 3 years older than jessica and then we have x minus 5 as jennifer's age again because jennifer is 5 years younger than jessica all right so in the next step we're asked to Set up an equation so how do we set up an equation here well again just like we

saw in the last problem we have the sum of the individual amounts that are going to be equal to something so if we look at the problem again we're told that their combined age is 34. so what that means is if i take x which is jessica's age and i add it to x plus 3 which is ashley's age And then i add it to x minus 5 which is jennifer's age i should get a value of 34. so we simplify this x plus x plus x is 3x and then plus 3 minus 5 would

be negative 2. so i can put plus negative 2 or i can just put minus 2 it doesn't really matter so this is equal to 34 and then i'll add 2 to both sides of the equation that's gone and we'll have 3x equals 36. all right so now we'll divide both sides Of the equation by 3 so that x can be by itself this will cancel with this and i'll have x is equal to 12. and you never thought that solving the equation would be the easy part but now it's become that so i have

x equals 12. but again i'm not done please don't stop and turn your test in with this answer on you will be wrong you have to translate this now into a sentence So x equals 12. x is jessica's age jessica is 12 years old ashley remember we represented her age with x plus 3 because she was three years older jessica is 12 years old while ashley is 15 years old and lastly jennifer [Music] She's five years younger than jessica right we represented her age with x minus five so x is 12 12 minus five is

7. so and jennifer is 7 years old so jessica is 12 years old while ashley is 15 years old and jennifer is 7 years old so go back to your problem and make sure that it makes sense make sure that the answer is reasonable so there are three girls in a family Jessica jennifer and ashley ashley is three years older than jessica we said jessica was 12. so ashley's 15. jennifer is 5 years younger than jessica well we said that jessica was 12 and so jennifer is 7. their combined age is 34. so we have

ashley's age which is 15. we have jessica's age which is 12 and we have jennifer's age which is 7. so if you were to sum those amounts 5 plus 2 is 7 7 plus 7 is 14. Put the 4 down carry a 1 1 plus 1 plus 1 is 3. so yeah that works out this does sum to 34. so we have the correct answer again jessica is 12 jennifer is 7 and ashley is 15. all right let's take a look at a very easy problem to kind of conclude our introduction to applications of linear

equations so we're talking about consecutive integer problems and you might see these In your book some books skip it because it's so easy but basically when two integers differ by one they are called consecutive integers so if i have the numbers two and three they're different by one or 14 and 15 or 123 and 124 that's what we mean by consecutive integers so let's take a look at a problem so two pages that face each other in a pamphlet have 33 as the sum of the page numbers What are the page numbers so we're trying

to find what are the page numbers so this is an example of a consecutive integer problem if you think about a book or pamphlet if you have a page here and here and they face each other so let's say this is one and this is two or it could be It could be let's say 567 and 568. you know so on and so forth so we're trying to find the individual page numbers given that the sum is 33. so again just like we saw the previous two problems we're going to let x equal one of

them okay one of the page numbers and then we'll represent the other page number in terms of that variable x so let's let the smaller okay let's let The smaller integer bx so let x equal the smaller integer so then since they differ by 1 we could say that x plus 1 is equal to the larger the larger integer so now it's easy to set up the equation because we know that the smaller page number plus the larger page number and i Put smaller integer larger integer you could put smaller page number larger page it

really doesn't matter as long as you understand what's going on really we're thinking that okay x now which again is the smaller page number the smaller integer plus x plus one which is the larger page number or the larger integer should sum to 33. so let's go ahead and set up that Equation so we would have that x plus x plus one is equal to 33. again this plus this gives me this okay so that's my equation so now to solve it simplify over here you'd have 2x plus 1 equals 33 subtract 1 from both

sides of the equation and we're going to have 2x is equal to 32 divide both sides of the equation by 2 and i'll have x is equal to 16. so again i'm not done very important that you go back up and figure out what that means so x is the smaller integer so that means that the smaller page number is 16 and the larger page number is 17. so for the page numbers we could say that the two page numbers are 16 and 17. right if you take 16 and you add 17 you're going to

get 33 right and the two pages do face each other in the book so if you had a book or pamphlet or whatever it is this is 16 this is 17 if you close the book the two pages would face each other hello and welcome to algebra 1 lesson 9. in this video we're going to continue to Look at applications of linear equations so in the last lesson we started talking about word problems and we talked about how you would go about solving a word problem so the first thing you always want to do you

want to read the problem carefully and determine what you are asked to find so generally you might highlight that or write down hey this is what i'm looking for that allows you to have a laser focus Throughout your problem the next thing you want to do is assign a variable to represent the unknown now in a lot of these problems you're going to have more than one unknown so you're going to assign a variable to represent one of the unknowns and then you're going to represent the other unknowns in terms of this variable all right

the next thing you're going to do is write out an equation and you're going to do that by reading through the Problem you'll have something equal to something else this is going to allow you to write an equation then solve the equation and then state the answer in terms of the problem so it's very important you don't just solve the equation with a word problem and then hand in your paper and say i'm done you have to state the answer in terms of the problem and then lastly you want to check the result and make

sure it is reasonable if You get asked for how many cars are on a bridge and you get an answer that's negative seven you probably got the wrong answer right you should have either zero or some positive amount in that scenario so you want to make sure that your answer is reasonable all right so we're going to start out today by talking about everybody's favorite which you will come to know them as mixture problems So these problems involve mixing two or more substances and then finding some unknown now the problems we're going to look at

today are going to be relatively simple but as we move forward they're going to get more and more complex so the first thing we need to understand with a mixture problem is how to find the pure amount of a substance in a mixture so here's a quick example so a chemist has 20 liters of a 25 Acid solution so what does that mean when i say a 25 acid solution it means that 25 percent of it is acid how much pure acid is in this solution so to figure this out you just take the amount

of the solution which is 20 liters so i'm just going to write 20. i don't need to write liters there and i would multiply it by the percentage of acid that's in it so if it's 25 acid solution 25 of it is acid Now i'm going to convert 25 percent to a decimal so that's 0.25 and this will tell me how much pure acid is in the solution so 20 times 0.25 is the same as 20 times 1 4 and i can do that in my head that's going to be 5. right it's basically like

taking 20 and dividing by 4. so there would be 5 liters 5 liters of pure Acid it's 20 liters of a 25 acid solution so how much pure acid is in the solution again 25 of 20 is 5 so that means there's five liters of pure acid all right we're going to look at another one that's simple and kind of gives us a little warm up as far as you know understanding how to find the pure amount of something okay very very important so this won't be a typical Mixture problem this will be a very

easy one so for her birthday party tina mixed together two liters of joe's fruit punch and 10 liters of babes fruit punch joe's fruit punch contains 15 fruit juice while babes contains 45 what is the percentage of fruit juice in the mixture so what is the percentage of fruit juice in the mixture how do i obtain that forget about the numbers for now If i want the percentage of something well i would want the amount that i'm looking for as a pure amount over the total so in this particular case we're looking for the amount

of pure fruit juice so the amount of pure fruit juice and that's going to be in liters right because we're talking about this mixtures in terms of liters Over just the amount of the mixture the amount of the mixture and we know that this amount here the amount of the mixture is larger and this amount is smaller so we're going to end up with something that's less than 1 greater than zero right it's going to be a decimal something like point something now i would then take that amount and i would convert it to a

Percentage and i could say hey this is the percentage of fruit juice in this mixture so let's fill in the information here how much pure fruit juice do we have well we have two liters of joe's fruit punch and 10 liters of babes so we have 2 liters this is joe's and we have 10 liters of babes now i can already figure out what the Total amount of the mixture is going to be because 2 liters plus 10 liters is 12 liters so i'm just going to write 12. now what's the concentration of each well

it says that joe's fruit punch is 15 fruit juice and babes is 45 so joe's is 15 15 and babes is 45 so if i multiply 2 which is the number of liters of joe's Times 15 or 0.15 in decimal form that would give me the amount of pure fruit juice in joe's fruit punch right the amount i added from that so 2 times 0.15 is 0.3 then if i do the same thing over here if i have 10 which is the number of liters from the babe's fruit punch times 0.45 which is 45 percent

as a decimal that's going to give me the amount of Pure fruit juice that i'm throwing in from babes fruit punch so this is going to be 4.5 so if i sum these two amounts together the amount of pure fruit juice that i'm putting in from joe's plus the amount of pure fruit juice that i'm putting in from babes i'd get 4.8 right 0.3 plus 4.5 is 4.8 so essentially i have 4.8 divided by 12. so let's erase this real quick We'll do our calculation so again what i figured out was the amount of pure

fruit juice i have 4.8 liters and i know the amount of the mixture that's 12 liters if i divide one by the other i'm going to get as a decimal 0.4 so as a percentage i just move this guy over two places to the right and i'd have 40 percent so that means that my mixture is going to be 40 fruit juice so i can come back up here What is the percentage of fruit juice in the mixture the mixture is 40 percent fruit juice and it's kind of easy to check that i mean this

is a very simple one so tina makes two liters of joe's fruit punch and 10 liters of babes fruit punch again 2 plus 10 is 12. so that's the total amount of the mixture Now joe's fruit punch contains 15 percent fruit juice so if i take 2 liters or just the number 2 and i multiply it by 15 or in decimal form 0.15 that means that i have 0.3 liters of pure fruit juice that's coming from joe's fruit punch now from babes fruit punch i've got 10 liters and it's 45 fruit juice so i'm getting

4.5 liters of pure fruit juice there combine these two together through Addition and what i'd have is i'd have 4.8 liters of pure fruit juice in the mixture that has a total of 12 liters so all you have to do is take 4.8 and divide it by 12 which is the total number of liters in the mixture that gives me 0.4 and so as a percentage that's 40. so 40 of this mixture is going to be pure fruit juice all right so let's take a look at a Harder one now one that's going to involve

the use of variables to solve so a metallurgist needs to make 10 milligrams of an alloy containing 82 percent silver so let's stop for a second some of you don't know what an alloy is so an alloy is a metal that's made by combining two or more metallic elements all right so then she is going to melt and combine one metal that is 28 silver Let me highlight that 28 silver with another that is 88 silver how much of each should she use so here's our question how much of each should she use meaning how

much in terms of milligrams should she use of the one metal that's 28 silver how much in terms of milligrams should she use of the other metal that is 88 percent silver so when we move to our next step and we want to assign a variable to represent the unknown We have those two unknowns so let's let x equal one of those so let's let it equal let's let x equal the amount of metal in milligrams that is 28 percent silver okay so the amount of metal in Milligrams that is 28 silver now to express

the other unknown in terms of x you kind of have to read through the problem this is a little tricky the first time so i have to kind of walk you through it here's the information that you might have missed a metallurgist needs to make 10 milligrams of an alloy containing 82 silver this is the first key 10 milligrams So we know that x is the amount of metal in milligrams that is 28 percent silver so whatever x is if i add the other amount to it the sum should be 10 right in terms of

milligrams right it should be 10 milligrams x milligrams plus some other amount of milligrams should equal 10 milligrams so now i can kind of work backwards and say okay well 10 is equal to X plus this amount here can only be 10 minus whatever x is and i know that's a little tricky at first but 10 minus whatever this is let's say x was 4 for example i'm not saying that's the answer but let's just say x was 4. so if this was 4 10 minus 4 would be 6. so 4 plus 6 would be

10. in other words whatever x is the other amount when added to that has to sum to 10. so that's why this would work out so if let's say x was eight then this amount would have to end up being two right because these two amounts have to sum to ten so that's where you're going to get this formula from so x plus 10 minus x is going to equal 10. now that gives us our relationship for the other metal so we could say then 10 Minus x is going to be the amount of metal

in milligrams that is 88 silver [Music] and again i know the first time you see that it's a little confusing but as you work through these problems that's going to become something you look for all the time All right so now we want to set up an equation so some of you might immediately make the mistake and say okay well i know the weights so my equation is x plus 10 minus x equals 10. and that's okay let's go through and solve this x minus x is 0 and then i have 10 equals 10. and

you might go okay what happened what's wrong here Well you don't have enough information what you need here is to relate the fact that okay you have a certain weight for x but that weight is a certain percentage of silver so if .28 right 28 percent as a decimal is 0.28 times x is used to represent the amount In milligrams of pure silver then we add this to now we have the amount of metal in milligrams that's 88 percent silver so 0.88 which is 88 as a decimal times this remember to use this parentheses here

10 minus x very important okay you can't just put it next to the 10 that won't work this is the same thing it's the amount in milligrams of pure silver So these two amounts have to be equal to what it told us in the problem and in the problem it says that the metallurgist needs to make 10 milligrams of an alloy containing 82 percent silver so 10 is the total weight and the percentage is 82 percent so 0.82 and so now we have an equation now we have something we can solve so amount in milligrams

of pure silver is equal to the amount In milligrams of pure silver one side is going to be equal to the other and we have something we can work with let's go ahead and solve the equation now so we'll have 0.28 x plus 0.88 times 10. i'm just going to move this one place to the right so that would be 8.8 then minus 0.88 times x is 0.88 x and equals 0.82 times 10 is 8.2 so if we combine like terms here 0.28 x minus 0.88 x is going to give me negative 0.6 x then

plus 8.8 equals 8.2 i'm going to subtract 8.8 away from each side of the equation that's gone i'm going to have negative .6x is equal to 8.2 minus 8.8 is going to be negative 0.6 divide both sides of the equation by Negative 0.6 the coefficient of x that's going to cancel with that and this is going to cancel out also and become 1. so we have x equals 1. all right so let's make sense of this remember that x is the amount of metal in milligrams that is 28 silver so that we said was 1

because x was equal to 1 when we solved our equation so we're going to say that she needs she needs one milligram Of the metal that is 28 percent silver and then along with along with 10 which is the total weight of this thing minus 1 which is the weight of the metal that is 28 silver so 9 so along with 9 milligrams remember we set that up as 10 minus x 10 minus 1 would be 9. so Along with 9 milligrams of the metal [Music] that is 88 silver so that's what she's going to

use to make this alloy that is 10 milligrams and 82 silver so check everything out if you go back up again a metal allergist needs to make 10 milligrams of an alloy containing 82 Silver so we know that one milligram plus 9 milligrams is 10 milligrams so that part checks out we just need to check the pure amount of silver in the substance so our goal is to have 82 percent it needs to be 82 percent so if we have one milligram that's 28 silver so times 0.28 and we add that to 9 milligrams That's

88 silver so times 0.88 this should be equal to 10 milligrams times 0.82 and this is the exact equation that we set up we just didn't know the values so 1 times 0.28 is 0.28 and then plus we have 9 times 0.88 that's 7.92 and again 10 times .82 would be 8.2 so this is what i'm looking for i'm looking for 8.2 milligrams of silver right in this 10 milligram alloy So what is 0.28 plus 7.92 well that is 8.2 right you get 8.2 equals 8.2 and so your answer here is correct right she needs

to use one milligram of the metal that is 28 silver she's gonna have .28 milligrams of silver from that and then along with 9 milligrams of the metal that is 88 silver she's going to get 7.92 milligrams of pure silver from that she combines them together and she gets 8.2 milligrams of pure silver for her alloy That is 82 silver right it's 8.2 milligrams out of a total of 10 milligrams all right so now we're going to talk a little bit about something known as motion problems these problems involve the distance formula and there's more

than one distance formula we're going to cover the other one in the last part of algebra one it's relates to how much distance is between two points on a graph we're not talking about that one We're talking about distance equals rate times time so rate of speed times the amount of time traveled at that speed and this formula is very very intuitive pretty much everybody has taken a road trip at some point and said if we go 60 miles per hour for three hours we're going to end up going 180 miles so in other words

if my rate of speed is 60 miles per hour and this is in terms of hours so Our time has to be given in terms of hours for this to make sense so 60 miles per hour times three hours how far are we going to go well forget about the units here we're going 60 times 3 that's 180. so we're going to end up going 180 miles right if i go 60 miles per hour for 3 hours we go 180 miles we all know that and that's what the distance formula does it tells us that

the amount Of distance traveled is equal to the rate of speed times the amount of time traveled alright so shannon left the science fair and traveled toward the recycling plant amanda left one hour later traveling 22 miles per hour faster in an effort to catch up to her after two hours of traveling amanda finally caught up what was shannon's average speed here's the key question here what was Shannon's average speed i can tell you this is kind of a difficult problem to solve if you've never done a motion word problem right you really have a

lot of things that you have to consider here so the first thing is let's revisit our distance formula so distance is equal to the rate of speed times the time Traveled and again if you ever forget that just think about okay i'm taking a road trip and i'm going for two hours that's my time times 50 miles per hour that's my rate of speed that gives me a distance so in that case 100 is equal to rate of speed of 50 times time traveled too and i left the units out just for simplicity but this

is 50 miles per hour this is 2 hours this is 100 miles now you see that one side equals the other so if i have a rate of speed and a time traveled i have a distance and i can calculate that now knowing this information here if we go back up to the problem kind of the first thing we're given is that shannon left the science fair and traveled towards the recycling plant it Doesn't give us a distance amanda left one hour later traveling 22 miles per hour faster in an effort to catch up to

her after two hours so this is key after two hours of traveling amanda finally caught up so one thing we know about the distance in this particular case amanda and shannon travel the same amount Or travel the same distance so their distances in the end will be equal now later on in algebra we use some different types of notation which won't come up for a while but i'm going to write d a capital d with a small a to relate the amount of distance this is The distance for amanda and this is equal to i'm

going to put a big d with a little s and this is the distance for shannon and this is just here to show you that these two distances are equal they both leave from the same place and they both go to the same place now they do it in a different amount of Time and so they do it at different rates as well so the time is different and the rate of speed is different but the distances are the same so it's the first thing so let's talk a little bit about time now so shannon

left the science fair and traveled toward the recycling plant amanda left one hour later so whatever amount of time that shannon drives for it's an hour more than amanda because Amanda delayed her trip by an hour so it says here after two hours of traveling amanda finally caught up if amanda drives for two hours and she drove for an hour less than shannon then shannon drove for three hours so when we talk about time let's put a big t and a little a and this is the time for amanda time for amanda Let's put a

big t and a little s and this is the time for shannon and we know the time for amanda is two hours we know the time for shannon is one more hour than that so three hours we'll just put two and then three just need to get some information about the rate of speed so shannon left the science fair and Traveled toward the recycling plant amanda left one hour later traveling 22 miles per hour faster in an effort to catch up to her now what i can do here since amanda is going 22 miles per

hour faster i don't know how fast shannon's going so i can use a variable to represent that so let's let x let's let x equal shannon's speed And miles per hour so then x plus 22 as it tells us in the problem she's going 22 miles per hour faster amanda is is going to equal amanda speed again this is in miles per hour so we have a little bit more information let me scroll down and get a little room we know from the distance formula again distance Is equal to rate of speed times time traveled

so for shannon what information do we have her distance we have her rate of speed which i assigned the variable x to and we have her time as 3 for 3 hours so distance for her equals 3x now for amanda her distance looks like this She has a time traveled of two hours and her rate of speed is x plus 22. remember x plus 22 is the rate of speed so the whole thing has to be multiplied by two if you don't use parentheses this just looks like 2x plus 22. that's wrong the 2 has

to be multiplied by the entire thing each term so we have we have a distance for each person individually and remember the two distances are the same So this is how we get our equation in this type of problem we know that shannon's distance okay remember i put d with a little s for shannon's distance is equal to amanda's distance d with a little a so i can take 3x which relates to shannon's distance 3x and i can set that equal to 2 Times the quantity x plus 22. remember if you give me a rate

times the time i have a distance so i have a rate times the time here that's a distance and it's equal to or the same as a rate times the time here which is another distance right these two distances are the same they look different because the time and the rates are different but in the end they're the same so what i can do now is just solve this equation and find out rates of speed So over here i'm going to simplify 2 times x is 2x plus 2 times 22 that's 44 over here i

have 3x subtract 2x away from each side very very simple equation we end up with x is equal to 44. remember 44 was the rate of speed for shannon let x equal shannon's speed in miles per hour then x plus 22 or 44 plus 22 or 66 is amanda's speed in miles per hour so Let's go ahead and write our answer so what was shannon's average speed is our official question so shannon's average speed was 44 miles per hour and we can check this to make sure that it's true so we know that shannon traveled

for Three hours so three hours think about your road trip three hours times 44 miles per hour how far am i going to go i'm going to go 132 miles now we also know that amanda leaves an hour after her so she only travels for two hours she's traveling at x plus 22 miles per hour x was 44. 44 plus 22 is 66 so 2 times 66. She drives for 2 hours at 66 miles per hour how far does she go she goes 132 miles right so these distances again are equal and so we know

we have the correct answer shannon's average speed here was 44 miles per hour hello and welcome to algebra 1 lesson 10. in this video we're going to learn about proportions so before we get into proportions let's recall that a ratio is a comparison of two quantities so one way you might see this written Let's say you had the ratio of a to b you might see it written with a colon so you might see a and then a colon and then b so this is the ratio of a to b you also might see it

written with the word two so you might see a to b and then lastly you might see it written as a fraction so you might see a over b so each of these would be a different Method to display the ratio a to b alright so as a quick example let's think about the ratio of boys to girls in a classroom so suppose there are 12 boys and three girls so the ratio of boys two girls looks like this you have 12 boys 12 boys and you have three girls so you can put 12 to

3 like that You could put 12 colon 3 like that or you could put 12 over 3. now remember a ratio can be reduced to its lowest terms the same way you would reduce a fraction so you look at kind of the first part the part that would be in the numerator and you look at the last part the part that would be in the denominator and you say hey what's my greatest common factor So when we look at 12 and we look at three the gcf would be 3. so if i divide 12 by

3 i get 4 if i divide 3 by 3 i get 1. so the ratio of 12 to 3 can be simplified and thought of as the ratio of four to one now a lot of times i know when working with fractions we might get rid of this one we might just write that as the Number four when we're working with ratios we don't do that so if i have the ratio of boys to girls i want to say it's four to one i don't want to just say it's four right so you want that

one there it's very important when you're working with the ratio to just leave it there because you're talking about a comparison of one thing to another all right let's talk a little bit about a rate so a rate is also a ratio And this is one in which the units are different so the units are different so typically we see these as unit rates okay unit rates and every time you go into a grocery store you're going to deal with this if you look at the little grocery labels next to the price it'll probably say

this is the price per ounce or this is the price per pound you know so on and so forth so how much of something per One unit of another so you think about a car drove 300 miles on six gallons of gas you can use this information to set up a unit rate and figure out the miles per gallon here so 300 miles 300 miles put that in the numerator and then i'm gonna put six gallons six gallons of gas in the denominator so we have 300 miles per you can think of your fraction bar

saying per Six gallons of gas and we could simplify this by looking at the numbers if i divide 300 by 6 i would get 50. so i would write 50 and the units are going to stay there miles per and then if i divide 6 by 6 i'm going to get 1. so 1 gallon of gas so 50 miles per 1 gallon of gas and That's very very common we're always thinking about our fuel economy so what is the number of miles i'm going to get per gallon of gas you know on average so if

you do a road trip you can track how many miles you do and on how many gallons of gas you can set up the same thing you can figure out your miles per gallon and one thing about a unit rate you're going to notice that the units do not cancel so with a typical ratio we have the same units and so we really Don't list them right you don't need to because you could think of them as kind of cancelling out here i have miles in the numerator and i have gallons of gas or gallon

of gas in the denominator so those we're going to write right for clarity but they don't cancel out so we just keep them there and basically the only thing you need to understand is you're always going to Take the number in the numerator which in this case is 300 and divide it by the number in the denominator which there is 6. that's going to give you some amount in the numerator per one amount in the denominator so it's always per a single unit you always want a single unit in the denominator that's why it's called

a unit rate so let's take a look at another one real quick so 12 eggs purchased for three dollars and sixty Cents so i can put three dollars and sixty cents in the numerator and i can put per 12 eggs in the denominator and what i'll do is i'll divide this 3.60 cents or just 3.60 by 12 and i'm going to get a unit rate which is a cost per 1 egg so if i divide 3.60 by 12 i get 0.3 so i would get 0.3 Dollars per one egg right so this is the unit

cost right the cost per one egg and again i take the number part from the numerator and i divide it by the number part in the denominator and that's going to give me my unit cost so this denominator will always be singular so instead of eggs it becomes egg and then the numerator whatever number you end up with it's going to be the Value per one unit so again this is a single a single unit and that's why we call it a unit rate all right so now that we've kind of reviewed the basics of

what a ratio is let's talk a little bit about a proportion so we talked about this in pre-algebra a proportion states that two ratios are equal and a lot of times we're just looking at two fractions right we're trying to see if those two fractions are equal and if they are we Say that we have a proportion so i want you to recall that we can determine if there's a proportion by using the equality test for fractions the equality test for fractions how do we do the equality test for fractions it's very very simple i

take the denominator of one fraction and i multiply it by the numerator of the other so 4 times 2 is 8 and i look at that Value i then take the denominator of the other fraction and i multiply it by the numerator of the other fraction 8 times 1 is 8. if i get the same value on each side then the fractions are equal or you could say the ratios are equal however you want to think about it and so you have a proportion so yes this is a proportion and it's important to note that

this is Called cross multiplying so it's always going to be the denominator of your first fraction times the numerator of the second one and then the denominator of the second one times the numerator of the first one let's take a look at this one we have 3 8 and 4 7 we want to know if we have a proportion so again we're going to cross multiply so i would take the denominator of this fraction here Multiply it by the numerator of this fraction here so 8 times 4 is 32 then i take the denominator of

this fraction here and i'd multiply it by the numerator of this fraction here 7 times 3 is 21. so we do not have the same value 21 is not the same as 32. so this is no it's not a proportion again in order to have a proportion this times this has to be the same as this times this have to get the same value so before we kind of move on To solving a proportion equation i want to kind of cover something you're going to see in your textbook so you're going to see that if

a over b is equal to c over d then it must be true that a d is equal to bc so what they're doing is they're cross multiplying they're saying okay b times c Is bc and that has to be equal to d times a which is a d or d a or however you want to write that it doesn't matter now you might think about where did this cross multiplication come from well if i kind of look at the denominators here i have b and i have d so let's say i multiplied both sides

of the equation by the lcd which is bd right b times d If i multiply this side by bd which is perfectly legal with an equation as long as i do it to this side we see here that this would cancel here and we also see that this would cancel here so what would i have left i'd have d times a which is the same as a d right i just switched the order and this is equal to c times b or bc right so if i multiply both sides of the equation by the lcd

i end up with this right here this is kind Of just a shortcut right i'm multiplying this by this and i'm multiplying this by this because the other part is going to cancel out so i end up with a d equals b c so we're going to be using this logic to solve a proportion equation so here we have 4 over 7 equals 2 over x now in order to solve for the unknown in a proportion equation we just remember that if it's true that 4 7 equals 2 over x that's 7 times 2 which

is 14 7 times 2 has to be equal to x times 4 or 4x otherwise we can't put an equal sign here so knowing that i just cross multiply and set up an equation like i did here this just becomes 14 is equal to 4x and we simply solve for x we divide both sides of the equation by 4 And we end up with 7 halves right if i divide 14 by 2 i get 7 if i divide 4 by 2 i get 2 so 7 halves equals x or in decimal form you could write

3.5 but typically if i'm working with fractions or ratios or however you want to think about that if i'm working with that i want to stay with that i don't want to switch to decimals so i get x equals 7 halves or i have 7 halves equals x it doesn't really matter How you say that and to check it you just plug it in and just plug in for x so what would i have i'd have 4 7 is equal to 2 remember i could write 2 as 2 over 1 divided by this x which

is 7 halves and then it makes it obvious that i'm going to end up what i'm going to take 2 over 1 and i'm going to multiply by the reciprocal of 7 halves which is 2 over 7. 2 times 2 is 4 over 1 times 7 which is 7. so this becomes 4 7 as well so we get that 4 7 equals 4 7. both sides are the same value and so we know our solution x equals 7 halves is correct all right let's take a look at another one we have 7 over y equals

4 8 or 4 over 8. so again i'm going to cross multiply and set up an equation y times 4 which is just 4y has to be equal to 8 times 7 8 times 7 is 56 so i can write 8 times 7 and do it in two steps so that's the first thing just cross multiply set up an equation so then we have 4y is equal to 56 right 8 times 7 is 56 and then we just divide both sides of the equation by 4 the coefficient of y 56 divided by 4 is 14.

so we get 14 here so y equals 14. all right so if i plug in a 14 for y you can mentally see that that's going To work out right because 4 is half of 8 and 7 is half of 14. so if i reduce each if i divide 4 by 4 i get 1 if i divide 8 by 4 i get 2. right so if i think about this just as taking the fractions and simplifying them 4 8 is one-half 7 14 is also one-half right if i divide 7 by 7 i get 1.

if i divide 14 by 7 i get 2. so one-half equals one-half is what you'd end up with and so y equals 14 is the correct solution all right now We want to take a look at x over 2 equals 7 6. so again i'm just going to cross multiply and set up an equation so 2 times 7 is 14 14 and that equals 6 times x that's 6x all right so once we've set up the equation we just divide both sides of the equation by the coefficient of x which is 6. that will cancel

with that just leaving with x What is 14 divided by 6 well i can use a decimal which again if i'm working with fractions or again ratios i just want to keep it in that form make it nice and simple so i'm just going to simplify this since i wouldn't get a whole number each is divisible by 2 so this divided by 2 is 7. this divided by 2 is 3 so x is going to equal 7 3. again plug in for x there so i'd have 7 3 Divided by 2. so divided by 2

and again for the purpose of working with fractions we'll say divided by 2 over one and so i know that that's basically seven thirds times the reciprocal of two over one so it's seven thirds times one half seven times one is seven which is what i have up here three times two is six which is what i have here so i would get 7 6 on the left as well as on the right So again x equals 7 3 is my correct solution all right let's take a look at one that's a little bit harder

and by harder i just mean more tedious so we have 5 over 3 is equal to k minus 2 over k minus 7. so i'm doing the same thing i'm going to take 3 and multiply it by this side now very very important here i am multiplying 3 by this whole quantity so i need to use parentheses around that Quantity i need to say that 3 is going to be multiplied by k minus 2. if i just put 3k minus 2 like that i'm wrong because i didn't multiply 3 by 2 right so i need

to make sure that 3 is multiplied by each term here and to do that again i use parentheses so 3 times the quantity k minus 2 is equal to I'm going to have this value multiplied by this so 5 times the quantity k minus 7. again you've got to use parentheses you're multiplying 5 by k and by negative 7. so you use parentheses for that please don't make that mistake of not putting parentheses and not multiplying that number by each term very very common and it will cost you an answer on your test All right

so now we simplify so 3 times k is 3k minus 3 times 2 is 6 and this equals 5 times k is 5k then 5 times negative 7 is negative 35. so now this is pretty easy to solve i'm going to subtract 5k away from both sides that way i just have a variable term on the left nothing on the right so this will go away and i'm going to end up with i'm just Going to kind of drag this over here 3k minus 5k is negative 2k and then minus 6 equals negative 35. i

want to isolate this on one side of the equation so i'm going to add 6 to both sides so this is gone and what i'm left with over here is negative 2k and on the right negative 35 plus 6 is going to be negative 29. okay so continuing i want k by itself so I just divide both sides of the equation by negative 2 the coefficient of k that cancels with that i have k over on the left on the right i have negative 29 divided by negative 2 and 29 is a prime number i

would just leave it as 29 over 2 right negative divided by negative is positive so you can remove the negative signs and list that it's positive but you really can't simplify it any further Now checking this will be a little tedious but we want to do it anyway all right so i would have five thirds equals we have 29 halves minus two i'm going to go ahead and write two as four over 2. so i have a common denominator and then this is over again for k we have 29 over 2 and then minus we

have 7. i'm going to go ahead and write that as 14 over 2 again so i have a common denominator so this is basically what a complex Fraction over here we already know how to simplify a complex fraction essentially there's two different ways we could multiply the numerator and the denominator complex fraction by the lcd which would be two or we could just simplify the numerator and denominator separately so i think it would be a little faster if we multiply the numerator and denominator here By 2. so let's do that so i'm going to multiply

this by 2 and i'm going to multiply this by 2. now you might be thinking oh this is an equation so i got to do this to the other side no remember if i multiply a fraction by the same number top and bottom i don't change the value of that i'm multiplying by two over two which is just basically i'm multiplying by one If i multiply by one over here i can multiply by one over here but multiplying by one doesn't change the value of something so i don't need to do that to both sides

i just need to do it to this side this 2 is basically going to be distributed to this and it would cancel this 2 here and then to this and we cancel this 2 here so that's gone same thing over here would be Distributed this cancels this cancels when it gets over here so that's gone so i can kind of erase the denominators here because they would completely go away so this is gone and essentially i'd have 29 minus 4 which is 25. so i could write this as 25 and then over 29 minus 14

which is 15. now i can simplify this and make it look Like this 25 is divisible by 5. 25 divided by 5 is 5. 15 is divisible by 5. 15 divided by 5 is 3. so when i simplify 25 15 i end up with five thirds so you get five thirds equals five thirds so yeah our solution k equals 29 halves is correct all right let's take a look at another one we have 7 8 equals v minus seven over v plus four so again set up your equation eight Times this quantity v minus seven

again have to use parentheses here so i'm gonna put eight times the quantity v minus seven is equal to this quantity v plus four times seven so again i gotta use parentheses seven times the quantity v plus four again eight has to be multiplied by v and by negative seven right don't just do this don't just say okay a times v Minus seven that doesn't work so a times v is eight v minus eight times seven that's 56 and this equals seven times v that's seven v plus seven times four that's 28. all right so

i'm going to move my number to the right here move my variable term to the left okay and i'm going to do that in one step so i'm going to add 56 to both sides of the Equation that way there's no longer anything other than a variable term on the left and i'm going to subtract 7v from both sides of the equation that way there's no longer a variable term on the right so on the left now i have 8v minus 7v which is just v on the right i have 28 plus 56 which is

84 okay So v equals 84 is our solution here all right so we have 7 8 is equal to v minus 7 and again v is going to be 84 so 84 minus 7 over v plus 4. so again 84 plus 4 and so we'll have 7 8 over here what is 84 minus 7 well that's going to be 77 what is 84 plus 4 that's going to be 88 so each here is divisible by 11. if i Divide 77 by 11 i get 7. if i divide 88 by 11 i get 8. so 7

8 equals 7 8 and so v equals 84 is the correct solution all right let's take a look at another one we have negative 3 6 equals x minus 5 over 2x plus 2. so again i'm going to set up my equation so 6 is going to be multiplied by this quantity x minus 5. again you have to use parentheses so 6 times the quantity x minus 5 This is equal to now i'm going to take 2x plus 2 multiply it by negative 3. okay so negative 3 times this quantity 2x plus 2. again you

have to use parentheses you might want to like put that next to each problem like a big star so that it just gets into your head that i have to use parentheses when i'm multiplying a single number times a quantity so 6 times x is 6x Then minus 6 times 5 that's 30. this equals negative 3 times 2x that's negative 6x then negative 3 times 2 would be minus 6. all right so i am going to add 6x to both sides of the equation and i'm also going to add 30 to both sides of the

equation so this will go away and so will this and what will i have left 6x plus 6x is 12x Negative 6 plus 30 is 24. divide both sides of the equation by 12 the coefficient of x and we get x is equal to 24 divided by 12 is 2. so we have negative 3 over 6 is equal to up here i'm going to plug a 2 in for x so i'd have 2 and then minus 5 up here and then down here i'd have 2 times x so 2 times 2 and then plus 2.

so i'd have negative 3 Over six is equal to two minus five is negative three then two times two is four plus two is six so we get negative three over six equals negative three over six and if we wanted to we could reduce this to negative one-half in each case right this would divide by three and become one this would divide by three and become two same thing over here so we could have negative one-half Equals negative one-half but or you could also leave it as negative three six equals negative 3 6. the main

point is that your solution here which was x equals 2 is correct hello and welcome to algebra 1 lesson 11. in this video we're going to start talking about how to solve linear inequalities so this lesson is going to be all about learning how to solve a linear Inequality in one variable now the procedure is basically the same as with solving a linear equation in one variable but with a few key differences so let's start out today by just reviewing some basics about inequalities so the first thing is that this symbol right here it points

to the left it's known as the less than symbol then this symbol right here that points to the left with a bar underneath it is the less than or equal to this type of Inequality is known as a strict inequality so something is strictly less than this type of inequality is known as a non-strict a non-strict inequality so it allows for the possibility that they're equal right it says or equal to right in this case it's strictly less than in this case it's less than or equal to now we also have the greater than this

points to the right and then we have the greater than Or equal to points to the right but has a bar underneath it and again this greater than is a strict inequality this is a strict inequality strictly greater than this type of inequality that's greater than or equal to is a non-strict is a non-strict that's because we allow for this or equal to so again the symbols less than and greater than are referred to as strict Strict inequalities then the symbols less than or equal to and greater than or equal to are known as non-strict

inequalities very important that you understand the difference between the two so in order to start thinking about inequalities in the right way i want you to first think about a basic equation let's say we have something like x equals 2. so if i was to graph x equals 2 using this basic number line What i would do is i would take and i would put a filled in dot or filled in circle however you want to think about that specifically at the number 2 on the number line if i had something different let's say i

had x equals 5 for example let me do this in a different color so that we don't get confused so let's say i had x equals five Well i would graph that by putting a filled in dot or a filled in circle at five that's my solution let's look at one more let's say we had x equals negative 3. well i'd put in again a filled in dot or filled in circle at negative 3. so in each case we're showing a visual representation of the solution to the equation and this is going to get much

more complex as we go throughout algebra For right now this is about as basic as it gets what about x is greater than 2 how would i graph that well we're going to come back to this but i just want you to realize that in the case of an equation we have an exact number it's kind of a laser focus it's x has to be two it's equal to two in this case where we have x is greater than two we're now thinking about a Range of values because x can really be anything larger than

2 and this is a true statement right if i replaced x with let's say 7 and i had 7 is greater than 2 this is true right this is true because this 7 is greater than this number 2. so that's true if i put something in that's 2 or lower than 2 it's going to be false so like for example if i put in 2 That's false 2 is not greater than 2. 2 is equal to 2. so this is false or if i put something much lower than 2 let's say i don't know negative

7 negative 7 is not greater than 2. again this is false so when we think about inequalities our solution is generally going to be a range of values so in order to deal with that we have something known as interval notation Let me just write that down we have interval and then notation and so this notation is going to specifically help us deal with a solution that's a range of values now typically the way you're going to write this you're going to have your smallest value your smallest value then comma your largest your largest Value

now it's not going to exactly work out this way because in a lot of cases you're going to see you can't exactly put the smallest value because it's impossible for you to list it so you're going to put something close to that and then you're going to use a special type of notation so let's kind of jump in and just think about x is greater than 2 this is what we saw before If i want to use interval notation what's the smallest possible value that x can take on well it can't take on 2 but

it could be anything larger than 2. if i think about what's the smallest number that you can think of that's larger than two well you might say okay 2.1 and then you could get smaller you could say okay well 2.0001 you could get smaller and say 1 2.0000001 you can keep doing that Forever and ever and ever so because you can't possibly list the smallest value that is larger than 2 you're just going to put 2 there and you're going to place a parenthesis to the left of that to say that the 2 is not

included so it's not included in the solution so when we look at this and we see it we say okay the smallest value is anything Larger than 2. now i'm going to put a comma now if x is greater than 2 what's the largest value that x can take on well in terms of numbers there's really not one because numbers continue forever whatever the largest number is that you can think of you can always add one and get to the next largest so to kind of notate this concept we use the infinity symbol so the

infinity symbol and it kind of Looks like a sideways eight and because infinity is not a number you're also going to put a parenthesis next to it to say that hey this is not included right it's not included now we're going to see in a minute that we're going to have a different type of notation when things are included so we see our interval notation for x is greater than 2. how do we use this to graph Well it's similar if i have x is greater than 2 i find 2 on the number line and

what i'm going to do is i'm going to put a parenthesis at 2 that faces in the direction of my solution so this is going to face to the right because everything larger than 2 is in the solution the parenthesis is telling me that 2 is not Included not included and then i'm going to shade everything to the right so i'm going to shade all this and when you're doing this on a test just make it clear that you're shading everything to the right and just shade that arrow there just to say hey this is

going to continue forever and ever and ever let's try another one Let's say we had something like x is greater than four so x is greater than four so in interval notation again you want the smallest value on the left now i can't possibly list the actual smallest value because again i can't get to it i don't anything that i could list that's the smallest value could always get smaller so i'm just going to go ahead and put 4 and i'm going to put a parenthesis next To 4 to say that it's not included it's

not included but hey anything larger than that is so then i put my comma and then my largest value and again because x can be anything larger than 4 there is no largest value so we're going to put the infinity symbol in again it looks like a sideways 8. so we put our infinity symbol in and again we put a parenthesis to say hey it's not included And then graphically we're looking for 4 on the number line we're going to put a parenthesis there at 4 to say hey 4 is not included that means not

included and then we're going to highlight all this and we're going to highlight that arrow to say hey this solution is going to continue forever and ever and ever in that right direction so what about Something like x is less than 5. well if i look at my number line here i see 5 is here so x can be anything less than 5. it can't be 5 itself but any value lower than that so if i was to notate this what is my smallest value what is my smallest value well there's not going to be

one right because i can keep going to the left forever and ever and ever so we're going To notate this using the concept of negative infinity right just like we had positive infinity to say hey numbers increase forever and ever and ever going to the right so with positive infinity whatever number i can think of i can always add 1 to get to the next largest with negative infinity whatever number i can think of i can always subtract 1 and get to the next smallest right numbers continue forever in the left direction as well When

we think about a number line so a negative infinity looks like this you put a negative and then you put your infinity symbol remember that's just a sideways looking 8. now because infinity is a concept infinity or negative infinity it's not really a number it's a concept you can't really include it so you're always going to use a parenthesis next to them Okay so i put a parenthesis there now what's my largest value what's my largest value i know that x can be anything up to but not including 5. so if i look at the

number line i can get as close to 5 as i want but i can never touch 5. so because i can't really list that as an actual number i'm just going to use the number 5 and then i'm going to put a parenthesis Next to it to again say that hey this is not included this is not included so anything smaller than five works but i can't actually use five and let me put this over here too that this is not included and then graphically it's basically the same thing now i'm using a parenthesis at

5 to say that 5 is not included but notice how now i'm going to point the parenthesis Towards the left because my solution goes to the left and now i'm just going to shade all of this and then make sure you shade that arrow in to say hey this is going to continue forever and ever and ever in the left direction all right so when we have a non-strict inequality we're going to use a bracket next to the number that's included so this is the biggest thing we need to remember so now we're looking at

x is greater than or equal to two Or equal to two very important you understand the difference so when i look at my number line x can be two now it can actually be equal to two that that can be a solution it can also be anything larger than 2 just like with the strict inequality so in this particular case when we're writing out our solution in interval notation the smallest possible value that x can take on is actually 2 And i put a bracket there to say that 2 is included now comma now what's

the largest value that x can take on well again it's going to be infinity right just because i have a non-strict inequality doesn't mean that both of them are going to have a bracket x is anything that is greater than or equal to two so x still goes out forever and ever and ever in the right direction so we're going to use infinity here To notate that and then you always put a parenthesis there and again this is not included okay so now graphically you're going to change things up a little bit so at 2

instead of a parenthesis you're putting a bracket and that bracket faces in the direction of the solution that's going to face this way and then i'm going to shade everything To the right including that arrow to say hey 2 is included is included and then also anything to the right of 2. okay so 2 itself and then anything to the right of 2 continuing out forever and ever and ever out to infinity what about x is less than or equal to 5. so again here's 5 on the number line so now x can be 5

or anything less than 5. So when we think about writing our solution in interval notation the smallest possible value is still the same the smallest possible value is still negative infinity right because we can have x as anything that's 5 or less that continues to the left forever and ever and ever so we still have negative infinity and we still use a parenthesis next to that to say hey this is not included Then we have comma now what's the largest value that x can take on well x can take on the value of 5 now

but nothing larger so i'm going to put the number 5 and i'm going to put a bracket next to 5 to say that 5 is included so basically what you need to take away from this is that a bracket tells you it's included parenthesis tells you it's not included and to graph this again very very simple You put a bracket at 5 and it's going to face in the direction of the solution region so 5 is included that's why we use the bracket we just shade everything to the left everything to the left shade the

arrow because this continues forever and ever and ever in that direction so again the main takeaway here is that if you're using a parenthesis you're Saying that the number is not included if you're using a bracket you're saying it is included now not to confuse you but i want to cover something that your teacher or your textbook might do so i want to erase this real quick just the graphing part your teacher your textbook your tutor whoever you're using to learn this information may do things a little bit differently when you graph things You can

do it in the way that i showed you perfectly legal perfectly acceptable or there's another way instead of using a bracket when something's included you use a filled in circle or filled in dot so this means is included if something's not included you can use an open circle or an open dot meaning it's it's empty it's not filled in so this is not included so for example if x is less than or equal to 5 I could show that graphically by kind of making a filled in circle here and just going all the way to

the left and highlighting this this is the same as putting a bracket facing to the left and shading to the left you may see this in your textbook as another example we go back and revisit x is less than 5. and so again instead of this parenthesis here to say 5 is not included What your teacher may do is put a circle at 5 that's not filled in you see how there's a gap there let me make it a little bigger let me make it a little bit bigger so let me just make that huge

and so you can see there's open or non-filled in area there and your teacher may show you that right so this means 5 is not included if i filled it in it means 5 is included so i just want to bring that to your Attention in case you're reading a book and it shows you a different way and you're like where'd that come from why is he doing it this way both ways are accepted and probably as you move higher the way that i'm showing you now is more accepted or used more often i wouldn't

say more accepted but used more often all right so now that we've gotten the basics out of the way let's cover a few things that you're going to need to know to be successful With solving a linear inequality in one variable so the first thing is called the addition property of inequality and you'll recall that we had an addition property of equality well this is just for inequalities now so it tells us we can add or subtract any number 2 or in the case of subtraction from both sides of an inequality without changing the solution

So that property that we use to solve equations where we added or subtracted the same number to or from both sides of the equation can be extended to inequalities so we can now add or subtract the same number to or from both sides of an inequality now here's the other one and this one is something you need to really pay attention to we have the multiplication property of Inequality so i'm gonna i'm gonna give you two different scenarios here we can multiply or divide both sides of an inequality by the same positive number there's a

reason i put this in red and bold so positive number without changing the solution so just like the multiplication property of equality the stipulation there was same non-zero Number right you could multiply both sides of an equation by the same non-zero number and you didn't change the solution here what i'm telling you is that you can multiply or divide both sides of an inequality by the same positive number and positive does not include zero for those of you who are thinking that zero is not positive or negative it's neither so by the same positive number

without Changing the solution here's where we run into problems and i can't tell you how many tests i've graded in my life and i've seen students just not get this if we use the multiplication property of inequality with a negative number okay something that is less than zero we must reverse the direction of the inequality to preserve the solution so if you multiply both sides of the inequality by The same negative number flip the inequality if you divide both sides of the inequality by the same negative number flip the inequality so before we move on

and start solving problems let's explore this concept more thoroughly so let's start off with something very very easy let's say i had 2 is less than 7. so at this point we should all know that this is a true statement right this is a true Statement and what makes it true well 2 is less than 7 right 2 is a smaller value than 7 2 is to the left of 7 on the number line if i had something that said 2 is greater than 7 this would be false right 2 is not greater than 7

so that's the difference between a true and a false statement when you're dealing with an inequality so beginning with a true statement like 2 is less than 7 we can begin to experiment a little bit let's say we think about our multiplication property of inequality and let's say we multiply both sides of the inequality by the same positive number recall that this should have no effect on the direction of the inequality right the statement should remain true so let's say i multiply both sides by the number 4. So i'll multiply the left by 4 and

also the right by 4 so 4 times 2 is 8 and this should be less than 7 times 4 which is 28. and we can see we still have a true statement right 8 is less than 28 right it's a smaller value if i have 8 dollars in my pocket i certainly have less than if i had 28 dollars in my pocket now let's think about something else again starting with this same true statement Let's multiply both sides of the inequality by a negative now so let's multiply both sides by negative four so times negative

four over here and times negative four over here now negative four times two is now negative eight and we're saying this is less than seven times negative 4 which is negative 28. now the statement is false and a lot of you who haven't worked with integers a lot might be a little Confused by that but think about this for a second negative 8 is actually a larger value than negative 28 right negative 8 is not less than negative 28. and you kind of think about this involving the number line negative 8 is to the right

of negative 28 on the number line negative 28 is to the left of negative 8 on the number line you kind of think about that either way and Remember if a number is larger than another it lies to the right of it on the number line and if a number is smaller than another it lies to the left of it on the number line so this is false in order for us to get a true statement again when we multiply both sides by that same negative value we have to flip the direction of the inequality

so if i would have flipped this here instead of a less than now i'd have a Greater than then we could say it's true then we could say it's true right negative 8 is greater than negative 28. and let me show you one more example of this real quick before we move on so again let's start out with a true statement let's say i had something like 10 is greater than 5 okay so at this point we all know this is true again if i multiply both sides by the same positive value we'll still have

a true statement So let's say we multiply both sides by positive 2. so positive 2 times 10 is greater than 5 times positive 2 and 2 times 10 is 20 and this is larger than or greater than 5 times 2 which is 10. yeah this is still true again if i multiply both sides by the same negative value in order for me to keep a true statement i've got to reverse the direction of the inequality so if we erase this and say okay so let's Multiply both sides by negative 3. so let's multiply by negative

3 over here and by negative 3 over here and we're just going to leave this as it is for now and i'll show you that it's false negative 3 times 10 is negative 30. we're saying this is greater than 5 times negative 3 which is negative 15. again now this is false right it doesn't work anymore and one of the reasons you can kind of think about this is okay if i'm starting Out with a bigger number when i multiply it by a negative i'm going to have a bigger negative and a bigger negative is

actually a smaller number right as numbers go this way as the numbers get bigger and bigger in the negative direction they're actually getting smaller and i know again if you haven't had a lot of experience kind of thinking about integers in that way it might be a little confusing but just think about Your number line right it looks like this you're zero and as numbers move this way right you're getting smaller and smaller values so in this case negative 30 is greater than negative 15 would be false in order for us to have gotten a

true statement when we multiplied both sides by the same negative value we have to flip and we have to flip and so this would be a less than and That would make it a true statement so just keep that in mind as you're working these problems again if you multiply or divide by the same negative value you've got to flip the direction of the inequality in order to get the correct answer all right let's take a look at some sample problems and we're going to use the same skill set we used from equations our goal

is always to isolate x whether we're solving an inequality or an Equation and again we use kind of the same techniques here so i have x plus 8 is less than 5. i want to isolate my variable x so i want to isolate this guy so all i need to do in this case is move eight away so to do that i just subtract eight from both sides of the inequality so minus eight minus eight very very simple remember when we're solving equations like x plus eight equals five i did the same thing i just

subtracted 8 away from each side and i Got a solution of x is equal to negative 3. well it's the same i'm just isolating x and i'm using pretty much the same techniques so this will cancel x is now isolated and it's less than 5 minus 8 is negative 3. so what we're saying is that x can be any number that is less than negative 3 and i'll get a true statement all right so let's write our solution using Interval notation and in most cases this is what your teacher is going to expect you to

do so the smallest value that x can take on is really negative infinity right because x can be anything less than negative 3. so we'll write negative infinity and remember we put a parenthesis next to that because it's not included comma and then what's the largest value x can Take on well it's got to be less than negative 3. so it can be any value up to but not including negative 3. so we'll put negative 3 there and then again we'll put a parenthesis there to say that negative 3 is not included so let's graph

this real quick i'm going to take these down here and graphically x is less than negative 3. you find negative 3 on your number line i'm going to put a parenthesis facing to the left And i'm going to shade everything this way including that arrow so this is x is less than negative 3 graphically now when we solved our linear equations in one variable we always checked to make sure that we got the right answer checking a linear inequality in one variable is a lot more work so the official way that you're going to check

it is basically to use three steps so the first thing you're going to do is you're going to replace your Inequality symbols with equal symbols so i'm going to change this x plus 8 is less than 5 to x plus 8 equals 5. and i'm going to change x is less than negative 3 to x equals negative 3. so i'm going to make sure that this is the solution okay so i'm going to take this negative 3 and plug it in for x so negative 3 plus 8 does equal 5. so that checks out so

this is referred to as our boundary now the boundary is going to separate The solution region from the non-solution region and graphically you can kind of see that negative 3 is the cutoff anything less than negative 3 anything over here works so this is your solution region then anything starting at negative 3 and beyond doesn't work so this is your non-solution region so what we need to do after we establish the boundary We need to take a number from the non-solution region and try it and then also a number from the solution region and try

it so from the non-solution region the easiest number to try is 0. so let's go back up here and we can plug in a 0 for x and we'd have 8 is less than 5. so that's in the non-solution region so it should be false and it is so it's okay that this is false that's what we want now The next thing we're going to do is we're going to pick a number from the solution region so that's over here that's anything less than negative 3. so i can pick anything i want so let's say

negative five so i'd plug a negative five in for x i'd have negative five plus eight is less than five negative five plus eight is three and three is less than five so that's true and that's exactly what we want because we picked a number from the solution Region so again a lot more complex to check your answer here so i'm not going to do it moving forward it's something that you can refer back to this procedure and you can check it on your own because it is extremely time consuming all right let's take a

look at the next one so we have x plus 2 is less than or equal to 1. so to solve this i'm just going to subtract 2 Away from both sides of the inequality this will cancel i'll have x by itself and i'll have less than or equal to 1 minus 2 is negative 1. so this is my solution x is less than or equal to negative 1 and to write this in interval notation again the smallest possible value we're dealing with a less than or equal to so the smallest possible value is negative infinity

right because there is none it continues Forever and ever in the left direction and then the largest possible value is negative 1. and because there's a non-strict inequality here negative 1 is included so i'm going to use a bracket there so graphically this is going to look like this at negative 1 i'm going to put a bracket facing to the left and i'm going to shade everything to the left including that arrow so this is x Is less than or equal to negative 1. all right let's take a look at 7x is less than 28.

so if i want x by itself i need to divide both sides of the inequality by 7. now remember i'm dividing by a positive number so i keep the direction of the inequality the same this 7 cancels with this 7 and i'm just left with x and that is less than 28 divided by Seven is four so that's my solution x is less than four so in interval notation it's a less than so the smallest possible value is negative infinity right because it continues forever and ever and ever in the left direction then the largest

possible value it's not quite 4 right because 4 is not included but we're going to put 4 and then we're going to put a Parenthesis next to that to say hey we can go all the way up to any amount that's close to 4 but we can't actually get to it so again this parenthesis is telling me that 4 is not included let me just make those even heights and let's graph x is less than 4. all right so i'd find 4 on the number line and that's right here i'd put a parenthesis facing to

the left and then i'm going to shade everything to the left Shade all this and shade the arrow to say hey x is less than 4. so it can't quite be 4 but it can be anything less all right let's take a look at negative 6 x is greater than 30. now i want x by itself and in order to get x by itself i need to divide both sides of the inequality by the coefficient of x which is negative 6. now again i'm dividing by a negative If i multiply or i divide by a

negative i've got to flip the direction of the inequality so do that first so you don't forget so right now it's a greater than it needs to become a less than so you flip it very important so this will cancel with this and i'll just have x and then 30 divided by negative 6 is negative 5. so x is less than negative 5. we think about this in interval notation we know there's no possible way to get a smallest value right so we're going to use negative infinity for that because the solution continues to go

to the left forever and then for the largest possible value i'm going to put negative 5 and then i'm going to put a parenthesis next to it to say hey negative 5 is not included now graphically when we look at this Solution so i'm going to find negative 5 i'm going to put a parenthesis there and i'm going to shade everything to the left including that arrow to say hey negative 5 is not included but everything to the left is and so x is less than negative 5 is the solution now i know it's a

little time consuming but i want to show you this very very important If i was to check this let's say that i did not flip the inequality let's say that i went through i didn't flip this and i ended up with x is greater than negative 5. let's say i said this is the solution it's not but i'm going to say i forgot to flip all right if you were checking this the first thing you would do again is you would Replace all your inequality symbols with equal symbols so in this case forget that this

exists right now this is gone just forget about it i would have negative 6x equals 30 as kind of the original thing then the solution is x equals negative 5. okay so if i plug a negative 5 into x i'd have negative 6 times negative 5 equals 30. so that works so i know that i have Negative 5 as my boundary and again what that means is it separates the solution region from the non-solution region graphically and again let me kind of erase this because we're kind of thinking about how to determine we have the

wrong answer graphically if i look at negative 5 if i came up with the wrong answer and i said x is greater than negative 5 i'd be thinking about okay anything larger than Negative 5 would be in the solution region and anything less than negative 5 would be in the non-solution region so if i try something like 0 which should be in the solution region i'm going to find that i'm going to get the wrong answer because again i forgot to flip and 0 is greater than negative 5. so that works out here but in

the original inequality which was negative 6 times x so i plug in a 0 there is Greater than 30 well negative 6 times 0 is 0 that's not greater than 30. this would be false so that would be your first indication that something went wrong i checked something that should have been in the solution region and it didn't work right it did not work so at that point i'd go up here and go okay this is false this doesn't work this doesn't work so i would go back and Say okay well i forgot to flip

the inequality so it's very time consuming to check but when you first start doing these you might want to just as to avoid making the errors and let me show you what happens if you do this correctly i know this is time consuming but i feel like it's very important so let's say i didn't forget to flip we go back to our original solution So x is less than negative 5 so we know our boundary is at x equals negative 5. so anything less than negative 5 should work as a solution right so if i

plugged in a negative 10 for x that's less than negative 5 that would be in my solution region so it should give me a true result so negative 6 times negative 10 should be greater than 30. negative 6 times negative 10 is 60 that is greater than 30 that's true so that works out That's exactly what we wanted so if i use something like positive 10 that should not be a solution so i'd have negative 6 times 10 is greater than 30 negative 6 times 10 is negative 60 that's not greater than 30 that's false

right so that's exactly what we want that's okay right again when you're checking this stuff and i have to reverse this because i gave you the scenario where we got the wrong answer when you graph this Anything less than negative 5 is in the solution region anything that's negative 5 or larger is in the non-solution region and again negative 5 is your boundary it separates the solution region from the non-solution region and that's why you establish that so you can say okay i'm going to pick something on the non-solution and see if it works it

shouldn't right because it shouldn't Be a solution and then i'm going to pick something on the solution region and that should work and when we plugged in 10 10 is in the non-solution region it didn't work that's what we want when we plugged in something from the solution region we tried negative 10 it did work and that's what we want all right i know this video is kind of long but when you first start out doing these it's very important to get all the basics down so it's kind of worth the time even if you

Have to break it up so we're going to look at one final problem here we have negative 1 3 x is greater than or equal to 2. so if i want to isolate x and i have a fractional coefficient i have a fractional coefficient we know what to do we multiply both sides of the inequality by the reciprocal so the reciprocal of negative one-third is negative three Over one and remember we're multiplying by a negative so we've got to flip this so this is going to go this way so instead of greater than or equal

to it's now less than or equal to this is going to cancel and i'm just going to have x is less than or equal to 2 times negative 3 is negative 6. so x is less than or equal to negative 6. let's do the interval notation for that So it's going to be negative infinity over here on the left right because we're not going to have a smallest possible value so we use negative infinity and then for the largest possible value we're going to use negative 6 and that's included because this is a non-strict inequality

so we're going to use a bracket next to negative 6. now graphically i'm going to find Negative 6 and i'm going to put a bracket there facing to the left and i'm going to shade everything to the left and i'm saying hey negative 6 is a solution so i got the bracket and then anything to the left is also a solution hello and welcome to algebra 1 lesson 12. in this video we're going to continue Learning about how to solve linear inequalities in one variable so we're just going to kind of jump in here we

already know the basics and so anything else that we're going to learn we're going to learn along the way so when i look at this i have negative 4 times the quantity x minus 4 is greater than or equal to negative 2 times the quantity x plus 1. now the same way i would do an equation i'm Going to simplify each side completely so i'm going to use my distributive property on the left negative 4 times x is negative 4x then i have negative 4 times negative 4 that's plus 16 and this is greater than

or equal to we have negative 2 times x that's negative 2x and then negative 2 times 1 that's minus 2. now just like with an equation i want all my variable terms on one side of the Inequality and all my numbers on the other so i can use my addition property of inequality to accomplish this i can add 2x to both sides of the inequality this will cancel and let's see what we have we have negative 4x plus 2x that's negative 2x so then plus 16 is greater than or equal to negative 2. so now

i just want to subtract 16 away from both sides of the inequality So this can go away and my variable term is isolated all right so now on the left i just have negative 2x and then is greater than or equal to on the right negative 2 minus 16 is negative 18. and i have one final step i just want to isolate my variable x and so i'm going to divide both sides of the inequality by negative 2. but remember we talked about this in the last lesson If you multiply or you divide by a

negative you've got to flip the direction of the inequality so right now this is a greater than or equal to we're going to make it a less than or equal to again because we divided by a negative all right so once we've flipped it we don't have to do anything else this cancels with this and we just have x and then on the right negative 18 divided by negative 2 is not so our solution here is x is less than Or equal to 9. now one thing we want to do when we're working with inequalities

we want to write our solution in what's known as interval notation so we spent a lot of time on that in the last lesson and so if you haven't covered that yet i want you to go back to the last lesson and learn how to do that so if we have x is less than or equal to Nine in interval notation remember it's less than or equal to nine so the smallest value there won't be one right because x could be 9 or anything less than that so we're going to use negative infinity in that

position and remember when we use positive infinity or negative infinity we always put a parenthesis next to that And then for the largest value we're going to use 9. now 9 is included in our solution and when something's included we're going to use a bracket right not a parenthesis so we have our solution in interval notation and then we just want to go ahead and graph it so we're going to locate 9 on our number line and 9 is right here and i want to put a bracket at 9 that's Facing to the left it's

always going to be facing in the direction of the solution so the bracket goes there and then i'm just going to shade everything to the left everything to the left shade all of that including the arrow so x could be 9 that's why the bracket's there or anything to the left of 9 on the number line all right let's take a look at another one we have the opposite of the quantity r plus 3 And this is less than or equal to negative 3 times the quantity r minus 1 and then minus r so i'm

going to use my distributive property here i'm going to distribute this negative to each term and remember you could put that as plus negative 1 and think about that as okay i'm multiplying negative 1 times r that's negative r and then negative 1 times 3 that's minus 3. you can also just change the sign of Each term inside the parentheses that will allow you to remove the parentheses remember if i see just a negative out in front of parentheses in order to remove the parentheses i change the sign of each term inside so negative or

minus 3 is less than or equal to negative 3 times r is negative 3r and then negative 3 times negative 1 is plus 3 and then minus r all right so let's continue simplifying on the left i Have negative r minus three and this is less than or equal to on the right i have negative three r minus r i can combine those like terms that'll give me negative four r and then plus 3. and now i want to move all the variable terms to one side and have the numbers on the other so what

i'm going to do is i'm going to add 4r to both sides of the inequality so this is going to go away And negative r plus 4r would be 3r so then minus 3 is less than or equal to 3. and now i'm simply going to add 3 to both sides of the inequality so that i can isolate this 3r so this will go away and on the left i'll just simply have 3r now this is less than or equal to 3 plus 3 is 6. now as a final step i'm going to divide Both

sides of the inequality by 3 which is the coefficient of r now again because i'm dividing by a positive number i don't need to do anything with the sign it's only if i'm doing it with a negative number so this will cancel with this i'll have r and then is less than or equal to 6 divided by 3 is 2. so r is less than or equal to 2. all right so in interval notation again we have r is less than or equal to 2. so that's going to look like this we'd have negative infinity

right here we put a parenthesis there to say that negative infinity is not included and then this is up to and including two right so we're going to use a bracket at two because r can be anything kind of starting out at two including two Anything moving to the left so the smallest possible value there's not one that's why we put negative infinity here the largest possible value is two and two is included so we put our bracket now again when we graph this we put a bracket at 2 facing to the left it's facing

towards the solution and then we're going to highlight or we're going to we're going to shade everything to the left including this Arrow to say hey 2 is included and everything to the left of 2 is included as a solution as well all right let's take a look at another one we have negative 8 fifths minus x is greater than negative 1 4 x minus 11 fifths so just like if we had an equation with fractions if we have an inequality with fractions we can go ahead and multiply both sides of the inequality by the

lcd so the lcd here would be 20 right five Times four so i can multiply 20 times the left side which is negative eight fifths minus x remember to use brackets or parentheses right you've got to multiply 20 by both terms this is greater than i'm going to use brackets over here negative 1 4 times x minus 11 fifths we're multiplying everything over here by 20 as well so 20 times negative eight fifths twenty Would cancel with five and give me four four times negative eight is negative 32 so then minus 20 times x is

20x and this is greater than over here 20 times negative 1 4 x so 20 divided by 4 is 5. 5 times negative 1 is negative 5. so this would be negative 5 x and then minus 20 divided by 5 would be 4 4 times 11 is 44. all right so i want to get my variable term on one side of the inequality and a Number on the other so i'm going to add 5x to both sides of the inequality that's going to go away and on the left side i'm going to have negative 32

negative 20x plus 5x is minus 15x this is greater than negative 44. so then the next step is to add 32 to both sides of the inequality so that'll cancel and i'll have negative 15x is greater than Negative 44 plus 32 is negative 12 and as a final step i'm going to divide both sides of the inequality by negative 15 which is the coefficient of x again again again when i divide by a negative and i'm working with inequalities i've got to flip the direction of the inequality so this is going to get flipped right

now it's a greater than we're going to make it a less than So this would be that x right because these are going to cancel is less than negative 12 over negative 15 that's positive really think about 12 as 4 times 3 and 15 is 5 times 3 so the gcf would be 3 cancel that between numerator and denominator and you're going to get 4 fifths so we'd have x is less than four fifths all right so let's write this in interval notation So since x is less than four-fifths really for the smallest value we're

gonna put negative infinity because this is gonna keep going forever and ever and ever in the left direction so we'll put a parenthesis next to that and then for the largest number we're going to put four fifths and remember that's not included because we have a strict inequality so we're going to use a parenthesis there as well now graphically we don't have a notch For four fifths but let's say it's somewhere right there let's say this is four fifths i'll put the notch there and so i'm gonna use a parenthesis at four fifths it's gonna

face to the left and i'm going to shade everything to the left including this arrow right here so that would be x is less than four-fifths all right let's take a look At another one with some fractions we have 33 tenths minus one fifth x is less than negative one half x plus one minus two x so the first thing i'm kind of looking at here is to clear the fractions the lcd is ten right i have 5 i have 2 and i have 10. 5 times 2 is 10 so the lcd is 10. so

let's multiply both sides of the inequality by 10 so i'd have 10 times remember to use brackets Or again you could use parentheses if you want so 33 over 10 minus one-fifth times x close the brackets and then this is less than open brackets negative one-half times x plus one minus two 2x close your brackets and this is multiplied by 10 as well so 10 times 33 over 10 the tens would cancel and you just have 33 Then minus 10 times one-fifth 10 divided by 5 is 2 so then 2 times 1 is 2 then

times x so you just have minus 2x and this is less than over here we have 10 times negative 1 half x so the 10 would cancel with the 2 and give me a 5 5 times negative 1 is negative 5 and then don't forget the x so negative 5x there then plus 10 times 1 is 10 Then minus 10 times 2x is 20x so i can combine like terms on the right side here negative 5x minus 20x is minus 25x so i'll put negative 25x plus 10. so over here nothing to combine i'm just

going to write negative 2x out in front plus 33. and now i want to get a variable term on one side of the inequality and a number on the other so in order to do that i'm going to add 25x to both sides of the inequality that's gone and negative 2x plus 25x would be 23x so then plus 33 is less than 10. i'm going to move this over here i'm going to simply subtract 33 from both sides of the inequality so this is gone and i'm going to end up with 23x is less than

10 minus 33 is negative 23. so as a final step i'm going to divide both sides of the inequality by 23 Which is the coefficient of x so divide this by 23 and this by 23 and i'm going to get x is less than negative 1. all right so an interval notation again the smallest possible value if you're dealing with a less than it's less than negative one so this would continue to the left forever and ever and ever so we write negative infinity here and we always use a parenthesis with infinity or negative infinity

and then comma for the largest possible Value we put negative one and again that's not included so we put a parenthesis there to show that now graphically i'm just going to put a parenthesis at negative one facing to the left facing towards the solution and then i'm going to shade everything to the left including that arrow and say hey negative 1 is not included so i got the parenthesis there but everything to the left of that would be so another Thing we're going to come across will be three part inequalities so you see here we

have five is less than x and x is less than seven or you could say x is greater than five and x is less than seven what's important to note here is that both conditions have to be true so in other words you can kind of think of these as separate problems i can kind of think of this as x is greater than five So if i was to graph that x is greater than 5 i would think about what i would think about putting a parenthesis here and i would graph everything to the right

of 5. and then for the other part of this problem x is less than seven so x is less than seven now if i graphed that i would put a parenthesis at seven and i would graph everything to the left everything to the left Now when we have a three-part inequality we're kind of combining the two and saying hey i only want the section that solves both so if we look at what's shaded by both and it's kind of hard to tell but it's going to be this area between 5 and 7. 5 is not

included and 7 is not included but anything between those two would be included if you think about it it should make Sense to you let me kind of erase this x can be any number that satisfies both conditions so if i plug in something between 5 and 7 like 6 6 is greater than 5 and 6 is also less than 7 so that works out if i plug in something that is outside of the solution region for one of them let's say i plug in x is zero well zero is not greater than five it's

less than seven but it's not greater Than 5. so it doesn't satisfy the inequality because one part fails and it's got to be a solution to both all right so let's graph this again from scratch you're going to start out at your smallest value and in this case it's going to be us starting at 5 with a parenthesis because 5 is not included and then we're going to look at our largest value and that's going to be us going to 7 and putting a parenthesis because 7 is not included And we shade all the area

between all the area between so anywhere in there is a solution i know we think about 6 because it's a nice neat whole number but really anything between 5 and 7. so 5.122228 would be a solution right because that's larger than five and less than seven now in interval notation it kind of follows a similar pattern for our smallest number we'd put five and a parenthesis next to it to say 5 is Not included then comma we'd put 7 for the largest and a parenthesis to say 7 is not included so this is your interval

notation and here is your graph so to solve a three-part inequality is no more difficult than solving a regular inequality now you can do it two ways and the most popular way the easiest way is kind of to do the same thing to each Part so let me kind of show you that way first so i'm trying to isolate the variable in the middle and in order to do that i just think about okay i have 7 here i'm going to subtract away 7 from each part and this is the tricky thing because we're used

to just doing it to two sides now i have this here and this here so everywhere i see that i've gotta subtract away from That part so subtract away seven from here here and here right all three parts so this will cancel and 22 minus 7 is 15 so i have 15 is less than or equal to negative 5x which is less than or equal to 47 minus 7 is 40. now to isolate x in the middle i have this negative 5. i'm going to divide each part by negative 5 each part by negative 5

Each part by negative 5. now remember when i divide by a negative i've got a flip and i've got two signs here i got to flip both of them so this is going to flip this way and this is going to flip this way as well so 15 divided by negative 5 is negative 3 negative 5 divided by negative 5 is 1 so this is just x 40 divided by negative 5 is negative 8. now i'm going to rewrite this i want my smallest number on the left so i'm going To put negative 8 and

it's less than or equal to x so i'm going to put less than or equal to x and then i'm going to put x is less than or equal to negative 3 so less than or equal to negative 3. so now this runs in the order of the number line and that's my solution x is greater than or equal to negative 8 and it's also less than or equal to negative 3. if i find negative 8 on the number line that's right here I would put a bracket there because negative 8 is included so that's

the smallest value that x can take on negative 3 is the largest value that x can take on so i'd put my bracket there and shade everything in between right so i'm saying that x can be negative 8 and larger up to and including negative 3. anywhere in those values would work if i picked negative 5. Well negative 5 is larger than negative 8 and it's smaller than negative 3. so that works if i picked something outside of this range in the non-solution region let's say 0 plug that in 0 is larger than negative 8

but it's not smaller than negative 3. so that doesn't work right so it's got to be something in this range of values now for my interval notation I would put negative 8 as the smallest possible value and i'd use a bracket there right because negative 8 is included and then i'd put negative 3 as my largest possible value and then a bracket next to it again because negative 3 is included all right so let's look at one more problem so we have that negative 5 minus 12x is greater than or equal to negative 5 and

less than 127. so my goal is to isolate The variable x in the middle here so i'm going to add 5 to each part remember you got to do it to each part so that's going to cancel and be 0. make sure you write 0 there you don't have anything else over there so this is less than or equal to and then this is going to cancel i'll just have negative 12x in the middle and then is less than we have 127 plus 5 that's 132. and now i'm going to divide each part by Negative

12. again because i'm dividing by a negative i've got to flip all the signs very important this will be a greater than or equal to this will be a greater than and so 0 divided by negative 12 is 0. this cancels and i'll have 1 times x or just x and then 132 divided by negative 12 is negative 11. so i'm going to rewrite this in the Order of the number line so x is greater than negative 11. so i'm going to put negative 11 here and x is greater than that and x is less

than or equal to zero so less than or equal to zero so x is greater than negative 11 and x is less than or equal to zero so in interval notation my smallest value i'm going to put negative 11 and i'm going to put a parenthesis next to it i have a strict inequality there x Has to be strictly greater than negative 11 but up to and including zero so my largest value is zero and that's a non-strict inequality so i'm going to use a bracket so that's my interval notation and then graphically i'm going

to find negative 11 and i'm going to put a parenthesis there i'm going to find 0 and i'm going to put a bracket there And then my solution region would be anything in between now 0 is included remember because i have a bracket so it could be 0 or anything less up to but not including negative 11. hello and welcome to algebra 1 less than 13. in this video we're going to learn about linear equations in two variables all right so the last chapter was all about linear equations in one variable and we Also talked

about linear inequalities in one variable but now we're going to move on and talk about linear equations in two variables okay in two variables so a linear equation in two variables is an equation of the form ax plus b y equals c where a which is the coefficient of x b which is the coefficient of y and c which is a constant term are real numbers And again i know i haven't given you an official definition for you know what a real number is but it's at this point any number you can think of so

i could put fractions there decimals you know negatives doesn't really matter anything you could think of at this point could go in there but there is a restriction we say that a and b are not both zero so these are the coefficients for our Variables so we can't have both of them zero at the same time i could have one of them be zero but the other one could not be zero so here are some examples we have 4x plus 3y equals 7. so you see you have two variables here you have an x variable

and a y variable we have negative 2x minus 4y equals 13. again two variables here you have an x variable and a y variable and then one fourth x plus 18 y equals 12. again x and then y so when we worked with a linear equation in one variable we generally had exactly one solution and i know we talked about a special case scenario where that's not going to be true but in general we had one solution so if i looked at something like 2x equals 8 we already know how to solve this we divide

both sides of the equation by 2 which is the coefficient of x This cancels with this and i just have x equals a divided by 2 is 4. so there's exactly one solution here there's only one number that when plugged in for x gives us a true statement and what we're going to see when we start working with linear equations in two variables is that there's an infinite number of x y combinations that are going to make the equation true so let's talk a little bit about how to write a solution for a Linear equation

in two variables so we typically write our x and y values using what is known as an ordered pair so you might want to write that down because it's an extremely important vocabulary term an ordered pair looks like this you have an x comma a y so all the way on the left is always your x value your x value and then on the right is always your y value These are not interchanged ever it's always the x value comma the y value so in other words if i give you an ordered pair 3 comma

negative 1 3 is your x value and negative 1 is your y value so again the x values on the left the y values on the right all right so now let me show you how to check an ordered pair and see if it's a solution to a linear equation in two variables so we have this 4x plus 2y equals 12. And we want to check the ordered pair 3 comma 0. so the important thing to remember is that again the left value is the x value and the right value is the y value very

very important you understand that so just like we did with a linear equation in one variable so for example if i have 2x equals 8 like we had earlier and i said x is equal to 4 right this was the solution I plug a 4 in for x so 2 times 4 equals 8 and we see that the left and the right side are equal 8 equals 8. that's how i know that x equals 4 is the solution well it's the same here all i'm going to do is take each number and plug it in

for its respective variable so 3 will get plugged in for x because it represents the x value 0 will get plugged in for y because it represents the y value so i would have 4 Times again i have an x here this is my x value so i'm going to put a 3 plus 2 times 0 is my y value so i'm putting in a 0 for y equals 12. very very simple i'm just plugging in just like i did with a linear equation in one variable right if i had the solution there i plugged

in for x i found out if the left and the right side were equal i'm doing the same thing here I'm just including another variable i now have two of them so 4 times 3 is 12 plus 2 times 0 is 0. so i can put plus 0 or i could just leave it as 12. and this is supposed to equal 12 and it does we get the same value on the left and the right side and so we can say this ordered pair 3 comma 0 is a solution for our equation all right let's

take a look at some problems now we have negative 4x minus y equals 16 and we're going to check each ordered Pair and see if it's a solution to the equation so we have our first ordered pair which is negative five comma four again the left value is the x value the x value the right value is the y value so we have negative 4x so instead of x i'm going to plug in a negative 5. this is what i'm plugging in for x so negative 4 times negative 5 minus Now i have a y

and i'm plugging in a 4 for y because that's my y value so minus 4 and this equals 16. so negative 4 times negative 5 is 20 and minus 4 equals 16. 20 minus 4 is 16. so you get 16 equals 16. so yes this ordered pair negative 5 comma 4 is a solution so i'm just going to put a check mark here all right the next ordered pair we're going to look at is 0 comma 2 And again we have our left value as our x value our right value is the y value and

that's not going to change okay you just memorize that the one on the left is the x the one on the right is the y so we have negative four times our x value is zero so i'm going to plug in a zero there minus our y value is two equals sixteen negative four times zero is 0. So then minus 2 equals 16. so we get negative 2 equals 16 and that's not true this is false so this does not work as a solution so this doesn't work all right let's try the ordered pair 0.5

comma negative 18. so again the left value is the x value the x value and the right value is the y value so we're just going to plug in we'll Have negative 4 times x and again x is going to be 0.5 or that's one-half whatever you want to put in there minus for y we have negative 18. and this is very important when you're plugging in for something this is negative 18 that i'm plugging in here for y there's a minus sign already out in front so this stays there and then you put minus

18 or negative 18. and minus a negative is plus a positive so i can write minus a negative 18 or i could simplify that and put plus positive 18 which i'm going to do in the next step so equals 16. so what's negative 4 times 0.5 again you could do that using negative 4 times 1 half and kind of mentally say ok well negative 4 times a half is negative 2. so this is negative 2 minus a negative 18 is plus 18 Equals 16. and so yeah this is going to be true negative 2 plus

18 is 16. 16 does equal 16 same value on each side of the equation and so we're going to check mark this and say yes it is a solution to the equation and one thing this would immediately draw your attention to is the fact that again we can have more than one ordered pair that works as a solution to A linear equation in two variables when we worked with a linear equation in one variable generally speaking we're going to have exactly one value for our variable that makes the equation true here we have multiple ordered

pairs or multiple solutions here that work and in fact you're going to find that with a linear equation in two variables there's an infinite number of ordered pair solutions all right so now we're going To look at negative 7x minus 3y equals 42 and we want to check the ordered pair 3 comma 8 the ordered pair negative 9 comma 7 and the ordered pair 2 comma negative 5. so let's start with this ordered pair three comma eight so this is your x value this is your y value and let's do some plugging in so we

have negative seven Times for my x value i'm putting in a 3 minus 3 times for my y value i'm putting in an 8 and this should be equal to 42 negative 7 times 3 is negative 21 negative 3 times 8 is negative 24 so minus 24 equals 42 negative 21 minus 24 is negative 45 and this is not equal to 42 this is false so this does not work out as a solution To make this a little quicker i'm just going to i'm just going to erase this part which is where we plugged in

right the rest of the equation is always going to be the same you'll have your negative 7 times your spot for x then minus 3 times your spot for y then equals 42. so i'm just going to erase those parts now i'm going to be plugging in a negative 9 for x this is my x value So negative 9 goes here right we have negative 7 times x so negative 7 times negative 9 and i'm going to plug in a 7 for y again here's my y value i have negative 3 times y so i'm

plugging in a 7 for y so i go negative 3 times 7. so negative 7 times negative 9 minus 3 times 7 equals 42 negative 7 times negative 9 is 63 minus 3 times 7 is 21 this should equal 42 and in fact it does 63 minus 21 is 42 so you get 42 equals 42. so that checks out and so we could say that negative 9 comma 7 this ordered pair is a solution all right let's do one more and again i'm just going to erase the part where we plug in and i have

a 2 that's going to represent my x value and a negative 5 that's going to represent my y value now so where's x we have negative 7 times x negative 7 times this right here so 2 Which is my value for x is going to go on right here and then my y value of negative 5 is going to go in right here right again i'm just plugging it in right there next to negative 3. so i have negative 7 times 2 minus 3 times negative 5 equals 42. negative 7 times 2 is negative 14.

negative 3 times negative 5 is 15 so plus 15 equals 42. negative 14 plus 15 is 1 And 1 is not equal to 42 this is false so this doesn't work out so for these type of equations we can pick a value for x and solve for y or vice versa so i could pick a value for y and solve for x and this is going to be a very important skill that we're going to use throughout this chapter all right so let's start out by looking at this equation we have x minus 3y equals

15. now i'm given some values for x and for Y here this is a table of values and what's going to happen is okay for example here i'm given a value of 0. so i would plug 0 in for y and i would solve for x so let me show you what that looks like so we have x minus 3 times y i have a given value of 0 for y so i'm going to plug that in equals 15. so then if i simplify this i'd have x Equals 15. right 3 times 0 is just

0. if i'm subtracting away 0 it's like i just have x all right so x equals 15. so that's the value for x that makes this equation true when y is 0. so i can go ahead and just fill in the missing value and put okay this is 15. so the ordered pair the ordered pair 15 comma 0 is a solution to this equation Right 15 is my value for x the value on the left 0 is my value for y the value on the right all right so now i'm given an x value of

0. so i'm going to plug a 0 in for x so instead of x minus 3y equals 15 i'm going to put 0 minus 3y equals 15. really the 0 out in front doesn't do anything so i can just get rid of it and i'm left with negative 3y equals 15. so to solve for y We already know how to do that i want to isolate y so i just divide both sides of the equation by negative 3. nothing we haven't done before this will cancel with this and i'll just have y and then it

equals 15 divided by negative 3 is negative 5. so if x is 0 y equals negative 5. let me kind of put that in a different color so it stands out so let's put negative 5 here So again this is another ordered pair that works as a solution to this equation we can put the ordered pair 0 comma negative 5 is a solution again where 0 is the x value and negative 5 is the y value all right now we're given a y value of negative 2 and we want to find the missing x value

so we have x minus i'm plugging in a negative 2 for y so i'm going to have 3 times negative 2 equals 15. So negative 3 times negative 2 is 6. so i would really have x plus 6 equals 15. and i want to solve for x and again we know how to do that if we want x by itself i just need to subtract 6 away from each side of the equation so that's going to cancel i'm going to have x is equal to 15 minus 6 is 9. so x equals 9. so let's

put that over here And i'm going to remember to use a different color this time and i'm going to put 9 there so that's another ordered pair that works as a solution so we can say the ordered pair 9 comma negative 2 again works as a solution for this equation let's do one more we're given an x value of negative 7 so i'd plug a negative 7 in for x so then minus 3y equals 15. want y by itself so i'm going to add 7 To each side of the equation that's gone i'll have negative

3y is equal to 22. if i divide both sides of the equation by negative 3 so that y can be by itself this will cancel with this and i'll have y equals negative 22 thirds and there's nothing to do to simplify that 22 is 11 times 2. i can't really cancel anything with the 3. we'll write that over here as negative 22 thirds And again that's another ordered pair that's a solution to our equation let's just write that real quick it would be negative 7 comma negative 22 thirds and i can go through and keep

generating as many ordered pair solutions as i want i can pick a value for x doesn't matter what it is i can plug it in and i can solve for y so that would give me the x value that i picked and the y value that i solved for Put those two together that gives me an ordered pair that's a solution i could do the opposite thing i could pick a value for y and solve for x all right let's take a look at another example of this so we have 2x plus 3y equals negative

24. so we're given an x value of 3. we want to solve for the unknown y value so if x equals 3 here i would have 2 times i'm plugging in a 3 for x Plus 3y equals negative 24. so 2 times 3 is 6. so i'd have 6 plus 3y equals negative 24. subtract 6 away from each side i'd have 3y equals negative 30 and then i would divide both sides by 3 right in order to isolate the variable y this cancels with this negative 30 divided by 3 is negative 10. so you get

y equals negative 10. so i can put a negative 10 there and so my ordered pair would be 3 for my X value comma negative 10 for my y value now let's look at we're given a y value of 8 we want to find the unknown x value so we have 2x plus 3. we're plugging in an 8 for y this equals negative 24. so we have 2x plus 3 times 8 is 24 equals negative 24. we subtract 24 away from each side that's going to cancel and i'm going to have 2x is equal to

negative 24 minus 24 Is negative 48 so now i can divide both sides of the equation by 2 in order to isolate x this will cancel with this and i'd have x is equal to negative 48 divided by 2 is negative 24. so we can come back up here and say that the missing value here is negative 24. right this would be another ordered pair solution negative 24 for the x value 8 for the y value all right for the next one i have an x value of 4 And i want to find the unknown

y value so i'll have 2 times i have x i'm plugging in a 4 for x plus 3 y equals negative 24. so 2 times 4 is 8 so we have 8 plus 3y equals negative 24. we subtract 8 away from each side we'll have 3y is equal to negative 32 divide both sides of the equation by 3 so that i can isolate y And i'll have y is equal to negative 32 thirds and there's nothing i can do to simplify that if i think about 32 that's 2 to the fifth power right i can't

cancel any of those factors of 2 with this 3 here so this is simplified so our missing value for y would be negative 32 thirds All right so my ordered pair solution there would be 4 comma negative 32 thirds all right for the last one we're going to look at we're given a y value of negative 10. so we have 2x plus 3 times i'm plugging in a negative 10 for y is equal to negative 24. so we then we'd have 2x 3 times negative 10 is negative 30 so minus 30 Equals negative 24. i'm

going to add 30 to both sides of the equation that's going to cancel i'll have 2x is equal to negative 24 plus 30 is going to be 6 and i can divide both sides of the equation by 2 to isolate that variable x so divide by 2 divide by 2 and i'm going to get that x is equal to 3. all right so i'm going to write that x is 3 it's our unknown value And again i'll write this as an ordered pair just so you get a little practice doing it and seeing it so

the ordered pair would be 3 for the x value comma negative 10 for the y value and again you can pick any value you want for x and solve for y or you could pick any value you want for y and solve for x so for example let's say that i wanted to pick i don't know a value of 0 for x i could say what is the resulting y Value that makes the equation true when x is 0. so i just plug in a 0 for x so i'd have 2 times 0 plus 3y

equals negative 24. so 2 times 0 is 0. so i just have 3y equals negative 24. and this is just a linear equation in one variable we know how to solve this already we're just trying to isolate y at this point so i divide both sides of the equation by and i get y equals negative eight right So if x is zero y is negative eight right this is the ordered pair zero comma negative eight and it's a solution to this equation two x plus three y equals negative 24. and again if you want to

go back and prove that to yourself plug both of them in at the same time so plug in a 0 for x 2 times 0 plus 3 times plug in a negative 8 for y so 3 times negative 8 Equals negative 24. 2 times 0 is 0. so this is gone so we just have 3 times negative 8 that's negative 24 and that does equal negative 24. you can see that this ordered pair again 0 comma negative 8 is a solution for this equation hello and welcome to algebra 1 lesson 14. in this video we're

going to learn about plotting ordered pairs all right so in the last lesson we learned that a linear equation in two variables has an Infinite number of solutions now we display a solution as an ordered pair so you have an x value comma a y value and it's always in that order the x values on the left the y values on the right so for example in this equation we have 3x minus 3y equals 18. and we have two ordered pairs here that are proposed solutions so if this ordered pair 1 comma negative 5 is

a solution to this equation that means that i can take this value Here and plug it in for x this value here and plug it in for y and i should get a true statement right the left and the right side should be equal so let's see if it is so i would take a 1 and plug it in for x so 3 times 1 minus i take a negative 5 and plug it in for y so i'd have 3 times negative 5 and this should be equal to eighteen three times one is three Negative

three times negative five is positive fifteen so plus fifteen this should equal eighteen and it does eighteen equals eighteen so that works out so we can say that this ordered pair 1 comma negative 5 is a solution for this equation again because when i plug in a 1 or x and a negative 5 for y i get a true statement let's do the same thing with this ordered pair 2 comma negative 4. All right so we'd have 3 times x our value for x is 2 minus 3 times y our value for y is negative

4 and this equals 18. so 3 times 2 is 6. and we have negative 3 times negative 4 that's 12. so plus 12 equals 18 and it does 6 plus 12 is 18. so you get 18 equals 18. And so we could say that this ordered pair 2 comma negative 4 is also a solution for this equation 3x minus 3y equals 18. so you might be asking yourself what are we going to do with these ordered pairs well in the next section we're going to learn how to graph a linear equation in two variables but

in order to do that we first have to be able to plot an ordered pair all right so in order to talk about plotting Ordered pairs we're going to first begin by talking about something known as the cartesian coordinate system the again cartesian coordinate system now this might be a source of some confusion for you because this has many names and probably in your textbook it'll be called cartesian coordinate system and then after that they start calling it The rectangular coordinate system and sometimes they just call it the coordinate plane and a lot of students

will be like well what is this now what is this they're all the same thing okay it's named after its inventor and you'll probably read the story as a side note in your textbook where it says you know he's lying in bed and he's watching a fly move and he comes up with this this coordinate plane and so long story short it's named after him and it's Called the cartesian coordinate system but you might hear it called something else most of the time i'm just going to call it a coordinate plane this is nice and

simple what you need to know is that it consists of two number lines one's horizontal and one's vertical so let's take a look at this thing all right so here's an example of the coordinate plane now if you just look at this right here Starting here and going to the right this is the number line we've been working with for a long time now it's a horizontal number line and you'll notice that if i were to tell you this is the point 0 everything to the right of 0 is a positive number we know that

already everything to the left of zero is a negative number we know that already but new to us is this vertical Number line okay this vertical number line and the way this works is that everything above zero is positive everything below zero is negative so a couple of important points here the first thing that you need to know is that the horizontal axis this axis that's going left to right this is the x-axis this is the x axis and actually let me write this let me write this right here x axis Okay so it's going

left to right now the y axis is going up and down so let me kind of write this like this this is the y axis now the place where they intersect you'll see it right here right this comes down and meets this right here see where i have the zero there that's called the origin that's the place of intersection so this is the origin All right so the last little piece of information i'm going to give you about this the coordinate plane has what we call quadrants so you can kind of box this off into

[Music] four areas so this is an area this is an area this is an area and this is an area so this first area all the way to the right the top right this is quadrant Number one and let me erase this so i can show you something let me erase all these in quadrant one the x values are positive so you see that we exist in a plane of space where all the x values are to the right of zero and all the y values are above zero so in quadrant one your ordered pair

is going to have an x value that's positive And a y value that's positive so your x is positive your y is positive now that's important for you to remember because that might come up on a test now we move counterclockwise so we go kind of this direction so the next quadrant is over here this is quadrant two and in this quadrant you'll notice that my x values are to the left of zero so they're going to be negative my y values are still above 0 so they're Going to be positive so this is my

x this is my y so x is negative here y is positive all right now we continue going this way now this is quadrant three now in this quadrant we're to the left of zero so our x values are negative and we're below zero so our y values are also negative so again this is my x values these are My y values all right then the final quadrant we go this way again it's counterclockwise so it just goes this way we end up here and this is quadrant four and in quadrant four the x values

are to the right of zero so they're positive the y values are below zero so they're negative so this is the x values these are the y values so something you might want to write down You might have a test and your teacher might say what are the x values in quadrant four what are the y values in quadrant four are they positive or negative or she might give you an ordered pair and say what quadrant does it lie in and if you've memorized the signs for each quadrant you'll be able to tell right away

for example if i gave you an ordered pair and that ordered pair was let's say 6 comma negative 3 well i know my x value has to be positive so that can only be in quadrant 1 or 4 and the y value has to be negative that can only occur in 4 or 3. now what fits both well the x value is positive the y value is negative so that's quadrant four x value's positive y value is negative so this would be in quadrant four let's say i gave you Negative 2 comma negative 3. i

said what quadrant is that in well if we look here this is negative and this is negative so that occurs in the third quadrant right negative and negative so then this is in the third quadrant so not very difficult just some basic memorization going on there and something you might be tested on so you might want to take note of that all right so now let's talk about the Main thing here how exactly do we plot an ordered pair well we simply find the meaning of the x location and y location and i'm going to

show you this through some examples it's very very easy once you do it a few times you pretty much have the hang of it right away so we want to plot each ordered pair and determine which quadrant it lies in and i'm going to start out with 3 comma 2. so we'll come down to our coordinate Plane let me just write our ordered pair off to the side 3 comma 2. and i just want to call your attention to a couple of things i want you to recall that this is the origin okay this is

the origin the horizontal axis the one that's going sideways is the x-axis this is your x-axis your vertical axis is the y axis so i'll label this one as x this one is y Now with your ordered pair remember the left value is the x value the right value is the y value so what i can do is i can start at the origin if i have an x value of 3 that tells me to go 3 units to the right so 1 2 3 units to the right this is 3 and then a y

value of 2 means i'm going up 2 units so i'm going up one and then two so this would be the location three Comma two right that's where that ordered pair would lie on the coordinate plane now you could have done that a different way and i'll just fill in that space kind of take your pencil or your pen and just kind of you know fill in a circle just like if you're filling in the answer on a scantron right that's the same thing just make a filled in circle filled in dot that's a Little

too big let me make it a little bit smaller something like that now also what i could have done is started at a y value of two so going up one two and then went over to the right three so go over one two three and again i end up at the same spot so whichever one you do first you end up at the same spot this is 3 comma 2. and then it asks which Quadrant does it lie in well you remember the quadrants go like this this is the first quadrant and it goes

counterclockwise so it goes this way so this is the second this is the third and then this is the fourth so this lies in the first quadrant so this is quadrant number one all right so our next ordered pair is six comma five so now we're looking at six comma five Again this is your x value this is your y value so i can start out at the origin and go 6 units to the right this is 6 on the x axis and i want 5 on the y axis so i would just go 5

units up so 1 2 3 4 5. so that's right there let me use a different color here so that would be 6 comma 5. and again if you wanted to you could start off at your y location so i could start off by Starting at the origin say i'm going up to 5 on the y axis so 1 2 3 4 5. so that's right here and then i could go over to the right 6 units so 1 2 3 4 5 6. and i'm just counting for the sake of completeness you can visually

see where the intersection of 5 on the y-axis and 6 on the x-axis is you can kind of pinpoint that but again when you first start you want to be completely sure so you might want to say Okay well this is where 6 is on the x-axis this is where 5 is on the y-axis the intersection of the two you can kind of draw a little dotted line and that wasn't straight right and that's where they intersect so this would be the point six comma five and then what quadrant is it in well again this

is the first quadrant everything is positive and it goes this way this is the second quadrant the Third quadrant and the fourth quadrant all right the next one is negative 8 comma negative one so again this is your x location this is your y location and so i can start at the origin and i can go to the left eight units to find negative eight or i can kind of make it quicker and just say okay the x location is negative eight i can just start there and my y location is negative one so i

would basically just go down one And i would go down one that's what i'm looking for here is negative one on the y-axis so you're looking again for where they meet so you can kind of draw a line for each this is the meeting point this is negative eight comma negative one let me do that in different color negative eight comma negative one and what quadrant does it lie in well again This is the first move this way this is the second so then this would be the third right it lies in the third quadrant

and then for the sake of completeness this is the fourth quadrant all right what about one comma five again my x location my y location so an x location of one and a y location of five so let's kind of reverse it let's start with the y location of five So i'd find that here and an x location of one is here so if i'm already up five i just need to go to the right one again you can kind of draw a little line to see where they would intersect and it's right there all

right so this is one comma five and this lies in the first quadrant right one two three four all right what about negative 2 comma negative 7 What about negative 2 comma negative 7. so this is the x value this is the y value so an x value of negative 2 if we're on the x axis here's negative two a y value of negative seven is here so essentially i'm going left two and down seven right or i'm going down seven and left two it doesn't matter which order you kind of do this or you

could start at negative two and just go down Seven again a lot of different ways to do it so if i want to to make it completely clear let me just draw a line there draw a line here and that's my point right this is negative 2 comma negative 7 and which quadrant does a lion again this is the first one this is the second one this is the third one so this is lying in the third quadrant And this is the fourth one right it's counterclockwise just remember that it starts positive positive so it

starts here and it goes this way one two three four all right so the next one is negative 10 comma five all right so negative 10 comma 5 this is an x value this is a y value All right so i'm going to find negative 10 on the x axis that's all the way over here i'm going to find 5 on the y axis that's right here i'm just looking for their point of intersection so i can go to the left 10 and up 5 or i can go up 5 into the left end again

it doesn't matter all right i could start out at 5 on the y axis and go over 10 or i could start out at negative 10 and go up five however you do it you're Going to arrive at the same location which is negative 10 comma 5. and again if i want to just kind of draw a line like that and then another one like that to make it clear where they intersect and then what quadrant is this in well this is the first quadrant and it goes counterclockwise so this is number two this is

number three and again this is Number four so it's in the second quadrant all right so now we have 4 comma negative 3 so 4 comma negative 3. again this is the x value this is the y value so let's start out at negative 3 on the y axis so let's start out here let's move four units to the right on the x-axis so one two three four so that's right here and so this is gonna be our point four Comma negative three again if i want to draw kind of some lines to show the

intersection again this is four comma negative three and then what quadrant does it lie in again this is number one it goes counterclockwise this is two three and then four so it lies in the fourth quadrant all right for the next one we have negative nine comma negative one So negative nine comma negative one and this is your x value this is your y value so where is negative nine at so negative 9 on the x-axis is over here and negative 1 on the y-axis is here so i can go to the left 9 and

down 1 i could go down 1 to the left 9 i could start out at negative 9 and go down 1 i could start out at negative 1 and go over to the left 9. again it Doesn't matter any way you kind of think about this any way you kind of understand it you can do it because you're always going to end up in the same location so this is the intersection of those two let me just draw that right there and it kind of messes up my negative nine so let me put this one

little thing there and i'm just going to notate that this is negative nine comma Negative 1. and then what quadrant does it lie in again this is 1 this is positive positive and we go this way so this is 2 and then this is 3. so lies in the third quadrant and then this is 4. so 2 3 and then 4. hello and welcome to algebra 1 lesson 15. in this video we're going to learn about graphing linear equations in two Variables so in the previous two lessons we developed the skills that we needed to

be able to graph a linear equation in two variables but one of the questions that might be on your mind why would we want to graph a linear equation in two variables well recall that linear equations in two variables have an infinite number of solutions so a graph gives us a visual representation of all solutions of the equation Now because we're graphing a linear equation in two variables the graph is going to be a line that's why they're called linear equations now each and every point on that line represents an ordered pair an x and

a y that is a solution for that equation meaning i can take the x plug it in for x in the equation take the y plug it in for y in the equation and the left and the right side will be equal all right so let's get started on this topic and It's not very difficult unfortunately the way that it's taught in this kind of section here you get a slow tedious method to kind of execute this process a few lessons from now when we start talking about slope and we start talking about solving for

y and all these other things you're going to get a much faster way to graph a linear equation in two variables in fact If you solve the equation for y it's very very very quick but for right now we're going to kind of start out slow and i'm going to teach you this method that uses a table of values so we have our equation 2x minus 3y equals negative 12. and i have a table right here with x and with y so in a previous lesson i gave you an x value and you solved for

y where i gave You a y value and you solved for x here you can come up with whatever x and solve for y or come up with whatever y and solve for x that you want keep in mind that you have to plot these on a coordinate plane so you want to keep them nice and small and you want to work with integers if possible because your scale on the coordinate plane is generally going to be in integers so let's go ahead and start out by Picking a value for x so you can choose

numbers at random but you're going to see later on that it usually works best to work with the number 0. so let's plug in a 0 for x and i would have 2 times 0 minus 3y equals negative 12. so if x is 0 what's y going to be well this is going to go away 2 times 0 is 0. and i'll have negative 3y equals negative 12. [Music] Divide both sides of the equation by negative three and i'll get y is equal to four so if x is zero y is four so that's one

ordered pair that we would have now how many do we need we're going to find out in your textbook or from somebody that two ordered pairs or two points as we call them make a line but when you're doing this you want to Always use three because the third point is going to check to make sure that you have a line so in other words if i get a point here and let's say here yeah i can draw a line connecting those points but let's say a third point would have been down here well that's

not a line i must have made a mistake if i'm trying to draw a graph there it'd be something like this right i mean that's that's not a lot so if you get a third point that's not in line Meaning it doesn't look like this let's say this point was right here then you know you're okay so that's why that third point is used as a check so now i'm going to get two additional points again keep in mind you want to choose some nice small integer values so i'm going to choose something for x

let's say i choose the number 1. so 2 times 1 minus 3y equals negative 12. 2 times 1 is 2 so i would get 2 minus 3y equals negative 12. subtract 2 away from each side of the equation i'd have negative 3y is equal to negative 14 divide both sides of the equation by negative 3 and i'm going to get that y is equal to 14 thirds now if i'm using a computer or something i can plot the point 1 comma 14 thirds but if i'm doing this on paper do i really want to work

with something like that no so if i get a non-integer value i just kind of move on and try something else i don't want to work with that because it's too hard to plot it on the coordinate plane so let's think about something for a second before we move on we have 2x minus 3y equals negative 12. If i think about this right here whatever i plug in for x i'm going to multiply it by 2 then i'm going to subtract away or add depending on whether it's positive or negative to both sides of the

equation so that's going to end up being over here whatever value that is over here it's got to be divisible by negative 3 otherwise it's not going to be an integer so keep that in mind so One thing i can do is i can say okay if i plug in a 3 here i know that 2 times 3 is 6 so 2 times 3 is 6 and if i subtract 6 away from each side of the equation that's gone i'll have negative 3y is equal to negative 18. well negative 18 is divisible by negative 3.

so if x is 3 remember i plugged in a 3 for x y would be divide both sides by negative 3 and i would get y equals positive 6. so there's an ordered pair 3 comma 6. And i could also do this with negative 3. so 2 times negative 3 minus 3y equals negative 12. 2 times negative 3 is negative 6 minus 3y equals negative 12. i would add 6 to both sides of the equation that's gone i'd have negative 3y is equal to negative 12 plus 6 is negative 6. divide both sides by negative

3 and you're going to get that y is equal to negative 6 over negative 3 is 2. so If x is negative 3 y is 2. all right so now once i've done that kind of scratch paper work i'm ready to move on and actually graph my equation so how do i do that well i said that i needed three ordered pairs really two make a line but the third is for the check and so i take my ordered pairs down to the coordinate plane and i plot them Just like i did in the last

lesson only now i'm going to draw a straight line through them when i'm done let's take this down to the coordinate plane all right so we have the ordered pair zero comma four so let's plot that so zero comma four is right here then we have the ordered pair three comma six so i wanna go 3 units to the right and 6 units up that's going to be right here Then i have the ordered pair negative 3 comma 2. so i'm going to go 3 units to the left and 2 units up and you can

eyeball that and see that you could draw a straight line through that if one of your points was somewhere way over here you have a problem you need to go back and look at it and check your math you did something wrong okay so now all i want to do is draw a Line through the points now in most cases you're not going to be able to draw a perfect line you've got a ruler at best and you've got a piece of graph paper hopefully but some of us are going to be using just a

piece of loose leaf and no ruler so just do the best you can make it obvious to the teacher that you know you gave an effort to draw a straight line through the points and in most cases they're going to be nice and give you credit but Unless you're using a computer it's almost impossible to draw a completely perfect line so then one of the things you can do when you're done you can label it so this is the graph of the equation 2x minus 3y is equal to negative 12. and notice that each and

every point on that line again is an ordered pair that is a solution to the equation so for example it looks like this point Right here is on the line hopefully i drew it straight enough to where it is so that would be the point six comma eight so six comma eight let's see if that works as a solution so let's test an x value of six a y value of 8. let's see if that works so 2 times 6 minus 3 times 8 should be equal to negative 12. 2 times 6 is 12 Minus

3 times 8 that's 24 equals negative 12. you can see that works out 12 minus 24 is negative 12 so you get negative 12 equals negative 12. so yeah that's a solution to the equation so that's the general idea behind graphing a linear equation in two variables each and every point on that line is an ordered pair that is a solution to that equation so it's a visual representation of all your solutions now One of the things that i didn't explain but i should have went into depth on is that i drew arrows at each

end these arrows are the same as the arrows i'm drawing right here here here and then down here what is that telling us when we have an arrow tells us that it continues forever and ever and ever in that direction so this line will continue forever forever and ever and ever going in this Direction and forever and ever and ever going in this direction so that's why we put arrows at each end all right now let's talk about a few things real quick that are very very important i encourage you to take notes for this

section so the x-intercept you're going to hear about this a lot the x-intercept is the location on the coordinate plane where our graph crosses the x-axis let's go back up here and let's look at the x-axis so i have a Lot of stuff highlighted right now let me unhighlight some stuff and i'm just going to highlight the x-axis so here's the x-axis and i want you to look at the graph and i want you to see that the line crosses through right here this is where it crosses the x-axis so this is known as the

x intercept there's a point where it intercepts the X axis and if you look at the graph it looks like it's going to be at negative 6 comma 0. so negative 6 comma 0. this is the x intercept and you can tell that does work out i'd have 2 times negative 6 that's negative 12. and this negative 3y would go away because i plug in a 0 for y negative 3 times 0 is just 0. so that's gone so you get negative 12 equals negative 12. now i might want to bring your attention To the

fact that the y location is zero why do you think that is well if i'm on the x axis if i'm crossing through there that means that vertically i have to be at 0 on the y-axis because that's where the x-axis lies right this is 0 on the y-axis so to cross the x-axis i have to be at 0 on the y-axis now similarly we have this thing called the Y-intercept this is the location on the coordinate plane where our graph crosses the y-axis so let me un-highlight all this stuff so the y-axis is here

this is the y-axis so where does the graph cross there well crosses right here at 0 for the x location and 4 for the y location so this is your y-intercept okay And this occurs at 0 comma 4. this is the y-intercept and does that work out well yeah if i plug in a 0 for x this would go away i'd have negative 3 times 4 that is negative 12. so i get negative 12 equals negative 12. now one thing again i want to call your attention to is notice how your x coordinate is zero

because if i look at this horizontal axis the x-axis it has to be zero For me to impact the y-axis so this leads us to another method and this is called the intercept method you'll probably hear about this in your class so the intercept method is usually much faster than choosing random numbers to plug into the equation so we begin by finding the x and y-intercepts and then we also like one additional point for a check so how do we find the x-intercept well again i told you if you want to Cross the x-axis y

has to be zero right if i look at the coordinate plane y has to be zero to cross the x-axis so for the x-intercept y is 0 for the y-intercept x is 0. so that's how you find your two intercepts again if i want the x-intercept the opposite one y is 0. if i want the y-intercept the opposite one x is 0. so this would be the x-intercept And this would be the y-intercept and then we'll get a third point as a check so let's go ahead and plug in a 0 for y you'd have

2x plus 5 times 0 is just 0 so i can just have 2x this equals negative 20. divide both sides by 2 and you get x equals negative 10. so if y is 0 x is negative 10. and look how easy that was when you work With 0 you plug it in it makes things disappear makes your work a lot simpler that's why this intercept method is usually a little faster all right now the other one we're going to plug a 0 in for x so this is the y-intercept so 2 times 0 plus 5y

equals negative 20. and again i can just get rid of this so 5y equals negative 20 divide both sides by 5 and you get y equals negative 4. So now i've got two ordered pairs but again i like a third one for a check so what am i going to plug in for x well if i look at this equation 2x plus 5y equals negative 20 again if i think about this whatever i plug in for x i'm going to end up moving this over here right now negative 20 is divisible by 5. so if

i got to pick something that i plug in here and move it over here to where it's still going to be divisible By 5 otherwise i'm not going to end up with an integer so one thing you can think about is if i multiply 2 times 5 2 times 5 that would give me 10. now if i subtract 10 away from negative 20 i'm going to get negative 30 and that's divisible by 5. so that would work out so again you want to kind of think about that before you start plugging in random numbers because

if you end up with a non-integer you Just wasted your time right it's very hard to plot that so you plug in a 5 here you would have 2 times 5 or 10 plus 5y equals negative 20. i would subtract 10 away from each side of the equation that's gone i'll have 5y is equal to negative 30. divide both sides by five i'm going to get that y is equal to negative six all right so i have three ordered pairs i have negative 10 comma zero i have zero comma negative four And i have five

comma negative six so let's plot these points on the coordinate plane and then we'll draw a straight line through them all right so let's begin with negative 10 comma zero so that's all the way to the left on the x-axis so way over here this is negative 10 comma zero then i have zero comma negative four so that's down four on the y axis and then i have five comma negative six So i go to the right five and down six and you can see that when i draw a straight line through it the line's

gonna cross through here which the x location is zero so it's crossing through the y axis this is the y intercept right x location is zero at that point and then notice how it's also going to impact the x-axis here where y is going to be 0. okay y is going to be 0. that's your x-intercept all right so let's draw a straight line through these points now okay so put arrows at each end and the line's not perfect but it'll do for what we're trying to accomplish here again look at where this line crosses

the y-axis call your attention to that right here at the point zero comma negative four so the x location is zero and notice how on your horizontal axis You're in the center at 0. so that's your y-intercept this is the y intercept and again x is 0 at that spot then when we look at our x-intercept again that's the spot where our line crosses the x-axis that's going to be right here that's at negative 10 comma zero notice how your vertical location your location on the y-axis is zero there and let me just label this

real quick as the x Intercept and let me just label this line as the graph for 2x plus 5y equals negative 20. and again long story short the main thing i want you to understand here is that for your x-intercept y needs to be 0. so just plug in a 0 for y solve for x that's your x-intercept for the y-intercept x needs to be 0. so plug in a 0 for x solve for y that's how You get your y intercept so most of the examples you're going to see especially at first are going

to be like the previous two but occasionally you're going to have to do a little bit more work because you're going to get a line that passes through the origin you can tell right away if your equation looks like this ax plus b y equals 0. so some number times x plus some number times y equals 0. so we have an example here 4x minus 3y equals 0. so the reason it's additional work is because if you try to use your intercept method here the x and the y-intercept occur at the same point which is

0 comma 0. right the line passes through that because that's the origin if i plugged in a 0 for x you can think about this term just going away you'd have negative 3y equals 0. well the only way negative 3y can equal 0 is if y is 0. so 0 comma 0 is a point on that line but now i can also see okay if y is 0 x is 0 but if x is 0 y is also 0 so that's that's the only intercepts i'm going to get so now i have to pick two

other points and i have 4x minus 3y equals 0. so to make this a little easy on myself i'm just going to add 3y to both sides Of the equation [Music] and the reason i'm doing that is i want to make it a little easier to see when i choose something for x or choose something for y the result of that is not going to be divisible by something to where i get an integer right so in other words if i plug in let's say a 3 here I know that i'm going to get 12

and i know 12 is divisible by 3 so that will give me an integer so i can plug in a 3 here and so this would be 12 12 equals 3y divide both sides by 3 and i get 4 is equal to y now to make this really easy again i had 4x was equal to 3y i could just use negative 3 because i know that's going to work out also so what about negative 3 being plugged in For x i'd have negative 12 equals 3y divide both sides by 3 and i'll get negative 4

equals y all right so i have my three ordered pairs i have zero comma zero i have three comma four and i have negative three comma negative four okay so let's plot these points so zero comma zero that's at the origin it's going to be right here three comma four so go to the right Three go up four that's right here negative three comma negative four so go to the left three and down four that's going to be right here we're going to put your arrows at each end and i'm going to go ahead and

label this again this is the equation 4x minus 3y equals 0. and you can see the only place that this line crosses Through the y axis is at zero comma zero and the only place this line crosses through the x axis is again at zero comma zero so this is a line through the origin where both the x and the y-intercept occur at that origin 0 comma 0. all right so we've kind of talked about the basic scenarios that you're going to come across and at this point you should be Confident enough to be able

to graph most of the linear equations and two variables that you're going to encounter it's just something you need to practice at this point but i want to wrap up our lesson by talking about two special case scenarios and if you see these you're going to kind of be confused about what to do so the first one is a vertical line and the second one is a horizontal line so a vertical line is one that has a fixed Value for x no matter what the value is for y so this type of line is parallel

to the y axis so we have x equals negative three so what we're saying here is that no matter what the value is for y x is equal to negative three so if you want you can choose points and plot those points and draw a line so in other words i can pick some values for y x is always negative 3 so i'd have Negative 3 comma let's say negative 5 and then i could do negative 3 comma 0 and i could do negative 3 comma positive 5. no matter what i choose for y doesn't

matter what it is could be 1 billion x would be negative 3. so you can go through and plot these points like that so we have negative 3 comma negative 5. So negative 3 negative 5 that's right here let me use a different color and then negative 3 comma 0 that's right here and then negative 3 comma positive five so that's going to be right here so you can see you're going straight up and down right this would be a vertical line so the trick to this is i don't need to make any points you

just kind of do that at first to get your Mind wrapped around what's going on if you have x equals negative 3 you just find negative 3 on the x axis and you just create a vertical line that's all you need to do so i know that's not perfectly straight but we'll just pretend that it is and we'll put our arrows at each end and it's just that easy and remember no matter what the value of y is let's say y is seven x would be negative three so This would be a point on that

line i know the line is kind of a little crooked so it doesn't hit there but that's where it should be or let's say y was negative six well x is negative three so it's right there so if i gave you another one let's say i gave you the problem x is equal to positive six well you just find six on the x-axis and you draw a vertical line that's all You need to do [Music] okay and again i know that's not straight but do the best that i can so put my arrows at each

end and that's x equals 6. so no matter what the value is for y x will be six so it's just a vertical line now mathematically you might be asking how is this a linear equation in two variables let's go back to the definition i gave You when we first started talking about it i said that we had x plus b y equals c where a b and c were real numbers and a and b were not both zero so a and b are not both zero but one of them could be so if i

have 1x plus 0y equals some number let's say negative 3 for example when i simplify this i end up with 1x or just x Plus 0 times y is 0 so i can just kind of get rid of that so it'll really be x just equals negative 3. so this is a way to write it as a linear equation in two variables and it's also a way to check your ordered pairs right to mathematically show that makes sense if i have the ordered pair negative 3 comma 5 well i can plug it into this equation

and it would be true so i plug in a negative 3 for x and i Plug in a 5 for y 1 times negative three plus zero times five should be equal to negative three one times negative three is negative three zero times five is zero so this would be negative three equals negative three so that checks out and you could do that as long as x is negative three you could do that with any value of y because you're multiplying it by 0 so it's just going To drop out so the next concept is

a horizontal line and this is very very similar to what we just looked at this type of equation has a fixed value for y no matter what the value is for x so this type of line is going to be parallel to the x axis so let's look at y equals negative 2 to start and again you could go through and make some points but you don't need to if y equals negative 2 find negative 2 on the y axis it's right There so no matter what the value is for x let's say the value

is 2 for x y is negative 2. so the value is 6 for x y is negative 2. so the value is 10 for x y is negative two say the value is negative four for x y is negative two so it's just a horizontal line you just find negative two and draw a horizontal line okay all right so that would be y equals negative two y equals negative two Now if i said y equals five i would just find five on the y-axis draw a horizontal line if i said y equals nine find nine

on the y-axis draw a horizontal line so these types of problems are very very easy and again mathematically we can write it as 0x plus 1y equals in that case it would be negative 2. this is the same thing as y equals Negative 2. but we wrote it with two variables so we can say it's a linear equation in two variables and if i give you an ordered pair like let's say negative six comma negative two it's going to work plug in negative six for x zero times negative six is zero so i'm just left

with one times y plugging in a negative 2 for y get negative 2 equals negative 2. and so when you see an equation in this format y equals some number find that number on the y axis And draw a horizontal line hello and welcome to algebra 1 lesson 16. in this video we're going to learn how to find the slope of a line so in our last lesson we learned how to graph a linear equation in two variables kind of the next thing we want to learn how to do is take this graph right it's

always of a line that's why we call it a linear equation and analyze the steepness right the steepness is referred to as the Slope so i have here that a very important characteristic of a line is its slope let me highlight that word for you it's slope or steepness and you might not have run across this yet in your math class but you've seen it in your everyday life you can think about being on the interstate and seeing a 10 downgrade sign right what does that mean well it means for every 10 miles you travel

the road has dropped by a Mile right that's what a 10 percent downgrade means your road is going to look something like this it's slanted downward right i know that's not a perfectly straight line but let me draw a picture of a little car here and just for the record i am a terrible drawer and i know that so this is my car here and put a little exhaust going out of There and i'm driving and for every 10 miles i go the road is dropping by a mile so that's an example of a slope

all right so we measure slope by comparing the change in y values this is the vertical change remember the y axis is a vertical axis to the change in x values the horizontal change remember the x axis is the horizontal axis while moving along our line from one Point to another all right so how do we actually calculate slope well lucky for us we have something known as the slope formula that we can use and then we can also do it using the graph which we'll cover in a little while so calculating slope and this

is using the slope formula you're going to pick any two points on the line doesn't matter which two points You pick so this means any two ordered pairs and you're going to label the points as x sub 1 y sub 1 and x sub 2 y sub 2. now let me stop i'm introducing some new notation so i want to clarify this right here this one that's kind of sitting next to the x is read as sub 1. this 2 that's sitting next to the x is read as sub 2. this is a way for

us to notate that we have two different x values that we're Thinking about here without actually kind of changing variables right and in a lot of math i can just pick a variable a different letter every time i have something else i'm representing but here because we're specifically thinking about okay the x value is the value that we're thinking about that corresponds to the horizontal axis so i've got to be consistent there i can't say okay i have x and then i have Some variable named z that wouldn't make any sense so we have to

have two different x values and so to separate them we use sub 1 and sub 2. if i was working with something that had three different x values i would have a point that would be x sub 3 y sub 3. right so on and so forth and you're going to see that as you move higher math when it calls for that but for right now we just need 2. so that's where this is coming from so you have x Sub 1 y sub 1 and x sub 2 y sub 2. now the next source

of major confusion is which point do i label x sub 1 y sub 1 and which point do i label x sub 2 y sub 2 and the answer is it does not matter okay and i'm going to show you that so that you understand because i can't tell you how many people come up to me and say well how do i know which one the label is which it does not matter you can switch them around you're Going to get the same exact answer so once you've labeled them all you need to do is

plug it into the slope formula and i advise you to copy this down we have that m which stands for slope this stands for slope and i don't know why m stands for slope but it does and this is equal to we have y sub 2 minus y sub 1. so this is referred to as your change Your change in y values and then this is over we can think about it as dividing by x sub 2 minus x sub 1 which is what it's the change in x values okay the change in x values

so we have the change in y values over the change in x values and then one important note we can never have the case where x sub 2 equals x sub 1. why because we can't divide by zero if i Have the same number here let's say 2 and 2 2 minus 2 is 0 i can't divide by 0. so that's a problem so x sub 2 cannot equal x sub 1. and another way we might see this written is that m our slope is equal to the rise over the run and this is just

a way to remember that it's the change in y values or the rise you can think about this is going up and Down over the change in x values or the run you can think about running as going you know sideways or horizontally being on that horizontal x axis all right so let's start out with two points this is a typical example of something you see on your homework or test you're given two points and you're asked to calculate the slope so we have the point negative 1 comma 1 and 20 comma negative 20. So

the slope formula is m equals y sub 2 minus y sub 1 over x sub 2 minus x sub 1 and i'm not going to write this every time but remember that x sub 2 is not allowed to equal x sub 1. it cannot all we need to do is label one of these points as x sub 1 y sub 1 and the other x sub 2 y sub 2 and then just plug in again it does not matter which point you choose to be x sub 1 y sub 1 and which point you choose

To be x sub 2 y sub 2. it's completely irrelevant you get the same answer either way so i'm going to pick this to be x sub 1 y sub 1 and i'm going to pick this to be x sub 2 y sub 2. one thing that you cannot do okay you cannot mess this up in terms of saying okay well this is x sub 1 y sub 2 and this is x sub 2 y sub 1. that will Give you the wrong answer do not do that if this is x sub 1 this has

to be y sub 1. if this is x sub 2 this has to be y sub 2 right you cannot mix those numbers up like that now once you have labeled everything go ahead and just plug into the slope formula it's really just that simple so i have y sub 2 here so i'm going to take negative 20 and plug that in so let Me just erase this and just put a negative 20 here in its place and then i have minus i have y sub 1. so that's 1. let's erase this and put 1.

and then i have x sub 2 down here that's 20. so i'll erase this and put 20. and then for x sub 1 i have negative 1. so let's erase this and put a negative 1 in now again again again it's very important to Pay attention to your signs when you're plugging in for something i already had a negative out in front i'm plugging in a negative 1 for x sub 1. so it's minus a negative one don't make the silly mistake of just seeing that out in front and just putting minus one you get

the wrong answer so then all i need to do now is just simplify i have negative 20 minus 1 Which is negative 21 and this is over 20 minus a negative 1 which is the same thing as 20 plus 1 which is 21. so i end up with negative 21 over 21 which is the same thing as negative 1 right negative 1. so m again my slope let me label that real quick my slope is equal to negative 1. so let's see if we can make a little sense out of this slope thing So we

saw that m is equal to negative 1. so what does that mean well again if you kind of look at the rise over run explanation so rise over run we know that we can write negative 1 as negative 1 over 1 right that's the same thing if you divide by one it's just itself so i can put this is equal to negative One over one so i look at what's in the numerator i have rise here and i have negative one here so that means i'm going to rise by negative one or fall by one

unit for every run which we think of as going sideways by one unit okay now because this is positive down in the run that means i'm going to the right so positive in terms of the x-axis means you go right negative means you go left Positives in terms of the y-axis means you go up negative means you go down so if i started out at a point on this line so we know that negative 1 comma 1 which is going to be right here so that's a point on the line i could actually use this

to graph the line because i know that i'm going to fall by one unit so i'm gonna fall by one unit thinking about the rise For every one unit i go to the right so i'm gonna go down one and over to the right one so this point would be on that line and i can keep doing that i can go down one to the right one and go down one to the right one down one to the right one down one to the right one down one to the right one or another thing i

can do remember when we're dividing and we think about having A negative in the numerator but a positive in the denominator i can legally switch that and i can legally say that this is 1 divided by negative 1. now how would we read that well that means that we would rise one unit for every unit we go to the left on the x-axis so starting out at this point i would go to the left one unit and then i would rise a unit we see we end up at a Point on the line left one

up one left one up one left one up one left one up one left one up one i can keep going left one up one here's another point left one up one here's another point you can see that i could easily draw a completely straight line through those points and let me do that real quick [Music] not perfectly straight but you get the idea And later on we're going to learn how to take this information that we have and find the equation for the line so this is actually y equals negative x and i know

that you don't know how to do that yet that's okay we're going to learn it very very soon in the future so quickly i want to show you that if i switched which ordered pair i labeled as x sub 1 y sub 1 with the ordered pair i labeled as x sub 2 y sub 2 that it would make no Difference and your slope would still be negative 1. let's do that real quick before we move on all right so let's just switch the way these are labeled and so instead of this being x sub

1 y sub 1 it's going to be x sub 2 y sub 2 and this is going to be x sub 1 y sub 1. and let me erase this real quick so the slope formula again it's m is equal to y sub 2 minus y sub 1 over x sub 2 minus x sub 1. And let's just crank this out real quick so y sub 2 is now 1. so i'm going to plug that in and minus y sub 1 y sub 1 is negative 20. again pay close attention you're plugging in a negative

20. so that's going in there and then this is over x sub 2 which is now negative 1 minus x sub 1 which is 20 minus 20. okay so what are we going to get we have 1 minus a negative 20. that's the same thing as 1 plus 20 that's 21. this is over here we have negative 1 minus 20 which is negative 21. so again you see it's going to be the same thing 21 divided by negative 21 is negative 1. so this is negative 1 or again in terms of slope we can think

about this as negative 1 over one or we could write it as one over negative one right again as it relates to rise Over run we've already seen that we can use that in two different ways all right let's take a look at the next problem so we have the ordered pair negative 7 comma negative 11 and 5 comma 5. so these two points are what we're given and we want to calculate the slope so m again our slope is equal to y sub 2 minus y sub 1 over x sub 2 minus x sub

1. and again x Sub 2 cannot be equal to x sub 1. so we're going to label our points and a general practice for me and probably for you is just to label the first the first point as x sub 1 y sub 1. and again as i just showed you it doesn't matter you get the same answer either way and this one's going to be x sub 2 y sub 2. okay so y sub 2 is 5 so let me erase that and Put a 5 minus y sub 1 that's negative 11. let me

erase this and put negative 11. this is over we have x sub 2 that's going to be 5 minus x sub 1 which is negative 7. so minus a negative 7. so in each case here we have to pay close attention to our signs because we have minus a negative so 5 minus a negative 11 is the same thing as 5 plus 11 which is 16. 5 minus a negative 7 is the same thing as 5 plus 7 and that's 12. now we can reduce this both are divisible by 4 16 divided by 4 is

4 12 divided by 4 is 3. so we get 4 thirds here or 4 over 3. now we think about this in terms of again rise over run so rise over run is equal to 4 over three so that means that if i'm at a given point on a line I can rise four units or go up four units and then run three so go to the right three and i'll end up at another point on that line so a point on the line is five comma five so that's going to be right here again

our slope m is equal to four thirds so in terms of rise over run this is four over three so If i wanted to find another point on this line i could go up four so up one two three four and then i could run three okay when you talk about run again positive means you go to the right so i'm going to the right three one two three so here's another point on that line now let's say you are out of space going up on your graph like i am my kind of top number

here is 10. well no problem you could start at 5 comma 5 and you can Go in this direction you might say well how do you do that well one cool trick is to remember that a negative divided by a negative is a positive so really i could write this as negative 4 over negative 3. this has the same mathematical value as 4 3 because if i divide a negative by a negative i get a positive which is what i have here so using that thought process i could Start at this point and i could

fall 4 units right that's what negative 4 means in terms of rise so i would go 1 2 3 4 and then i could go to the left 3 units that's what negative 3 means in terms of run right i'm going to the left so 1 2 and then 3. so here's another point in the line and i could do that again i'm going to fall 4 1 2 3 4. i'm going to run to the left 3 1 2 three so here's that and then i could graph this line if i Wanted to okay

and again not perfect but kind of the best i can do all right and then what's the equation for this line again i know that you don't know how to figure that out yet but i'm just going to go ahead and give it to you and so this would be the graph for y equals 4 3 x minus five thirds and again really shortly we're Going to learn how to do that all right so as you may or may not have figured out at this point if you know one point on the line and the

slope we can easily graph any equation this is kind of the preferred method to do it you saw kind of in the last lesson we already go through and make these ordered pairs and even though with the intercept method it's a little bit easier This method of graphing is just so so fast so my given point is negative two comma four and my slope is negative two so again m is negative two and my given point is negative two comma four so here's negative two here's four so that would be negative two comma four and

if my slope is negative two again that means that my rise over run Is equal to negative two over one so that means i'm falling two units for every one unit i go to the right so starting here i will go one two down and then one to the right there's a point one two down and one to the right there's a point and once you have three points really two you can draw the line right but three again is nice to get a nice clean line and i like to do a few more Actually

if i'm free hand so one two down one over one two down one over to the right one two down one over to the right so kind of a faster way to graph a line okay so another thing we want to talk about let's say you get to your homework and you're presented with a graph and so the graph of this line they have some clearly indicated points and it asks for the slope of the line well using the rise over run kind of Thought process so rise over run you can calculate it so let's

say you start out at one of the points so this point right here looks like negative six comma 8. so negative 6 comma 8. all right so you think what would i need to do to get to the next point so how much do i rise so to get over here i've got to fall Right so i'm rising by a negative amount i'm going down one two units so right here is not the next point but it's the same y location as the next point so that means i'm falling by two units and then i'm

running over to the right which is positive i'm going over 1 2 3 units to the right so my slope is negative two thirds and then you can check that again starting at the next point down 2 1 2 go over to The right 3. one two three and i'm at the next point down two one two over to the right three one two three i'm at the next point again you can keep doing that so our slope in this case is negative two thirds and then one of the other ways you could check this

you could look at one of the other points so this point right here is negative three comma six so negative three comma six and you could use your Slope formula and see if you get a slope of negative two-thirds let's do that real quick so remember your slope formula m m is the slope this is equal to y sub 2 minus y sub 1 over x sub 2 minus x sub 1 again where x sub 2 does not equal x sub 1. i'm just going to go ahead and label this first point as x sub

1 y sub 1 and the second one as x sub 2 y sub 2 and again if you wanted to change that Around you get the same answer so y sub 2 is 6. so i'm going to erase this and put a 6. y sub 1 is 8. i'm going to erase this and put an 8. x sub 2 is negative 3. so all races put negative 3 and x sub 1 is negative 6. so i'm going to erase this and put minus negative 6 or you can just do this right now and just say

this is plus 6. So 6 minus 8 is negative 2 and negative three plus six is three so you get negative two thirds as your slope and again that's exactly what you calculated here your rise over run is equal to negative two thirds or negative two over three all right so now that we have the basics down let's wrap up our lesson with a few basic things that you need to know about slope so the first thing is that a line with a Positive slope rises as we move to the right so that kind of

looks like a line that does this right so as we move to the right so as x increases remember x increases going to the right y increases so as we go to the right the line is increasing in value so similarly a line with a negative slope falls as we move to the right So in this case as x increases y is now going to decrease so that line would look something like this so as x is increasing remember x increases as we move to the right y is now decreasing your vertical values are moving

down as we go to the right all right let's talk about some special case scenarios the slope of any horizontal line is always zero so you have something like y equals seven so You'll recall graphically i just find seven on the y axis and i draw a horizontal line that's how i graph that okay so this could be my graph for y equals 7. now if we think about this in terms of again rise over run so rise over run just think about moving one unit to the right so let's say i'm at this point

right here and i go one unit to the right so my run is one What's my rise it's zero and i'm not moving at all vertically so i would have zero over one which is just zero right take zero and divide by any non-zero number and you get zero so this is going to be the case every time we see a horizontal line so a horizontal line has a slope that is zero so lastly a vertical line has an undefined slope this relates back to when i gave you the slope formula and i said x

sub 2 could Not equal x sub 1. well again you subtract the same number from itself and you get 0. division by 0 is not allowed so the change in x values will be 0 and division by 0 is undefined so if i have a vertical line like x equals 2 you just say that the slope is undefined so if i saw x equals 2 on a graph remember i'd find 2 on the x axis so again this is x equals two and if you think about this in terms of again rise over run Just

started at any of these points and all of them are going to have the x location of 2 so let's say i started out at this point so this is 2 comma 3. and i go to the next point above it which is 2 comma 4. well my rise is one unit right because i increased on the y axis by one but my run is zero right my x location didn't change at all so that's zero and we're not allowed to divide by zero this is undefined this is Undefined so when i have a vertical

line and they ask you for the slope you just reply that it's undefined because if you go to try to calculate it you're going to end up with division by zero and once again that's undefined hello and welcome to algebra 1 lesson 17. in this video we're going to learn about equations of a line so as we continue in this chapter on linear equations and two Variables we're going to now see that there are ways to manipulate a linear equation in two variables and each form of the line has a specific use so kind of

the most immediate would be slope intercept form and this form of the line is very very useful to us when we look at something in slope-intercept form we know the slope and the y-intercept and that's why it's Called slope intercept form so if i have a linear equation in two variables and i solve it for y meaning y is on one side of the equation by itself like we have here then i immediately know that m the coefficient of x is the slope this is the slope and you'll recall that in the last lesson when

i taught you the slope formula i denoted slope using m so it's no coincidence that we put m here next to x Because again m stands for slope then plus b now this part right here is denoting our y-intercept so this is your y intercept and that's going to occur at 0 comma b so whatever b is that's the coordinate for your y-intercept right that's the y part of it you'll have 0 always as the x-coordinate when you're Looking at your y-intercept and then b will be the y-coordinate for your y-intercept so as a quick

example let me just give you a generic linear equation in two variables let's say i had something like 4x plus 2y equals negative 10. really really simple so if i want to get y by itself if i want to solve it for y like i have up Here the first thing i need to do is just isolate this term 2y so i would subtract 4x away from each side of the equation and so this would go away and on the left i'd have just 2y on the right i'd have negative 4x minus 10. now to

isolate y to get in this format i would just divide both sides of the equation by 2. So i would divide this side by 2 and a lot of students will just do this they'll divide this by two but remember when you have division like this you're allowed to break it up and say okay i'm dividing this part by two and this part by two right it's like having a common denominator so you can do it that way it makes a little easier to simplify so this cancels with this and i'll just have y and

then negative 4 divided by 2 is Negative 2 so let's write that as negative 2 and then negative 10 divided by 2 is negative 5. so kind of looking up here i have y equals m the slope times x plus b which is our y-intercept so y equals negative 2x so in this case negative 2 is the slope this is your slope or your m and then minus five so minus five you could think of as plus negative five So your y-intercept would occur at zero comma negative five this is your y intercept all right

so let's take a look at kind of an official example and the reason for this example is i want to just convey to you in the past or kind of we'll just think about it in our last lesson we had to do a lot of work to find slope you know if we're given an equation we Had to calculate two points then go through and plug them into the slope formula and then now all we have to do is solve for y much much easier but i'm going to show you the difference between kind of

the old way and the new way so we have our equation 5x minus 4y equals negative 20. so old way i'd go through and get two points so let's say i look for the y-intercept so that's me plugging in a 0 for x So i'd have 5 times 0 minus 4y equals negative 20. so this would go away and i'd have negative 4y equals negative 20. a lot of you can just look at that and say okay i know the answer is 5 but we'll officially say okay divide by negative 4 on each side this

cancels with this and you'll get y is equal to 5. so if x is 0 y is 5. okay now let's go ahead and use the x-intercept so that means y would be zero So i'd have five x is equal to negative 20 right because if i plug in a zero there negative four times zero would just be zero so that would just go away so for this i would just solve by dividing both sides by five so then x would equal negative four okay very very simple so now i have two points so one

point is zero comma five and the other is negative four comma Zero but now i need to plug these into the slope formula so remember m which is the slope is equal to y sub 2 minus y sub 1 over x sub 2 minus x sub 1. all right so let's label this one as x sub 1 y sub 1 let's label this one as x sub 2 y sub 2. and let's see what happens when we plug in So for y sub 2 i have a 0 so plug that in for y sub 1

i have a 5. so plug that in for x sub 2 i have a negative 4 plug that in and for x sub 1 i have a 0. so plug that in so 0 minus 5 is negative 5 over negative 4 minus zero which is negative four so you get negative five over negative four which is five fourths So that is your slope and let's write that right here that m equals five fourths now i'm going to show you something in two different ways i mean first off you saw how long and how complicated that

process was just to figure out the slope i mean it's it's probably a minute minute and a half two minutes almost just to go through and get the first piece of information then you got to go back and plug it into the Slope formula and get that so it's very very time consuming let's erase this real quick i'm going to show you a few different things so the first thing i just want to show you and this is kind of a side note if you know the y-intercept which we figured that out it occurred at

zero comma five and you know the slope you can write the equation in slope-intercept form you have all the information you need it's y Equals m the slope which is 5 4 times x plus it's the y-coordinate from the y-intercept so plus 5. and you'll hear that they just say this is the y-intercept this is your y-intercept and then this is the slope now i'm going to show you that i can take this equation right here and transform it into this one all i need to Do is some basic algebra so we have 5x minus

4y is equal to negative 20. let me scroll down a little bit just to get some room going so again if i want to isolate y the first thing i need to do is isolate the negative 4y i can isolate that whole term first and to do that i would just subtract 5x away from both sides of the equation nice and simple so that's gone we'll have negative 4y is equal to negative 5x Minus 20. the next thing i want to do is get y by itself so i'm going to divide by negative 4. divide

over here by negative 4 and divide over here by negative 4. so what are we going to have let's scroll down this will cancel with this and i'll have y is equal to negative over negative is positive so that's 5 4 then times x and then you can think of this as plus Negative 20. so let's put a plus here negative 20 over negative 4 is 5. so what did we end up with we end up with y equals 5 4 x plus 5 and that is exactly what we had up here just by the

given information right y equals 5 4 x plus 5. so if you have a linear equation in two variables you can solve it for y and you can immediately figure out the slope and the y intercept without going through getting points and then plugging into the slope formula and You know so on and so forth the other thing is if you're given the y-intercept and you're given the slope you can just write the equation right again we're going to do that here in a little while all right so here's another example we have x plus

2y equals 16 and we don't need to go through any tedious process here we're just going to solve this for y and we're going to figure out what our slope is and we're also going to know what our y intercept will be So to solve for y i would subtract x away from each side right because i want to isolate this guy right here so i would have 2y is equal to negative x plus 16 and then i would divide each side of the equation by 2. so what would i have that would cancel i'd

have y is equal to you think about negative x as negative 1 Times x right we don't typically write that but to make it clear you have negative one-half times x there so negative one-half times x plus 16 over two is eight so i have my equation solved for y and now i know that the slope or m is negative one-half and i know that the y-intercept will occur at 0 comma 8. so here's another very common textbook type example something you'll Probably get on your test so we want to write the slope-intercept form of

each line given the slope and y-intercept and we just did this a little while ago so we have that m equals 9 and again m is standing for slope so the slope is 9 and the y-intercept is given as the point 0 comma 4. so how do i write the equation i have y equals mx plus b if you have a problem like that just Write that formula out to begin with you need to memorize y equals mx plus b and then just duplicate it so i have y equals i know what m is that's

9 then times x then plus b and again what is b it's the y coordinate when we think about the y intercept so the y intercept occurs at zero comma four so we just put plus four so y equals nine x plus four all right so the next one i'm given is a slope m equals negative seven halves y Intercept occurs at 0 comma negative 5. so again start out by writing that generic formula which is y equals mx plus b and just plug in so y equals my m is negative 7 halves so negative

seven halves times x and then plus b so in this case your b is going to be negative five so i'm gonna put minus five you could also put plus negative five either way it's the same You know both y equals negative seven halves x minus five all right what about if m is equal to five thirds okay so a slope of five thirds and a y intercept that occurs at zero comma negative four so again write your generic formula y equals mx plus b so y equals what's m that's five thirds so five thirds

times x and then your b here is going to be Negative four so again you can put minus four you can put plus negative 4 doesn't matter so i'm just going to put minus 4. so y equals 5 3 x minus 4. all right so we're going to kind of wrap up talking about slope intercept form with this last point so a very useful application of slope intercept form is graphing this is probably the quickest way to graph a linear equation in two variables so if we'd had this you know a few lessons ago When

we first started talking about it you're like man this is really easy so if you have something in slope intercept form you automatically have the slope and you have a point on the line you have the y intercept so it's super super easy to graph it so if i have y equals negative 1 4 x minus 3 i know that the slope m is equal to negative 1 4 and i know that the y-intercept the Y-intercept is going to occur at 0 comma negative 3. so to graph this equation i have that point on the

line 0 comma negative 3. that's where i'm going to start again that's my y-intercept so that's going to be right here right here and again the slope m is negative 1 4. now recall when we first started talking about slope we refer to it as rise Over run so this equals negative 1 over 4 or you could also think about it as 1 over negative four these are the same remember if i have a negative divided by a positive or positive divided by a negative net net net in the end it's a negative right so

you can write that either way so starting from this point i can fall one and go over four right that's rise over run so if i rise Negative one i'm falling one if i run four i'm going to the right four so i'm going one two three four so this would be the next point on the line and i'll do that again fall one over one two three four and you can get some points over here too by kind of reversing that and saying okay i can also rise one and go to the left four

right if i have negative 4 in the position of 1 that means i'm going to the left 4. so 1 2 3 4. do that again up 1 over to the left 1 2 3 4. so there are some points and we just draw a line through them to graph it okay so now we'll put some arrows at each end and so this is the equation y is equal to negative 1 4 x minus 3. and let's test some of these points just in case you don't believe me so let's Test this point right here

which is 4 comma negative 4. so we'll go back up here and what i'm going to do is i'm going to test the point 4 comma negative 4. let's see if that's the solution to this equation so y in this case is negative 4 so negative 4 is equal to negative 1 4 times positive 4 minus 3. let's see if each side is equal to the Other so this would cancel with this leaving me with negative 1 minus 3 negative 1 minus 3 is negative 4. you get negative 4 equals negative 4. so yeah that

works out all right so let's try this other one now we have y equals 3x minus 3. so again it's in slope-intercept form and we know that the slope m is equal to 3. right this is our this is our slope and then we know that the y-intercept Occurs the y-intercept occurs at 0 comma negative 3. again super super easy to graph this so our y-intercept is the same as in the last problem it occurs at zero comma negative three so let's go there and then our slope was positive three so the rise over run

here remember three can be written as a fraction as three over one so what that's telling me is that I rise three and i go to the right one so i start here i go up one two three to the right one that's a point rise one two three over to the right one rise one two three over to the right one you could also think about this as negative three over negative one negative over negative is positive so these two are the same so i could start here and fall three one two three go

to the left one go to the left one and if i start from here and follow the Original one go up one two three to the right one i'm back on the line so let's graph this guy again this is y equals 3x minus 3. all right so the next thing we're going to talk about is called point slope form this is another thing you talk a lot about in class and it comes from the slope formula so you have that m the slope is equal to y minus y sub 1 over x minus x

sub 1. now there's a little bit of a difference From the slope formula that you're used to you're used to seeing y sub 2 minus y sub 1 over x sub 2 minus x sub 1. now let me explain why there's a difference so in a linear equation in two variables when you normally encounter it you don't have a given x or a y you don't know what those are you have something like 5x plus 2y equals 7. so these variables here represent unknowns when we use the Slope formula like this we have two points

that were given in most situations and some that's not the case but in most situations we're given two points and we plug the given values in right we plug in the y sub two we plug in the y sub one we plug in the x sub two we plug in the x sub one and we do some general arithmetic and we pop out a slope well with point-slope form what happens is I have a known point so one known point and i know the slope and the slope is known so this right here this x

and this y that don't have any other notation that represents the regular x and y that we're thinking of where we don't know what they are this x sub 1 and this y sub 1 that's what you're plugging in for with your values from the known point Okay so you're using point-slope form when you have one known point and the slope and you're going to see some examples where it's not exactly like this but in most cases you have a known point and you know the slope so what does point slope form look like well

we just take and multiply both sides Of the equation by this quantity x minus x sub 1. so i'll multiply this over here and what's going to happen is this is going to cancel with this right it's the same thing divided by itself so that's going to cancel out and it's going to be over here x minus x sub 1. remember to use parentheses so i can erase this at this point i can erase this at this point now it looks a little different in your Textbook in your textbook it's going to look like this

y minus y sub 1 is equal to m times the quantity x minus x sub 1. and you can write it like that it doesn't matter i can take this and move it over to the left and take this and move it over to the right i can put the m out in front really all they did was just rewrite it nothing is different right so you need to realize that immediately so that's where point-slope form comes from so once you memorize This kind of generic formula here you're basically ready to go so let's look

at an example so we have a point and we know the slope so we're going to use our point slope formula so we have through the point negative 4 comma 1 with a slope of negative 1. so if i have my point slope formula y minus y sub 1 is equal to m times the quantity x minus x sub 1. this known point we label that as x sub 1 y sub 1. So i'm just plugging in so i have y minus my y sub 1 is 1 equals my m is negative 1 times the

quantity x minus x sub 1 we know that's negative 4 minus the negative 4 is plus 4. and so once you have this this is point-slope form in every situation they're going to want you to solve it for y and put it in slope-intercept form if the book says Write it in point-slope form you've done that this is point slope form so you would be done right because you've mirrored this right here that's the same right i've plugged in for m i've plugged in for x sub 1 and i've plugged in for y sub 1.

so i've done my job but in most cases they want you to solve it for y and put it in slope intercept form so let's go ahead and do that So kind of continuing here i would have y minus 1 is equal to negative 1 times x is just negative x negative 1 times 4 is just minus 4. and i would add 1 to both sides of the equation so that would cancel and i would have y is equal to i'll have negative x negative 4 plus 1 is negative 3. so minus 3. and this

is slope intercept form this is slope Intercept form i know my slope it's negative 1 right if i have a negative out in front of x that's the same as negative 1 and my y-intercept occurs at 0 comma negative 3. so that's slope intercept form all right so what if we have one that's through the point 5 comma negative 1 and the slope is given to us as negative three-fifths so again i would label this as x sub 1 y Sub 1 and my generic point slope formula is y minus y sub 1 is equal

to m times the quantity x minus x sub 1. again especially when you first start out you don't know this by heart write it out so you don't make a simple mistake and then you just plug in it's very very easy so for y sub 1 i have negative 1. so i can just erase this and put minus a negative 1 is the same as plus 1. For m i have negative 3 fifths so negative three fifths and then we have x minus x sub one so for x sub one i have five so this

is point slope form again if your book asked for that you could stop there but in most situations they want you to solve for y and put it in slope intercept form so we would continue simplifying so y plus one equals negative three fifths times x is negative three fifths x And then negative three fifths times negative five you think about that separately over here this would cancel with this and i'd have 1. so basically negative 3 times negative 1 is positive 3. so positive 3 or plus 3 here and so now all i have

to do is subtract one away from each side so that y can be by itself so i would have y is equal to negative three fifths x Three minus one is two so plus two again this is in slope intercept form i know that my slope is negative three-fifths and we were told that in the beginning and my y-intercept occurs at zero comma two zero comma two all right so a little bit later on in this kind of section where you're talking about equations of a line they're going to throw kind of a wrench in

your plans so you'll see this Randomly and you'll be given two points and they'll say go ahead and write the equation of this line in point-slope form and also in slope-intercept form now if i'm just given two points how am i going to do that well remember to use point slope form i need one point one point and i need the slope so i have two points right i have more than i need in terms of points but i Don't know what the slope is so how would you suggest we get the slope remember if

i have two points i can generate the slope using the slope formula so this is something you kind of have to figure out because a lot of textbooks just leave it out they throw the problem in there and they want you to do some critical thinking so essentially what you're going to do is calculate your slope so m equals y sub 2 minus y sub 1 Over x sub 2 minus x sub 1 and then go back through again any one can be x sub 2 and this could be y sub 2 and then this

is going to be x sub 1 this will be y sub 1 and we just plug in so y sub 2 is negative 4. so negative 4 and y sub 1 is 3. x sub 2 is negative 4 also and then x sub 1 is 1. so negative four minus three is negative Seven negative four minus one is negative five so negative seven over negative five is seven fifths so i have my slope now so m is seven-fifths so once i figure that out i have all the information to write an equation in point-slope form

i have one point and i have a slope but some of you will get confused now because you have two points to choose from and you'll say well which point do I need to use for point-slope form well the answer is whichever one you want it doesn't matter and i'm going to do it using both points just to show you that so let's start out by using this one and saying this is x sub 1 y sub 1 and the point slope formula is y minus y sub 1 is equal to m times the quantity

x minus x sub 1. now i will plug in a negative 4 for y sub 1. I will plug in a negative 4 for x sub 1. so let's simplify this so we'll have y plus 4 right minus the negative four is plus four equals for m remember that's seven fifths times the quantity x minus a negative four so that's plus four all right so then we'd have y plus four is equal to seven fifths times x plus seven fifths times four four times seven is 28 so this would be 28 fifths So now continuing

i would subtract four away from each side of the equation and i'm going to go ahead and write 4 as 20 over 5. so i have a common denominator there that's gone and i'm going to have y is equal to 7 fifths x plus 28 minus 20 is eight over the denominator of five so y equals seven fifths x plus eight fifths so again this is slope intercept Form i have my slope which is seven seven-fifths i have my y-intercept that occurs at zero comma eight fifths very very easy now i wanna show you using

the other point i get the same answer in the end so now let's say i had chosen to use this as my point so this is x sub 1 y sub 1 so i'm just plugging in so for Y sub 1 i'm plugging in a 3 for x sub 1 i'm plugging in a 1. so let's crank this out so i'd have y minus y sub 1 is 3 equals m which we know is 7 5 times the quantity x minus 1. so at this point when we have it in point slope form it looks

a little different you might say oh well that's not the same but when you solve it for y you're going to get the same thing so you'd have y minus 3 is equal to 7 5 Times x is just seven fifths x minus seven fifths times one is just seven so fifths would add 3 to both sides of the equation i'm going to go ahead and write 3 as 15 over 5. so i have a common denominator so that's gone y is equal to 7 fifths x so we have negative seven plus fifteen What is

that that's eight right so plus eight over that common denominator of five so we get the exact same thing y equals seven fifths x plus eight fifths so that is using a different point right so as long as that point is on that line you're good to go you can use whatever point you'd like as long as you know the slope so if you have one point on the line and the slope you're going to use Point-slope formula if you have two points and you're not given a slope you're going to use point-slope formula just

in that particular case you got to do a little extra work right you got to take the two points generate a slope okay using the slope formula then go back and plug in using one of the points it doesn't matter which one so go back and plug into the point-slope Formula that you have and you can solve it for y and get it in slope intercept form all right so lastly we want to discuss something known as standard form so the definition for standard form i'm going to tell you this right off the bat is

going to vary between textbooks so i'm going to give you a generic definition and then i'm going to give you the one that you're probably going to get when you're in high school so for standard form Kind of starting out i would say in college algebra you'd say ax plus b y equals c where a b and c are real numbers and a and b are not both 0. so a b and c can be anything you want they can be negative they can be fractions they can be anything you want doesn't matter now there's

a stricter version of this which you're probably going to be yelling at the screen going that's not What my teacher said no i it's got to be injured so let me show you that one so the stricter version the one that you generally learn in an algebra 1 or algebra 2 class in high school will be this ax plus b y equals c where a b and c are integers a and b are not both 0 and they also want a to be zero or positive right it's greater than or equal to zero as another

condition for a so With this definition you generally have a little bit more work to do but it's all about manipulating the equation you can kind of do whatever you need to do to fulfill this definition so let's look at some quick examples here so again write the standard form of the equation of the line we're going to use that stricter definition here because most of you are in high school and you kind of need to Know how to manipulate the equation if you're watching this and you're in a higher level mathematics class and you

know it doesn't matter then you know you kind of just get to practice anyway on how to manipulate an equation so again we want a to be greater than or equal to zero that's the coefficient of x and we want a b and c to be integers We know that a and b can't both be zero and so it looks like this ax plus b y equals c so how would i get y equals negative 3x minus 6 to look like that well i'd start by adding 3x to both sides of the equation and i'd

have 3x plus y is equal to negative 6. so at this point i'm actually done i've satisfied all the requirements a and b are not both 0 none of them are 0. A b and c are all integers 3 is an integer this you can think of as 1 so that's an integer and negative 6 is an integer and a is greater than or equal to 0 a is the coefficient on x so we've satisfied our requirement very very easy all i had to do is add 3x to both sides of the equation that's kind

of an easier example what if i got something like this 0.2 y equals 0.8 x minus 5. again the format i want is ax Plus b y equals c a is greater than or equal to zero a b and c are all integers and a and b are not both zero so again the first thing i'm going to do is subtract 0.8 x from both sides of the equation so i'd have negative 0.8 x and then plus 0.2 y is equal to negative 5. now i have two problems here i have the Problem that this

is negative this is negative and i have the problem that it's a decimal so i can fix both of them relatively easily to convert this from a decimal to a whole number or to an integer i would need to multiply by the appropriate power of 10. so in each case we have one decimal place so all i would have to do is multiply both sides of the equation by 10. very Very easy so if i do that so 10 times negative 0.8 x essentially i'm just moving this one place to the right right the decimal

point so this would end up being negative 8x then plus 10 times 0.2 y again i'm just moving this one place to the right so this would be 2y this equals 10 times negative 5 is Negative 50. so i've solved the part where i want integers but a is not greater than or equal to zero right now i have a negative there so how can i fix that well i can multiply both sides of the equation by negative one so if i multiply by negative one negative 1 times negative 8x is 8x negative 1 times

2y is minus 2y and negative 1 times negative 50 is positive 50. So now this is in standard form in the stricter form of standard form where the coefficient of x is actually positive so we have 8x minus 2y equals 50. so a b and c are all integers a is greater than or equal to zero and a and b are not zero so all of the conditions are met now some of you are going to say hey you could have done that in one step i could have multiplied up here by negative 10 and

i would have got This right away well yes i could have done that but i wanted to make it a little easier for you to understand what we're doing so i did it in two steps but kind of moving forward if you're looking to save time that's one of the things you might consider all right let's take a look at another so we have y equals 1 5 x plus 5. so again i'm looking for ax plus b y equals c and again i want a to be greater than or equal to 0 i want

a b And c to be integers and i want a and b to not both be 0. i'm going to start out by subtracting 1 5 x away from both sides of the equation so i'd have y minus 1 5 x is equal to 5. right i just subtracted 1 5 x away from here that would make it cancel i subtract away from here i have it over here in this particular case let's think about this let's think about Doing it in one step i have the fact that this is negative right i have a

negative coefficient for x i don't want that right a has to specifically be greater than or equal to zero and i have a fraction right so i don't want that either i want an integer so what can i multiply one-fifth by to make it a whole number i can multiply by five right if i multiply 5 by 1 5 i get 1. i multiply a number by its reciprocal you get 1. Additionally if i threw in the fact that i multiplied by negative 5 i would change the sign here because right now we have a

negative one-fifth so if i multiply negative one-fifth by negative five i solve both my problems there again to make that legal though i've got to do it by both sides of the equation so multiply over here by negative five and also over here by negative 5 negative 5 times y is negative 5y Negative 5 times negative 1 5x is going to be plus x you could put plus 1x if you want it doesn't really matter and this is going to equal 5 times negative 5 which is negative 25. now you might say well this doesn't

look like that all you have to do is switch the order here remember addition is commutative so i could rewrite this as x minus 5y is equal to negative 25 And i've written this in standard form using the stricter definition right my a term here is one so it's greater than or equal to zero my b term is negative five my c term is negative 25. so a b and c are integers a is greater than or equal to zero and a and b are not both zero right neither of them are zero hello and

welcome to algebra 1 lesson 18. in this video we're going to learn about parallel and perpendicular lines so parallel lines Are any two lines on a plane that will never intersect we can determine if two lines are parallel by examining the slope of each so non-vertical parallel lines have slopes that are equal let me highlight that non-vertical parallel lines have slopes that are equal okay very important that you write that down now vertical lines remember they have slopes that are undefined so that's why we specify and say that Non-vertical parallel lines have slopes that are

equal but vertical lines have slopes that are undefined now when you first start looking at parallel lines what you're going to notice is that there are lines that kind of look the same so parallel lines look like this let's say i had a line that goes this way and another line that looked like that so i know these aren't perfect but you can see by looking at these lines that They're never going to intersect and if we're on a coordinate plane you'd also see that they have different y-intercepts so essentially you're looking for same slope

different y-intercepts those are going to be parallel lines so then the other thing we want to talk about are perpendicular lines and perpendicular lines are lines that intersect at a 90 degree angle okay they intersect at a 90 degree angle So two non-vertical perpendicular lines have slopes whose product okay whose product is negative one so specifically if i take two non-vertical perpendicular lines and i grab their slopes i multiply them together the result is always going to be negative 1. so we're going to look at a few examples here not really a difficult concept Whatsoever

so we want to determine if each pair of lines are parallel perpendicular or neither okay so let's start out with x plus 2y equals 4 and then 3x plus 6y equals 48. so when you get problems like this for your homework or for a test the first thing you want to do is find the slope of each line now we found out in the last lesson that an easy way to do that was to put your Line in what's called slope intercept and basically that means you're going to put your line in the format of

y is equal to m the slope times x plus b the y-intercept right this is the slope and this is the y-intercept okay that's why it's called slope-intercept form so essentially what you have to do is Just take each equation and solve it for y that's all you need to do so i'll start out with this x plus 2y is equal to 4 and i'm going to solve this for y so in order to do that i'm going to subtract x away from each side of the equation so that's going to go away from over

here and then on the left i'm just left with 2y and this is equal to i now have negative x on the right and then plus 4. i want y by itself and what's stopping me is i have this 2 that's multiplying y so i need to divide by 2 on this side of the equation and to make it legal i need to do it to this side of the equation as well well remember we're allowed to break this up And divide each by two all right so then this would cancel with this leaving me

with just y over here over here i have negative x over two remember i can always put a 1 here and say that's negative 1 times x right it's kind of an invisible coefficient if you just have a negative out in front or nothing out in front you could say hey if this is a negative out in front it's really negative 1. if there's nothing out in front it's just a one right so i Can think of this as negative one half times x plus four over two is just two so there is my first

equation in slope intercept form so what we can tell is that our slope which we denote as m is going to be negative one-half and our y-intercept our y-intercept occurs at the point zero comma two okay let's erase everything and go back up to the top i'm just going to keep this Information right here so for this first one again in slope-intercept form we end up with y equals negative one half x plus two and again we said that this corresponds to a slope which is m equal to negative one half and a y-intercept a

y-intercept that occurs at the point 0 comma 2. All right let's take a look at this equation now and again all we're going to do is solve it for y so we have 3x plus 6y equals 48. okay so if i want y by itself i've got to first isolate the term 6y so i would subtract 3x away from both sides of the equation and this is going to cancel on the left now i just have 6y this is equal to negative 3x plus 48. in order to get y by itself i need to Divide

both sides of the equation by 6 which is the coefficient of y so i'm going to divide this by 6 and also this by 6. all right so this is going to cancel with this and negative 3 over 6 is the same thing as negative 1 half so i'll cancel this with this i'll put a one here and a two here and then 48 divided by six we know that's going to be eight So i'll kind of cancel this with this and just put an eight and a one so we're going to rewrite this as

y equals negative one half x plus eight what you can immediately see is that the slope again which is m is the same as in the other equation it's negative one-half the y-intercept however is different so the y-intercept is going to occur at 0 comma 8. so this is your y-intercept It occurs at 0 comma 8 right that point on the graph okay so let's erase this and go back up okay so for this line again in slope-intercept form it's y equals negative one-half x plus eight okay and again m in this situation is also

negative one-half and then the y-intercept the y-intercept let me kind of erase that real quick [Music] is going to occur at zero comma eight So what we can immediately tell is that we have the same slope in each case right m equals negative one-half m equals negative one-half so same slope different y-intercepts different y-intercepts we know that we have parallel lines so these lines are parallel so one thing i want to show you really quickly is what this visually looks like using a coordinate plane So remember we had y equals negative one half x plus

two and we had y equals negative one-half x plus eight in slope-intercept form these are our two lines now if we know a point on the line and the slope we can easily graph right remember we've kind of transcended the point where we have to go through and make ordered pairs and do all this stuff we can graph really quickly now So i know for this first one the y-intercept occurs at zero comma two so on the coordinate plane that's going to be right here and then my slope is negative one half so remember it's

rise over run it's rise over run so this equals negative one over two or you could put one over negative 2. it doesn't really matter how you kind of think about that so if my rise is negative 1 that means i fall 1 and if my Run is 2 positive 2 that means i go over to the right 2. so 1 2. so there's my next point down one over one two there's my next point down one over one two there's my next point down one over one two there's my next point or like i

said i could have also done one over negative two so if i started here i would rise 1 and if i have negative 2 in the Denominator that means i go to the left 2. so i go 1 2 to the left up 1 to the left 1 2. and obviously you don't need all these points to make a line but i like to kind of do that when we're in our first stages just to kind of get our brain used to graphing a line this way okay all right for the next one here we

have y equals negative one half x plus eight And again we know a point on this line it's at zero comma eight so that's going to be right here and then to graph it again we just use that slope negative one half so i can fall one and go to the right two fall one go to the right two ball one go to the right two fall one go to the right two and i'll go ahead and stop there or another thing i can do again i could use this i could go up one into

the left Two so let's go ahead and graph this some arrows in and let me kind of erase this part real quick all right so it's not perfect and no graph that you ever freehand draw will be i mean some of them are pretty good but you know if you were to put these equations in on you know a computer graphing program you could really see but even just looking at this simple Hand drawing here you can tell that these lines pretty much look the same right they have the same steepness or the same slope

what's really different about them is that they are on different parts of your coordinate plane right they have different y-intercepts this one has a y-intercept that's at zero comma two this one has a y-intercept that it's at zero comma eight but if you were to extend these lines Out indefinitely in each direction because they're parallel to each other they will never ever ever touch okay so that's what we're trying to get across here so again two non-vertical parallel lines will look like this they have the same slope but different y-intercepts all right let's take a

look at another one we have 4x minus 3y equals negative 12 and we have 6x plus 8y equals negative 16. Okay so again i'm going to solve each for y i'm going to do this a little quicker this time so we'll move this over here and say okay i have 4x minus 3y equals negative 12. subtract 4x away from each side of the equation so that's gone i'd have a negative 3y is equal to negative 4x minus 12. all right so the next thing i want to do is divide both sides of the equation by

negative 3 The coefficient of y so that is going to cancel with that and i'll have y by itself and this is equal to negative four over negative three is just four over three or four thirds right the negatives would cancel times x and then the negatives would cancel again so plus 12 over three is four so i know for this one that the slope which is m Is equal to 4 3 and that the y intercept the y intercept is going to occur at 0 comma 4. all right for the next one again i'm

trying to solve this for y so 6x plus 8y equals negative 16. i'm going to subtract 6x away from each side of the equation so this is going to cancel and i'm going to have 8y is equal to negative 6x minus 16. Divide each side by 8. so i'm going to cancel this 8 with this 8 and i'm going to have y is equal to so 6 divided by 2 would be 3. 8 divided by 2 would be 4. so this would be negative 3 4 times x and then minus 16 over eight is two

so my slope here is negative three fourths my y intercept occurs at zero comma negative two so m is equal to negative three fourths And again y intercept is at 0 comma negative 2. so kind of the first question you're going to ask yourself here is are these lines parallel well you see that the slopes are clearly different numbers right this is four-thirds this is negative three-fourths if they're different values you don't have parallel lines now the next thing you would ask yourself is are these lines Perpendicular well perpendicular lines have a property that tells

us that the product of the slopes would be negative 1. so in other words if i took four thirds which is the slope for this equation here and i multiplied it by negative 3 4 which is the slope for this equation here i should get negative 1 if these two lines are perpendicular so you can see that you would right Because this 4 would cancel with this four four over four is one this three would cancel with this three three over three is one but i'm left with this negative here right so i'd end up

with one times negative one or just really everything canceling except for negative one and that would be my solution here right the product of the two slopes would in fact be negative one so we do have perpendicular lines [Music] and again to kind of look at this on a coordinate plane let's take a look at our two equations in slope intercept form again we had y was equal to negative three fourths x minus two and we also had y was equal to four thirds x plus four so again really really easy to graph these We

know the slope and we know the y-intercept so we have a point on the line in the slope so for the first one the y-intercept's going to occur at 0 comma negative 2. that's right there my slope is negative 3 4. so again with rise over run rise over run of negative three fourths or again you can think about this as three over negative four doesn't matter In the first scenario i would fall three i would fall one two three and i would run to the right 1 2 3 4. and i can make another

point but that's going to kind of go too far so let me let me kind of backtrack over here and let me use this now let me rise three one two three let me go to the left four one two three four so that's another point and let me graph this real quick okay And now for the next one let me just erase this we have y equals four thirds x plus four so the y-intercept will occur at zero comma four and that's going to be right here and the slope is four thirds so again

with rise over run i can go up four so i can go up one two three four and to the right one two three And the other thing i can do remember negative over negative is also positive so i can do negative four over negative three so i can fall four i can fall one two three four and go to the left three one two three so that's right there and let me do one more point fall four one two three four to the left three one two three okay so let me graph this real

quick So one thing i want to draw your attention to if you look at that point of intersection you create a 90 degree angle there okay this is a 90 degree angle and you can really look at any of these and draw that so this is also a 90 degree angle this is also a 90 degree angle and this is also a 90 degree angle so it's all the way around let me make that a little better so again if you have two non-vertical Lines and you multiply the slopes together and you get negative one

as a result those two lines are perpendicular you'll have a situation just like this one all right let's take a look at another one we have x plus 5y equals negative 10 and we have x minus y equals negative 5. so again i'm going to solve each one for y so i'm going to start out with this one so we have x plus 5 y equals negative 10. I'm going to subtract x away from each side and i'll have 5y is equal to negative x minus 10. i will divide each side of the equation by

5. let me scroll down a little bit so this is going to cancel and what are we going to have we're going to have y is equal to we have negative x over 5. remember i can write a 1 there so it's the same as negative 1 5th negative one fifth Times x and then minus ten over five we could think of as minus two right or plus negative two because ten over five is two so then i have my equation in slope intercept form so i know the slope is negative one fifth the y-intercept

will occur at zero comma negative two let me erase all this so then m is equal to negative one-fifth and again the y-intercept the Y-intercept will occur at the point 0 comma negative 2. all right for this one pretty easy to solve for y we have x minus y equals negative 5. i just subtract x away from each side so that's gone i'll have negative y is equal to negative x minus 5. i just have a negative out in front of y that's keeping y from being by itself so i can think of this as

Negative 1 times y so in order to change the sign of that from negative to positive i can divide by negative 1 or i can multiply by negative 1. but i've got to do it to both sides of the equation so let's multiply both sides of the equation by negative 1. and so negative 1 times negative 1 is 1. so i'd have 1y or just y this is equal to negative 1 times negative x is x And then negative 1 times negative 5 is plus 5. so we get y equals x plus 5. so my

slope here is going to be 1 right because if i just have x by itself it's got an invisible 1 next to it so in this particular case m is equal to 1 and the y-intercept occurs at 0 comma 5. so what can we say about these two lines here are they parallel well the slopes are different I have a slope here that's negative one-fifth i have a slope here that's one that's not the same so they're not parallel are they perpendicular no you can tell that right away because if i multiply negative one-fifth times

one anything times one is just itself so negative five times 1 is just negative 1 5 right so these lines are also not going to be perpendicular so in this case we would write neither Right we're going to say neither all right so we want to write the standard form of the equation of the line described all right so as you get further into this section on parallel and perpendicular lines you're going to see a few problems like this where you're given a point on the line and you're given an equation that it's parallel to

or perpendicular to and you're expected to be able to write the Equation of the line and they'll tell you which form they want it and they'll say okay we want it in slope intercept form standard form whatever they want right so in this case we want to write the standard form of the equation of the line described so we're given this information here it's through the point 4 comma 5 and it's parallel to this equation here negative 6x plus 3y Equals 15. so remember if i have a slope and i have a point on the

line i can use my point slope form okay but before we get to that we don't know what the slope is until we figure out a little information with this equation here right we don't know the slope slope equals what so remember if two lines are parallel Okay they have the same slope so you're meant to look at this type of problem and say okay well if i solve this for y i can figure out the slope and then i know my slope and i know my point so i can use point slope form so

let's start out by just putting this in slope intercept form so negative 6x plus 3y equals 15. add 6x to both sides of the equation that's gone 3y equals 6x plus 15. Divide this by 3 divide this by three and this by three so i get y equals six over three is two so two x plus fifteen over three is five so y equals two x plus five remember in slope-intercept form y equals mx plus b m is your slope so 2 here is the slope so the slope is 2. now this is very important

it trips a lot of students up if you're a little confused About this topic you're not doing anything with this equation here i can line it out once i know what the slope is i don't need it it's irrelevant to me just get rid of it because it's going to confuse you you only need it to figure out what the slope is going to be in the equation that you're asked to write i now know what the slope is because it was parallel to this equation i took the equation figured out what the slope was

I don't need it anymore so i have my slope and i have my point and again when we have a slope and a point on the line we use point slope form which is what it's y minus y sub 1 is equal to m the slope times the quantity x minus x sub 1 okay where this y sub 1 and the x sub 1 are the given values from that point so this is x sub 1 this is y sub 1. and now we just plug in right we'd have Y minus y sub 1 which

is 5 is equal to m which is 2 right that's your slope times the quantity x minus x sub 1 x sub 1 is 4. now if they're asking for point slope form you're done but they didn't they actually asked for standard form so what we're going to do is we're going to solve this for y first so i'm going to have y minus 5 equals 2 times x is 2x And then 2 times negative 4 is negative 8. add 5 to both sides of the equation i'll get y equals 2x minus 3. now this

still isn't standard form again this is slope intercept form if it's solved for y this is slope intercept form very important you understand the difference between those Now for standard form again you want it to look like this ax plus b y equals c and i know everybody has a different definition for that but typically in high school you see it written like this and you'll say that a has to be greater than or equal to zero so it's either zero or some positive value and then a b and c need to be integers and

a and b can't both be zero at the same time so in order to make that happen if i Look at this y equals 2x minus 3 i know i want to move this over here so i would subtract it away from both sides so i'd have negative 2x plus y equals negative 3. but according to my definition that i'm using a has to be greater than or equal to 0. so to kind of clean this up i just don't want that to be negative so i can just multiply both sides of the equation by

negative 1. very very simple Fix for that so if i multiply this side by negative 1 and this side by negative 1 i'll end up with negative 1 times negative 2 that's 2 x and then negative 1 times y would be minus y and this equals negative 3 times negative 1 which is 3. so i get 2x minus y equals 3. so that would be the equation Of the line in standard form that passes through the point 4 comma 5 and is parallel to negative 6x plus 3y equals 15. all right we're going to take

a look at one more so this one passes through the point negative 4 comma negative 3 and it's now perpendicular to 12x plus 3y equals 15. so again i'm going to solve this for y so let me start out by just saying okay we have 12x plus 3y Equals 15. i want to subtract 12x away from each side that's going to go and i'll have 3y is equal to negative 12x plus 15 and we're going to divide each side of the equation by 3. and what are we going to have this will cancel i'll have

y is equal to negative 12 over 3 is negative 4 then times x and then plus 15 over 3 that's plus 5. All right so i'd have y equals negative 4x plus 5. now in the last example where we had parallel lines it was easy because the slopes would just be the same the slope for this line as we can see because it's in slope intercept form m is equal to negative 4. now with perpendicular lines it's a little bit more complex but really it's still kind of easy Perpendicular lines have a property that says

that the product of their slopes will be negative one so that means negative four times some unknown slope unknown slope right for the line we're trying to get to is going to be equal to negative 1. so i can use a variable here right i can always use a variable when i have something unknown so let's use z Now what i would do is if i had negative 4 times z equals negative 1 i would divide both sides of the equation right you kind of think about it this way as by negative 4. this would

cancel and z would equal negative 1 over negative 4 which is 1 4. so the slope i'm looking for is 1 4 because if i take 1 4 and i multiply it by negative 4 this would cancel with this i'd have 1 times negative 1 which would be negative 1. that's exactly the property that we know from perpendicular lines so now i know the slope of the line that i'm trying to put in standard form again it's 1 4. so my m is 1 4 and my point on the line is negative four comma negative

three now we are done with this equation here we don't need it anymore line it out forget it exist okay you've gotten the information you need from it and just again forget about it Now i'm going to use my point slope form point slope form again this is my x sub 1 this is my y sub 1 and for point slope form again it's y minus y sub 1 equals m times the quantity x minus x sub 1. i'm going to plug in i'm going to plug in a negative 3 right Here i'm going to

plug in a negative 4 right here so i'll have y minus a negative 3 so that's y plus 3 is equal to and i forgot to say that i'm going to plug in a 1 4 for m so is equal to 1 4 times this quantity x minus and we're subtracting away negative 4 so i'm going to put plus 4. all right so i'm going to solve this for y right now again this is point slope Form if it asks for that we'd stop there solving it for y is pretty easy we just have y

plus 3 is equal to 1 4 times x is just 1 4 x so you put x over 4 and then plus 1 4 times 4 is 1. so then i'd subtract 3 away from each side i'll go ahead and just do this like this y equals x over 4 plus and basically it's going to be negative 2 right you can put minus 2. So now i have the equation in slope intercept form y equals and you could write it as x over 4 or you could write 1 4 times x minus 2. it doesn't

matter and i want to put it in standard form and again let me kind of cover that definition again standard form looks like this ax plus b y equals c again where a b and c are integers a and b are not both 0 and a is greater than Or equal to 0. now what i want to do again i have y equals 1 4 x minus 2. i've kind of got two steps that i need to do i need to first start by subtracting this away from both sides of the equation so if i

do that what's going to happen is i'm going to have negative 1 4 x plus y is equal to negative 2. so once i've Done that again we want a to be greater than or equal to 0 and we also want a b and c to be integers so one of the things i can do is i can multiply by a negative and that takes care of this being negative right it'll make it positive but another thing i want to do since i want integers i want to clear this denominator of 4. so to do

that i would just multiply by four and if i'm multiplying by negative one and i'm Multiplying by four i can combine those and just really multiply by negative four right so i would multiply both sides of this equation by negative four and so negative 4 times negative 1 4 the negatives would cancel and 4 over 4 is 1 so i basically just have x and then plus i have negative 4 times y that's plus negative 4y or i can just write minus 4y and this equals negative 2 times Negative 4 which is positive 8. so

i have x minus 4y equals 8 and i've fit the definition of standard form i have ax plus b y equals c where a which in this case is one is greater than or equal to zero a b and c are all integers so i have a one a negative four and eight that's perfectly fine and then a and b are not both 0. neither Are 0 in this case so we're good to go so we have x minus 4y equals 8. or again you could say 1x minus 4y equals 8 as our equation in

standard form hello and welcome to algebra 1 lesson 19. in this video we're going to learn about graphing linear inequalities in two variables so a linear inequality in two variables is of the form ax plus b y is greater than c where our greater than here can also be a less than a less than or equal to or It can be a greater than or equal to now some specific things about this form we have a a b and a c a is a coefficient of x b is the coefficient of y and c which

is a constant so a b and c are real numbers and a and b cannot both be zero meaning a could be zero or b could be 0 but they can't both be 0 at the same time all right so graphing a linear inequality in two variables is very similar to graphing a linear equation in two Variables if you can graph a linear equation and two variables at this point which you should be able to then graphing a linear inequality in two variables is going to be no more difficult there's really just one additional step

so the first thing that you want to do is draw something known as a boundary line so this is a very important concept for this topic and we talked about it a Little bit when we talked about checking our solution for a linear inequality in one variable but the boundary line basically separates the solution region from the non-solution region so in other words if i have a line let's say it looks like this and it's on my coordinate plane well on one side of it the ordered pairs or the points would work let's say this

is my solution region So if i take values from over here ordered pairs or you could say points and i plug them into the original inequality i get a true statement then over here let's say this is the non-solution region the non-solution region so ordered pairs or points over here are not going to work so this line is what's going to separate the two this is called again the boundary line the boundary Line so we obtain our boundary line by replacing the inequality symbol with an equal sign and graphing the line so essentially all you're

doing is you're starting out by taking your linear inequality and two variables and transforming it into a linear equation two variables and just graphing it but there's one thing you really need to know before you do that so there's two scenarios So a strict inequality so strictly less than or strictly greater than that gives us a dashed or broken boundary line and why do you think that is well if you have a strict inequality the boundary line is not part of your solution and i want you to recall at this point when we talked about

a linear inequality in one variable let's say i saw something like x is less than Three this was strict so we had notation when we graph this on the number line so let's say this is 0 this is 1 two three four let's say over here this is negative one negative two negative three negative four So when we graphed x is less than three we found three and we said 3 is not included so what we did was we put a parenthesis there facing to the left i know some of you put an open circle

but it's the same thing and then we shaded everything to the left so this is very similar when we use a dashed or broken boundary line it's kind of like the parenthesis was here we're saying it's not included as part of the solution so then kind of the next Scenario would be a non-strict inequality and for this one you're going to have a solid boundary line okay a solid boundary line and the reason for that is it's included so let's say i had something like x is less than or now equal to three instead of

the parenthesis i had a bracket right so it's similar to that because the bracket told me that 3 was included so that's what we're doing here with the solid battery line that Line is included as part of the solution all right so let's start out with 2x plus y is less than or equal to 1. so i'm going to do this the slow way to start i'm going to put an equal sign here i'm going to put 2x plus y equals 1. i'm going to graph it the slow way so i'm going to make some

ordered pairs and i can see immediately if i put a 0 in for x y would be 1. that's the y-intercept so that's easy And then to get some other points going i can't put a 0 in for y because if i do i'm not going to end up with an integer right plug in a 0 for y you'd end up with x equals one-half so that's not something that i want to deal with on my coordinate plane so let's try if x was 1. so 2 times 1 plus y equals 1. so this would

be 2 plus y equals 1. and so subtract 2 from both sides of the equation that's gone i'd have y is equal To negative 1. so if x is 1 y is negative 1. let's try a value of x equals 2. so if x is 2 this would be 4 plus y equals 1. subtract 4 from each side of the equation that's gone y is equal to negative 3. all right so i have three ordered pairs let's plot those ordered pairs and let's draw a line this line is going to be our boundary line now

What do we want we have a non-strict inequality this is non-strict so if we have a non-strict inequality the boundary line is part of the solution so we want a solid line so let's go draw a solid line so i want the point zero comma one so that's right here i want the point one comma negative one that's going to be right here and i want the point two comma negative three So two over to the right three down that's right there and again we're going to draw a solid line okay so once you've graphed

the line go back up to the top now this is your one additional step you're going to read in your book something about a test point a test point Okay and the test point works like this you can test any point on either side of the line if it works that means that side of the line is the solution region and therefore you're going to shade that side of the line if it doesn't work you're in the non-solution region and you're going to shade the other side of the line so let's go back down here

We look at we're not going to choose any point on the line because we already know that that's going to work as a solution the easiest point to choose if it's not on your line is 0 0 right because zero is easy to work with so let's go back up and we're gonna plug in a zero for x and a zero for y so my test point is zero comma zero so i would have two times zero plus zero is less than or equal to one So two times zero is zero plus zero is zero

so zero is less than or equal to one that's true so what that tells me is that this point right here zero comma zero is on the solution region so that means that the solution region is below the line so in other words i would shade everything below the line and it's hard to do a good job of Shading so i just kind of do it this way i know there's some computer programs that will shade this perfectly but you're doing this freehand this is about as good as we're gonna do okay so in other

words this is the solution region and over here anything kind of above this line is the non-solution non-solution region And just try it out for yourself pick some points over here pick 6 comma 0 for example that is in the non-solution region so it shouldn't work if i plug it in i should get a false statement so if i plug in a 6 for x and a 0 for y should get a false statement 2 times 6 is 12 12 plus 0 is not going to be less than or equal to 1 right 12 is

not less than or equal to 1. that's false so that's kind of the first method you Can use to do that now let me show you the fast way so some books skip this all together and that should kind of make you a little mad because this is way way way faster so alternatively we can solve the inequality for y draw the boundary line same way just replace the inequality symbol with an equal symbol and then you shade above the line for greater than or greater than or equal to You shade below the line for

less than or less than or equal to so in the previous example i could have started out i didn't need to get any ordered pairs i could have just solved this guy for y forget about this test point don't need that anymore so solve it for y we have y is less than or equal to negative 2x plus 1. now for the boundary line i would replace this with equals so i'd have y equals negative 2x plus 1. I know this is in slope-intercept form so it's super easy to graph it the y-intercept occurs at

zero comma one the slope is negative two so you can see i already have that point here zero comma one and my slope is negative two so i go down one two over one down one two over one so i would graph the same line so i'd come about that a little bit quicker now the next thing is the fact that it's less than or equal to so if it's a less Than whether it's a less than or equal to or strictly a less than you shade below the line and that's what we did here

so we would just shade below the line okay below the line and once you get into the habit of doing it this way it's much much much quicker you don't need to worry about getting a test point and all this other stuff and going through and getting ordered pairs it's way quicker so you're basically putting it into slope intercept form to graph it Once you've done that you look at the inequality that you're dealing with if it's a less than or it's a less than or equal to it's below the line that you're going to

shade if it's a greater than or greater than or equal to you just shade above the line so for the next one we look at 7x minus 4y is greater than negative 8. so we're going to do this the quick way let's just rewrite this real fast we have 7x Minus 4y is greater than negative 8. solve it for y so i'd subtract 7x from both sides of the inequality i'd get negative 4y is greater than negative 7x minus 8. then i would divide both sides of the inequality by negative 4. but remember if i

divide by a negative i've got to flip the inequality so instead of this being a greater than inequality i'm going to have a less than right so that symbol there is going to Flip from greater than to less than so this is y is less than negative 7 divided by negative 4 is 7 4 then times x negative 8 divided by negative 4 is plus 2. so if i replace this less than with an equals i'd have y equals 7 4 x plus 2. so this is my boundary line and you'll notice that this inequality

is a strictly less than okay so i want to make sure that i have a Broken or dashed line so y equals seven fourths x plus two that's what i'm going to graph so my y intercept occurs at zero comma two it's right there and my rise over run is 7 over 4 so i go up 7 1 2 3 4 5 6 7 to the right 4 1 2 3 4. and i can do that using negative 7 over negative 4 as well negative over negative is positive so i can fall 7 1 2

3 4 5 6 7 and to the left 4 1 2 3 4. so i start out by just drawing a solid line and then i take my eraser i just kind of break it up so just break it up good enough to where your teacher or whoever's grading your work and see if that line is broken or dashed okay so this is a an example of a dashed line let me kind of break that up a little bit too So that tells me that an ordered pair that exists on that line is not going

to be involved in the solution so for example the point 4 comma 9 will not be a solution and we'll test that in a minute let me just write that we're going to test 4 comma 9. before we do that remember this is a strictly less than so that tells me i want to shade below the line So when i shade this i'm going to shade below the line so all this that's all you need to do is give your teacher a representation of where the solution region is going to be just shade all of

this okay so let's test a few things so let's test 4 comma 9 in the original inequality i'm going to erase all of this i don't need it we'll show you this doesn't work so it lies on that boundary Line so 7 times 4 minus 4 times 9 is greater than negative 8. 7 times 4 is 28 minus 4 times 9 is 36 this is greater than negative 8. so 28 minus 36 is negative 8 that's not greater than negative 8. but remember if i had a greater than or equal to here this would end

up being true so that's the difference between having a strict inequality and a non-strict inequality again If you have a strict inequality like we have here we do not allow okay we do not allow for anything on that boundary line to be part of the solution for that reason that i just showed you so we break it up to say hey nothing on this line works with a non-strict inequality the things on the line do work as solutions so you got to draw a solid line for your boundary line when you Have a non-strict inequality

and a broken or dashed line for a strict inequality now one more thing i want to just bring your attention to if we use the test point here we could still see that we have the right answer you could pick something in the solution region again zero comma zero the origin is very very easy to use so that's something that should work So let's go back up and test it let's see if zero comma zero works as it should plug in a zero here and here zero minus zero which is zero should be greater than

negative eight that is true okay so zero comma zero should be in the solution region and it is right because it does work so you can kind of use that as a little check to show that you graphed the correct region all right let's take a look at a Few special case scenarios so something like y is greater than or equal to 2 well we saw that if we had something like y is equal to 2 this is a horizontal line right so we would find 2 on the y axis and we would draw a

horizontal line we're essentially going to do the same thing for something like this so y is greater than or equal to 2 it's a non-strict inequality so my boundary line will be a solid line i'm Going to find 2 on my y-axis so that's right here okay so this would be the line y equals two again that's our boundary line in this case now because y was a greater than or equal to i'm just going to shade above this line because y is anything that is greater than or equal to 2. again no matter what

the x value is it's going to work because the y value that's Given will be larger or equal to 2. so if i pick something like i don't know let's say an x value of 5 and a y value of 10. well again i could represent this by saying i have 0x plus y is greater than or equal to 2. so whatever i plug in for x just goes away so i plug in the value for y i plug in a 10 10 is greater than or equal to 2. Yeah 10 is greater than 2

so this is true right so you can see how that's a very easy scenario to deal with all right so another easy one would be something like x is greater than 4. so we saw that x equals 4 is a vertical line so i would find 4 on the x-axis that's right here and i would draw a vertical line but here's the key Because we have a strict inequality this is strict i can't include that line as part of the solution so i've got to kind of break it up or make it dashed [Music] and

now i'm going to take my eraser just take chunks out of it okay so that's a broken up or dashed line again just do enough to where your teacher can tell what you're doing and Then x is greater than four so greater than in terms of the x-axis means you're going to the right remember if you're increasing on the x-axis you're going to the right if you're decreasing you're going to the left so if it's a greater than you're shading to the right so i want to shade in this direction now because again any x

value in this direction is greater than the value of 4. if i pick an x value of 6 doesn't Matter what y is that value for x is greater than 4. so anything to the right of 4 is going to work that's why we shade all this over to the right hello and welcome to algebra 1 lesson 20. in this video we're going to have an introduction to functions so for most of you in algebra 1 you've never seen or heard about functions before you can tell you that it's a topic that does confuse a

lot of students but you really don't focus a lot on it In algebra one you kind of briefly talk about it in your chapter on linear equations in two variables and then in most cases you won't see it again until algebra 2. so if you don't get it here don't be too worried you're going to have a lot of opportunities in the future to kind of get really good at working with functions all right so let's start out with a basic scenario so i want you to suppose you go to the store and it cost

two Dollars per one ounce of granola so this means for two ounces the cost is four dollars for three six dollars and so on to mathematically explain this relationship between the number of ounces of granola purchase and the cost we could set up a little equation right we could set up a linear equation in two variables so let's let y be equal to the total cost in dollars And we're going to let x be equal to the number of ounces of granola that's purchased so how would we take these two variables and set up an

equation that describes our situation well i know that two or two dollars specifically i'm going to leave off the dollar symbol i know that two is the cost per ounce of granola So this is the cost per ounce and i know that x is representing the number of ounces of granola purchased so if i multiply the two together if i take 2 the cost per ounce and multiply by x the number of ounces of granola purchased that's going to give me the total cost which is y right this is the total cost in dollars So

there's an equation to represent the scenario that i just gave you now the ordered pairs that satisfy this equation so for example one ordered pair would be zero comma zero if i plug in a zero for x meaning i don't buy any ounces of granola i would get a zero for y i didn't buy any ounces of granola i don't spend any money another one would be one comma two I buy one ounce of granola i spend two dollars another one would be two comma four i buy two ounces of granola i spend four dollars

right so on and so forth so i can kind of put a comma here and three dots because this pattern would continue forever after two for an x value i would get three and i would just double it to get my answer three times two is six Then four times two is eight then 5 times 2 is 10. so on and so forth if i plugged in 27 for x i just multiply it by 2 and get 54 my y value one thing that you might not have noticed here is that there's a restriction on

x and a lot of times when we work with these real world problems you have restrictions why is there a restriction on x can you think about what x can't be equal to well x specifically can't be negative it Can be zero or it can be positive so x is greater than or equal to zero here specifically because i can't walk into a grocery store and say hey i'd like to buy negative 18 ounces of granola right maybe that's you selling granola or something like that but you know we're not going to get into any

complicated scenarios or thought processes like that we're just going to say that hey we can only buy zero ounces of granola or some Positive amount so there's a restriction here there's a restriction and we're going to talk a lot about this moving forward where we're limiting a variable to certain values in this case we're going to say hey it can only be 0 or some positive value so the ordered pairs here that satisfy this equation have a special name when we're talking about functions they're known as a relation they're known as a relation This is

definitely a definition you want to write down it's something you're going to talk about when you study functions so a relation is just any set any set of ordered pairs and we haven't really talked about what a set is either and we'll talk a lot about sets in algebra two but for right now just understand that a set is a collection of things okay and when we Work with sets we put the elements of a set and that's what's contained in a set inside of what's known as set braces so for example here with these

ordered pairs i'll put them inside of set braces okay so this would be the set of ordered pairs that satisfy this equation and i can't possibly list all of the ordered pairs so again i have the three dots to say that the pattern here where i Increase x by one and y increases by two continues forever and ever and ever now one thing i want to do really quickly is just graph this equation y equals 2x where again x is greater than or equal to zero so i know the y-intercept occurs at zero comma zero

right it occurs at the origin this is in slope intercept form i could put plus 0 behind this if i wanted to or you could just realize again that's a Line that goes through the origin now i can't have any points that correspond with an x value that's to the left of 0. so i'm only going to plot points up here because again x has to be greater than or equal to 0. so if i use my slope i'd go up 2 i go up 1 2 and over 1. up 2 and over 1 up

2 and over 1. so let's graph this guy so this would be y equals 2x with x is Greater than or equal to 0. again notice how we don't have anything down here like we normally would and this shows the relationship between again the x-axis is telling us how many ounces of granola that we're purchasing the y-axis is telling us how much it costs in dollars so in other words if i buy one ounce of granola it's gonna cost me two dollars if i buy five ounces of granola it's gonna cost me 10 right so

on and so forth all right so Let's talk a little bit more about some basic vocabulary so we already know that a relation is any set of ordered pairs and the ordered pairs are again x comma y so in any ordered pair x and y are known as the components the components of the ordered pair something that you want to write down because again when you start talking about functions you get a lot of these new kind of definitions that are going To pop up it might be put on your test so just write down

again in any ordered pair x and y are known as the components of the ordered pair now highly important here the set or the list of all first components so basically they're saying x values in the ordered pairs of relation is called the domain you're going to hear a lot about the domain Moving forward what's the domain find the domain right so on and so forth now the set or the list of all second components again these will be the y values and the ordered pairs of a relation is called the range okay so you

might get questions that say what's the domain of this function well you're looking for the set or the list of all possible x values you might get a question that says hey what's the range of this function well You're looking for again the list or the set of all possible y values all right so let's take a look at some basic examples right now and we have more stuff to explain we haven't even gotten to functions yet right now we're just working with some relations and we're getting down some basic vocabulary so let's take a

look at two examples we have this set of ordered pairs and again we have a set of ordered pairs it makes up a relation So the first ordered pair is three comma seven the second one is two comma one the third one is negative eight comma four and the fourth one is negative two comma four now it says what is the domain so the domain again is the set of all first components and again that's kind of very textbooky right you want to say the set of all first components meaning the set of all X's

so what are the x values in this relation well it's everything that occurs first right these are your x values that's all you're looking for if it asks you for the domain so it's the number three it's the number two it's the number negative eight and it's the number negative 2. so this is your domain all right so now we want to find the range so we have the same relation 3 comma 7 2 Comma 1 negative 8 comma 4 and negative 2 comma 4. so the range is the set of all second components or

y values so 7 1 4 and 4. so for the range we'll have and let me make that better we'll have seven we'll have one and we'll have four i don't need to list four twice when you're writing a set if you already Have something in there you don't need to list it again so just seven one and then four so let's take a look at another one we have a relation here which is made up of these four ordered pairs so we have negative nine comma three twelve comma seven negative nine comma four and

one comma eight so what's the domain again that's the set of all first components so negative 9 12 negative 9 and 1. So your domain your domain we've got negative 9 we've got 12 we don't list negative 9 again and then we've got 1. all right so now looking at the same relation we want to know what's the range so the range again is the set of all second components so y value so 3 7 4 and 8. so that's your range and again let me make that better so 3 7 four and eight all

right so now that we've gotten some of the basic kind of vocabulary out of the way let's start going a little deeper and talking about functions so a function is a set of ordered pairs in which each first component each first component or again that's the X value corresponds to exactly one second component or y value a lot of times you're going to hear for each x there can be only one y now in order to understand that let's take a look at this relation here so in this relation we have the ordered pair three

comma two two comma one eight comma four and nine comma 7. so in other words This is going to be a function and so not every relation is a function but all functions are relations you're going to hear that in your lecture from your teacher or maybe your tutor will say it or maybe you just hear me say it but again all functions are relations but not all relations are functions so if i look at each ordered pair here this 3 corresponds To two so in other words i have an x value that's three i

have a y value that's two here i have an x value of two it corresponds to a y value of one here i have an x value of 8 it corresponds to a y value of 4. here i have an x value of 9 it corresponds to a y value of 7. this is a function because for each x that i have it Corresponds to exactly one y value okay and i know that doesn't really make a lot of sense when you first hear it or see it but let me show you an example of

something that's not a function so maybe you'll see it that way so this is an example of a relation that is not a function so you've got a set of ordered pairs but the problem is okay your ordered pairs you have 7 comma negative 4 negative 1 comma 5 7 comma 3 and 2 comma Negative 8. so if you think about your x values what do you have you have 7 you have negative 1 you have 7 again so remember when we were listing our domain we didn't list it twice so i'm just going to

keep that 1 7 there and then you have 2. so what are the y values we have negative 4 you have 5 you have 3 and you have negative 8. so You can think about this as your x values or again your domain right your list of first components you think about this as your y values or again your range your list of second components now for each x there can be only one y or they can be an association with one y so 7 is corresponding to negative four but over here it corresponds to

three That is a huge problem now negative one corresponds to five that's fine and two corresponds to negative eight that's fine this is the problem right here this seven and when you have a function you have to be able to say okay for this given x value i know what the y value is going to be so if i have an x value of seven i need to have a clear association that i'm going to have a y value of this and i don't here i have two y values i Have a y value of

negative four and i have a y value of three so there's no clear association between this x value and one y value it's linked to two different ones right so that's a huge problem for us so we're just going to say this is not a function not a function and when you first start looking at problems in this section these are the type of problems you're going to get they're going to give you a relation Which is basically a set of ordered pairs and they're going to say is this a function is it not a

function and all you're basically going to do is look for duplicate x values so if i see an x value here that's 7 and another x value here that's 7 and it's corresponding to two different y values i don't have a function so here's another example we can look at we have this relation again just a set of ordered pairs we have 7 comma Negative 7 2 comma negative 3 6 comma 5 and negative 12 comma 8. so again if you see this type of problem all you're looking for is duplicate x values i've got

a 7 a 2 a 6 and a negative 12. nothing's duplicated so this is a function because for each x there's one y so every x i have so in other words if i said okay if i have an x value of seven what's the y value well it's negative seven so if i have an x value of seven I know that i get a y value of negative seven if i have an x value of two i know i get a y value of negative 3. if i have an x value of 6 i

know i get a y value of 5. if i have an x value of negative 12 i know i get a y value of 8. so this is a function if i was to replace one of these let's say this wasn't six let's say this was two so then in this case it would not be a function and why Well instead of six being linked to five i would have two that's linked to negative 3 and also 5. so in other words if i said what is the value of y given that x is 2

you really can't give me an answer you'd have to say well it could be negative 3 or it could be 5. i really don't know and the whole idea behind a function is that if i give you an x you can give me a y so if i say okay x is 2 you've got to be able to say okay y is negative 3 right If i go back to the original example if i give you an x of 6 y is 5. so in that case where i replaced this 6 with a 2 2

was now linked to two different y values and so it wouldn't be a function but as it's originally written as we have it here where seven is linked to a corresponds to negative seven two is linked to or corresponds to negative three six is linked to a corresponds to five and Negative twelve is length two or corresponds to eight we do have a function all right here's another example so again another relation just a set of ordered pairs we have negative one comma three two comma five eleven comma seven and two comma seven so right

away you should notice that you have duplicate x values you have a 2 here and here that's a problem because Again if i list this out and i say ok my x values and my y values so the domain the set of x values that have negative 1 2 and 11 only for the range or the y values that have three five seven and then seven again negative one is length to three two is linked to five 11 is linked to 7 but then 2 is also linked to 7. so that's where we have a

problem if i tell you what is the value of y given that x is 2 you don't know you'd have to say well it could be 5 or it could be 7. there has to be a crystal clear association so it has to be i give you an x and you can give me a y so again this is not a function all right so here's another example So we have a relation again a set of ordered pairs we have three comma five we have negative two comma five we have six comma eight and we

have one comma four so some of you right away will make the mistake of saying oh i have a 5 here and a 5 here not a function you would be wrong you only need to look for duplicate x values okay and i'm going to show you why that's the case in a second but essentially Again i have 3 negative 2 6 and 1 no duplicate x values so this is a function and if you kind of list it out like we've been doing so the x values are the domain we have 3 we have

negative 2 we have 6 and we have 1 and then the y values or the range we have 5 we have 5 again we have 8 and we have 4. now for each x there can be 1 y so for this x here 3 it corresponds to 5. so if i say hey What is the value of y in this relation given that x is 3 you can answer 5. right there's no other answer if i say what is the value of y if x is negative 2 your answer is also 5 but that's ok

two different x's can be linked to one y no problem because if you say hey what is the value of y given that x is negative 2 you have a clear answer it's 5. that's all right What you can't have let me erase this real quick we'll come back to it is this going to two different ones let's say i had those two ordered pairs in there let's say i had something with negative two comma five and negative two comma eight well that's an issue because i'd say hey what is the value of y given

that x is negative two well it could be five but it also could be eight there's no clear association between your x value of negative 2 and Some y value it's linked up to two different ones so that's where we have a problem so again this is a function because for each x we have one y three corresponds to five negative two corresponds to five six corresponds to eight and one corresponds to four so i wanna kind of revisit one that we know is not a function already and i wanna draw a picture that's probably

in your textbook it'll probably set up something that Looks like this so your domain this is your x values right your set of x values and we'll draw a little something that looks like this and we'll list the numbers that are in the domain so we have negative 1 we have 2 11 and then 2 again and then you'll see a picture with the range so here's the range these are the Y values and in the y values or in the land of y we have three we have five we have seven so the problem

or the reason this isn't a function we have negative one that's linked to three so if i leave from negative one i end up at three if i leave from 2 i end up at 5 but i also end up at 7. so there's no clear association there and that's the Big problem i need to know that when i leave from 2 i'm going to end up at one location only if i have a function i need to have something where i say okay x is 2 y is some given value here x is 2

and y could be 5 or it could be 7. so again there's no clear association and then for 11 that's also linked to 7 over here and here was an example that we did that was a function so let's do the same picture so we have our domain or the x values And let's draw a little picture for those so we have three we have negative two we have six we have one and then you have your range or your set of y values so you have five and that's duplicated so i just put it

once you have eight and you have four so again for each x there can be one y so if i leave from three i know i'm going to five So there's a clear association if i leave from negative 2 i know i'm going to 5. again it's okay that two different x's go to one single y that's all right because in each case i can say okay if i leave from negative 2 i know that i'm going to 5. if i leave from 3 i know i'm going to five if i leave from six i

know i'm going to eight if i leave from one i know i'm going to four it's only a problem when you leave from An x value and you go to multiple y values because then there's no clear association it's all about give you something for x you've got to give me one and only one value for y okay so there's much more to the topic of functions but kind of in an algebra one course we're going to just kind of wrap up our lesson and look at something else known as the vertical line test and

then we'll kind of leave this topic until we get to algebra 2 and then we'll Get all into the land of functions so the vertical line test if any vertical line intersects the graph at more than one point we do not have a function so let's look at a graph and see why that's the case so if i look at this coordinate plane i just have some ordered pairs here and i want you just to think about what the ordered pairs would be so we have 3 4 as an ordered pair we have 2 7

as an ordered pair we have negative 2 2 as an ordered pair and we have negative 2 7 as an ordered pair and we also have negative 4 4 as an ordered pair so you can already tell just by looking at the ordered pairs that this wouldn't be a function right because the x value of negative 2 occurs twice but you can also determine this Graphically by drawing a vertical line so i'll go through and just make a vertical line really quick if that vertical line hits the graph in more than one location you don't

have a function not a function and why is that the case well for each x value okay this x value is negative 2 it can only correspond to one y value this x value of negative 2 Corresponds to a y value of 2 and also corresponds to a y value of 7. so that's a problem for us and if you don't have a function you're going to have the scenario where a vertical line will cross at least two ordered pairs in your relation so what about something like this where we have the ordered pair negative

five comma negative two negative five comma negative two we have The ordered pair negative three comma negative three we have the ordered pair negative one comma four we have the ordered pair four comma one and we have the ordered pair 6 comma negative 2. so for this relation here we could see if we made vertical lines going down our coordinate plane it would never hit in more than one location meaning it would never cross more than One ordered pair and the reason for that is we have a function we have a function here [Music] so

let's say we came across the graph of a line would this be a function we'll go through and draw some vertical lines and see if it intersects the graph at any place more than once well it doesn't right i can draw vertical lines all through this thing and it's never going to hit more than Once that's because a line is a function this is a function what if i saw something like this would this be a function well no it's not if i look at it i can tell right away that for each x i'm

going to have more than one y if i just draw a line through the x value of negative 4 i can see an impact here and also here so this x value of negative 4 corresponds to two different Y values i can go through and make a bunch of different vertical lines you're going to see it's going to hit the graph more than once everywhere right so for each x here there's more than one y so this is not a function hello and welcome to algebra 1 lesson 21. in this video we're going to learn

about solving systems of linear equations by graphing So once we've kind of mastered linear equations in two variables we move on to kind of the next part of algebra one where we start talking about systems of linear equations so a system of linear equations consists of two or more linear equations with the same variables and mostly in algebra one we're going to have linear systems that just contain two equations two equations with two unknowns so the first example of this that you're Going to see we have 3x minus 4y equals 4 so it's kind of

our first equation and then we have x plus 4y equals 12. so this right here is known as a system of linear equations so now our solution is going to be slightly different because our solution to the system to the system will be an ordered pair an ordered pair that satisfies both equations of this system So if it's a solution to one but not a solution to the other it's not a solution to the system so we're looking for an ordered pair now that's going to satisfy both equations of that system or if you have

more than two let's say all three of the equations of the system if you had three equations simultaneously okay so that's very very important all right so here is an example again we have 3x minus 4y equals 4 and x plus 4y equals 12. so i'm giving you the solution the ordered pair 4 comma 2 is the solution so the way we check this is the same we plug a 4 in for every x we plug a 2 in for every y and we make sure the left and the right side are equal in both

equations so for the first one we have 3 plug in a 4 for x minus 4 plug in a 2 for y should equal 4. 3 times 4 is 12 Minus 4 times 2 that's 8 that should be 4 and it is you get 4 equals 4. so it works out here for the second equation you have x which i'm plugging in a 4 for x plus 4 times y plugging in a 2 for y is equal to 12. so i get 4 plus 4 times 2 is 8 equals 12 and of course 4 plus

8 is 12. so you get 12 equals 12 and so yes this works out as well So this ordered pair 4 comma 2 is the solution for our system which again contains the equation 3x minus 4y equals 4 and also x plus 4y equals 12. now let's say you had a proposed solution that worked in one but didn't work in the other say you had the ordered pair 0 comma negative 1. is this a solution for the system well it would work in the first equation 3 times 0 Minus 4 times negative 1 is equal

to 4. that goes away negative 4 times negative 1 is 4 you get 4 equals 4. so that works out but when you plug it into the second equation it doesn't work so you'd have a 0 for x plus 4 times negative 1 equals 12 4 times negative 1 is negative 4 negative 4 does not equal 12. that's false so it doesn't work in this one so this is not a solution to the system so That's what we mean we need an ordered pair that works in each equation it's got to simultaneously solve both of

them to be a solution for the system all right so how do we find the solution to a linear system in two variables well the first thing we're going to do is we're going to learn about graphing and we're going to talk about that today now graphing is very slow and it's very Inefficient and it's not something you're going to use the reason it's even taught is just so you get the concept down if you have very large or very small values or non-integer values it makes no sense whatsoever to graph and so you should

always use another method but again it's something we teach so that you understand the concept now the next method which we're going to learn right away is substitution then after that we're going to learn Elimination and then in algebra 2 we're going to learn some matrix methods so we're going to cover these three in algebra 1 and then we're going to save this one for algebra 2 and a lot of the matrix methods we're going to cover in college algebra we'll just kind of touch on them in algebra 2 and then we'll really get good

in college algebra because the matrix methods are what you're going to use kind of moving forward when you get into higher math to Deal with these really really big systems all right so i have here let's start by learning the least efficient method graphing all right so it's actually very very easy to solve a system of linear equations by graphing you just simply graph each equation and then the solution is the point of intersection okay again the solution is the point of intersection and why do you think that is well if we kind of just

think About this logically if i graph a linear equation i have a line let's say this is the line here every point on that line is a solution to the equation if i graph another linear equation let's say this is another line every point on this is also a solution for the other equation so that point of intersection right there Is a solution to both right it solves the first one and it solves the second one because that point is on each line and so therefore it is a solution to both equations so let's go

ahead and take a look at the first problem so we have x minus y equals negative 1 and we have x plus 3y equals 15. so very easy to graph at this point we know all the tricks we're going to solve each for y put it in slope intercept form and we're going to graph them very quickly So to solve this for for y i'd have x minus y equals negative 1. i'm going to subtract x away from each side then i have negative y is equal to negative x minus 1. multiply both sides of

the equation by negative 1. i'm going to get y is equal to x plus one okay for the next one again i want to solve this for y So i'm going to subtract i'm going to subtract x away from each side so i'd have 3y is equal to negative x plus 15 and then just divide each side by 3 and that's going to give me what let's erase this and i'd have negative 1 3x i'm going to write it like this plus 15 over 3 is 5. so y is equal to negative 1 3 x

Plus 5. so let's take these two down to the coordinate plane all right so we have y equals x plus one so we know the y-intercept will occur at zero comma one so that's right there and the slope is going to be one so i go up one and over one up one and over one up one and over one or i can go down one into the left one down one into the left one Down one into the left one okay so this is y equals x plus one all right now for y equals

negative one third x plus five the y-intercept is going to occur at 0 comma 5. so that's going to be right here and the slope is negative 1 3. so that means i would fall 1 and go to the right 1 2 3 and i've found my point of intersection already Fall 1 go to the right one two three fall one go to the right one two three and let's go ahead and graph this guy okay so this point right here is your point of intersection so that ordered pair lies on both lines so therefore

it's a solution to the system and so that's going to be the point 3 comma 4 3 comma 4. and that's another problem with graphing if you don't graph that Well if you don't have a ruler or you can't get an accurate graph going it's kind of hard to tell where it intersects i mean here it's pretty obvious i chose a problem that would be very easy but if you had a real world problem or you had a problem that's kind of going to come later on in your textbook it would be really really difficult

to get an answer so let's go back up and let's check that proposed solution of three comma four So we're saying that's our solution the ordered pair three comma four let's check it in the original equation so in the first equation i have x minus y equals negative one so that would be three minus 4 equals negative 1 of course that's true you get negative 1 equals negative 1. so it works there in the second equation i've got x plus 3 y equals 15. so i'm plugging in a 3 for x Plus 3 times for

y i'm plugging in a four and this equals fifteen so three plus three times four is twelve equals fifteen and you get fifteen equals fifteen so checks out there as well so three comma four that ordered pair is a solution for this system let's take a look at another example we have 3x minus 4y equals 24. we have 7x plus 4y equals 16. so again I'm going to solve each one for y so 3x minus 4y equals 24. i'm going to subtract 3x away from each side i will have negative 4y is equal to negative

3x plus 24. i'm going to divide each side by negative 4. and so this is going to cancel i'll just have y negative 3 over negative 4 is just 3 4. i'll get rid of the negatives Just write this as 3 4 times x and then plus 24 over negative 4 that's going to be minus 6. so this guy right here if i write it in slope intercept form is y equals 3 4 x minus 6. for the next one we're also going to solve for y so i'm going to subtract 7x away from each

side i'll have 4y is equal to negative 7x plus 16. divide each side by 4 And get y is equal to negative 7 4 x plus 4. so this is y is equal to negative 7 4 x plus 4. all right let's go ahead and take this down to the coordinate plane all right so we have each equation solved for y very easy to graph it the first one y equals three fourths x minus six the y intercept will occur at zero comma negative six The slope is three fourths so up three one two three

to the right four one two three four up three one two three to the right one two three four so this one is y equals 3 4 x minus 6. i'll just always label the first one once we put the second one in we don't really need to label it just try to not get confused so we now have y equals negative seven Fourths x plus four so that has a y-intercept at zero comma four and a slope of negative seven fourths so i'm gonna go down seven i'm gonna go down one two three four

five six seven to the right four one two three four and we see we have again a point of intersection already now if i go down seven again i'm gonna go to the very bottom part of my graph so i'm gonna go down one two three four Five six seven to the right one two three four [Music] so you can label the second one if you want let's go ahead and put y equals negative seven fourths x plus four you don't have to label these in the previous example i didn't even label the second one

sometimes it's a good idea just when you first start out to not get confused about which equation was which but the Ultimate goal here we already know how to graph we've already figured that out in kind of the last section of algebra 1. now we're just looking for the point of intersection between the two lines and in this case it's very easy to find it's at the ordered pair 4 for x and negative 3 for y so 4 comma negative 3. so let me put that down the ordered pair is 4 comma negative 3. again

let's go ahead and check it in the Original equations make sure that we have the right answer so 3 times 4 for x minus 4 times negative 3 for y equals 24 3 times 4 is 12 negative 4 times negative 3 is also 12 so plus 12. this equals 24 and it does right 12 plus 12 would be 24. so that checks out and then for the next equation we'd have 7 times plug in a 4 for x Plus 4 times plug in a negative 3 for y and this equals 16. so 7 times 4

is 28 plus 4 times negative 3 that's negative 12 this equals 16 and of course it does 28 minus 12 or 28 plus negative 12 is 16 so you get 16 equals 16. so yeah that's true 4 comma negative 3 is going to be the solution to this system all right so overall pretty easy i feel Like in the lesson we'll just do two examples you could do more examples if you wanted to we're going to do a practice set and some test exercises but really not very difficult basically because i chose easy problems if

you had harder problems you wouldn't even be able to find the solution from graphing because it would go off your coordinate plane you know if i had something with a solution that was the ordered pair Negative 3 8 comma i don't know two-tenths you're not going to be able to graph that or if you had something that was like negative 150 comma 200 you're not going to be able to graph that only with a computer so now let's kind of move on and talk about special cases and then in the next lesson we're going to

go to a much more efficient way to do this it's called substitution so the special cases you're going to run across You have systems that don't have a solution so that occurs when you have two parallel lines and two parallel lines as you'll recall will never intersect so again for this type of system there is no solution no solution and you'll recall when we did our lessons on slope we talked about parallel lines having the same slope or the same steepness so they kind of look like this So i know that's not perfect but this

would be an example where you just you don't have a situation where you're ever going to intersect so there's no solution because there's no ordered pair that lies on both lines so here is an example of one where you won't get a solution so x minus 2y equals 8 negative 3x plus 6y equals 6. so let's go ahead and solve each for y and you're going to see right away because they're in slope-intercept form That they're parallel lines so i'm going to solve this for y so we'll have negative 2y i'm just going to subtract

x away from each side is equal to negative x plus 8. divide both sides by negative 2. you'll get y is equal to one half x minus four okay that's easy enough and then let's go ahead and do the other one so if i solve this for y i'd add 3x to both sides so i'd have 6y is equal to 3x Plus 6. again i just added 3x to both sides that's all i did so that would cancel y equals three x which is now over here plus six okay so now we divide both sides

by six and this will cancel you'll get y is equal to three over six is one half so one half x plus one so y equals one half x plus one so immediately what you can tell is that these two Lines have the same slope right the coefficient for x in each case is one half but different y-intercepts this y-intercept would occur at zero comma negative four this one would occur at zero comma one so same slope different y-intercepts parallel lines they'll never intersect and so you don't have a solution now i'm going to go

ahead and graph this one but when we get these in the future You know we might graph them for completeness but basically if you got that on a test once you figure out that they're parallel lines go ahead and just say okay there's no solution there's no solution and move on all right so for the first one to graph it the y intercept is at zero comma negative four that's here and the slope is one half so up one over two up one over two Up one over two and i'll do down one to the

left two all right the next one has a y intercept that occurs at zero comma one that's right here and the slope again is one half so up one over two up one over two up one over two or down one to the left two down one to the left two so let's go ahead Of course they're not perfectly drawn but you can see that these would be parallel lines so there would never be a solution for the system because these two lines would never intersect and therefore there's not an ordered pair that exists that

would be on both of the lines so there's never going to be a solution so you see this and you put no solution no solution all right so let's also talk about another scenario that you run Across remember i said there were two special cases the second case is you have two equations that are equivalent so for this type of system there are an infinite number of solutions so essentially what happens is you solve both of the equations for y and you end up seeing that they're the exact same equation so here's an example of

that we have 10x minus 4y equals 6 and we have negative 5x plus 2y equals negative 3. so this is the exact same equation if You kind of look at it and i multiply both sides of the equation by negative 2 here i would get this equation so these equations are the same they're just manipulated algebraically to look different and so you're going to see an example of this on your test or your homework and so if you see that you spot it right away you can just say that there's an infinite number of solutions

let's go through another process kind of The quicker way to do it if you don't spot it right away is just to solve each for y put it in slope intercept form and you'll see you have the same equation so if i solve this guy for y i have 10x minus 4y equals 6 i would subtract 10x away from each side that's going to cancel i'll have negative 4y is equal to negative 10x i'll have negative 4y is equal to negative 10x Plus 6. and i want to divide each side by negative 4 now so

that's gone and so i'll have y is equal to negative 10 over negative 4 is 5 halves right negative over negative is positive 10 divided by 2 is 5. 4 divided by 2 is 2. so 5 halves then times x and then positive over negative is negative 6 over 4 we could think of as 3 halves Right because 6 divided by 2 is 3 4 divided by 2 is 2. so this one right here is y equals 5 halves x minus 3 halves now the next one when we solve it for y we're going to

see we get the exact same thing so again negative 5 x plus 2 y equals negative 3. add 5x to each side of the equation that's gone you'll have 2y is equal to 5x minus 3. divide both sides of the equation by 2. that's going to cancel and you'll have y equals 5 halves x minus 3 halves so again y equals five halves x minus three halves once you see that you have the exact same equation you just stop and say okay the system contains the exact same equation so there's an infinite number Of solutions

because whatever i plug in that works in this equation obviously works in this equation because they're the same equation so there's an infinite number of solutions because a linear equation of two variables is going to have an infinite number of solutions so there's an infinite number of Solutions all right so let's wrap up our lesson now just talk about there's three possible solutions so the system is called consistent the system is consistent and the equations are independent okay so you're going to hear that in your class so this means our system has exactly one solution

or one ordered pair that satisfies both equations so again when this occurs and this is 95 percent of the time your system is consistent and the equations are independent so that's most of the problems you're going to see now again you're going to get special cases so the system is inconsistent and this means that our lines are parallel so they throw those at you every once in a while so an intersection is not possible right parallel lines do not ever cross so there is no solution for this type of system all right then the final

type of System we have that the equations are dependent so the graphs are the same line so we just saw that in the previous problem right we solved each for y we saw we have the exact same equation so this type of system has an infinite number of solutions and again the reason for that is it's the same equation and so if that equation by itself has an infinite number of solutions and anything that works in the first Equation obviously works in the second one because they're the same right so then the system has an

infinite number of solutions hello and welcome to algebra 1 lesson 22. in this video we're going to learn about solving systems of linear equations by substitution so in the last lesson we talked about how we could solve a linear system in two variables using graphing and what we found was at the point of Intersection we had an ordered pair that was on both lines so therefore was a solution to both equations of the system and it was a solution for our system but there's some big problems with graphing graphing is extremely inefficient and it's extremely

hard to find a solution for we have large numbers small numbers and non-integer values so in other words if the solution was something like Negative three fourths comma one seventh something like that it's very hard to graph that i know you can use a graphing calculator or a computer program but if you're using a sheet of paper like most of you are going to do on a test it's not very efficient or very effective at all so kind of the next thing we do is turn to an algebraic method we got the concept down we

understand about a system of linear equations and what the solution means But we need another way to kind of attack this type of problem so with substitution i'm basically going to substitute in for one of the variables and create a linear equation in one variable so let me kind of go deeper into that let's say i have this problem 6x plus 7y equals negative 9 and negative x plus y equals negative 5. what i can do is i can solve one of the equations for one of the variables now It doesn't matter which variable you

pick and it doesn't matter which equation you start out with but it's generally going to be easier if you look for a variable that has a coefficient of one that's the easiest or the second easiest would be negative one if i look at this equation here what i see is that i have a coefficient on the y variable of one i have a coefficient on the x variable of negative one So i can use either of those it's going to be quickest for me to use the y variable because the coefficient is 1. so it's

very easy to solve that for y i have negative x plus y equals negative 5. just add x to both sides of the equation so that's gone and i have y is equal to x minus 5. so please understand at this point that these two equations are the exact same this is the same as this i just Transformed it it solved for y now what does that mean y equals x minus 5. i'm saying that y y is the same as the quantity x minus 5. those two are the same so in terms of money

that's a good way to explain it let's say that i had a one hundred dollar bill and let's say that i also had five Twenty dollar bills well they're the same in terms of value they just look different right so these are the same they just look different so because they're the same i can go into this other equation here and i can plug in for y i can plug in for y so where i see a y i'm going to plug in an x minus 5 because that's what y is equal to so I'll

have 6x plus 7 times again i'm plugging in for y and i'm plugging in x minus 5 that quantity so make sure you use parentheses so the quantity x minus 5 and this equals negative 9. so now i have a linear equation in one variable and i can solve for x and find out what that value is so i have 6x plus 7 times x is 7x and then minus 7 times 5 is 35 okay and this is going To equal negative 9. combine like terms on the left 6x plus 7x is 13x so this

is 13x minus 35 equals negative 9. let's add 35 to both sides of the equation so this is going to be 13x is equal to what's negative 9 plus 35 well that's going to be 26. so i get 13x equals 26 Divide both sides of the equation by 13 to get x by itself and i get that x is equal to 2. so at this point i have an answer for one of the variables in the system i know that x equals 2. so how do i get the value for y well all i need

to do is plug a 2 in for x in either of the original equations let me erase everything so again i know that x equals 2. so i could plug that in here or here because remember When i look at a system this value for x has to work as a solution in both equations so that means i can plug it into either one doesn't really matter looks like it's going to be easier to plug it into the second one so i'd have negative 2 right plugging in 2 for x plus y equals negative 5.

let's add 2 to both sides of the equation and i'm going to get y is equal to Negative 5 plus 2 is negative 3. so x equals 2 and y is equal to negative 3. as an ordered pair this would be 2 comma negative 3. so let's erase this and put 2 comma negative 3. so i'm going to erase everything and we're going to check it so remember for this to be a solution for the system it's got to work as a solution for both equations it can't just work for one it's got to work

for both So 6x we'll put 6 times 2 plugging in a 2 for x plus 7y plugging in a negative 3 for y equals negative 9. 6 times 2 is 12. and then 7 times negative 3 is negative 21 so minus 21 equals negative 9 and that's true 12 minus 21 is negative 9 so you get negative 9 equals negative 9. so that works out all right let's take a look at the second one so we'd plug in a 2 for x I have a negative out in front and then i'm going to plug a

2 in there plus for y i'm going to plug in a negative 3 and this should equal negative 5 and of course it does negative 2 plus negative 3 is negative 5. so you get negative five equals negative five so again that checks out as well so your solution for the system here is two comma negative three so some of you might be doubtful and you might say well If i had solved another one of the equations for another variable let's say like x i wouldn't got the same answer well you would have the reason

that i solved this equation for y is because y had a coefficient of one when you look at your section in your book on substitution they're going to teach you to look for variables that have a coefficient of 1 or negative 1 because they're really easy to solve an Equation for that but i'm going to go ahead and solve this first equation for x and restart the problem so 6x plus 7y equals negative 9 i would subtract 7y away from each side that would go away i'd have 6x is equal to negative 7y minus 9.

divide both sides by 6 again the coefficient of x and so i'd have x is equal to negative 7 6 Y minus for 9 over 6 they're both divisible by 3 so this would be 3 halves all right so x equals x equals negative 7 6 y minus 3 halves so once i know what x is equal to i can plug in for x in the other equation so i started with this one i'm going to plug into this one so instead of an x there i'm going to have a negative And then i'm going

to have negative 7 6 y minus 3 halves then plus y equals negative 5. so i have a linear equation in one variable and that variable is y so a negative out in front you can think of that as a negative one out in front just change the sign of each term inside the parentheses so instead of negative 7 6 i'd have 7 6 and y instead of minus 3 halves i'd have Plus 3 halves and then plus y equals negative 5. if i want i can multiply both sides of the equation by 6. to

clear the denominators so 6 times 7 6y would be 7y plus 6 times 3 halves the 6 would cancel with the 2 and give me a 3 3 times 3 is 9 plus 6 times y is 6y and this equals negative 5 times 6 that's negative 30. All right so 7y plus 6y is 13y so i'll have 13y here then plus 9 equals negative 30. i would subtract 9 away from each side so that's going to cancel and i'd have 13y is equal to negative 39. as a final step i'll divide both sides of the

equation by 13. and i'm going to get that y is equal to negative 3. let me erase everything so When we come back up here we see that we originally found that y was negative 3. a little bit less work because we had an easier equation to kind of work with at first but we found the same answer either way and then again you have negative 3 so you go ahead and plug it into either of the original equations and you will get 2 as a result for x so let's do that real quick so

6x plus 7 times negative 3 is negative 21. So plus negative 21 equals negative 9 and we'd add 21 to both sides of the equation so this would cancel you'd have 6x is equal to negative 9 plus 21 is going to give me 12. divide both sides of the equation by 6 and of course you get x equals 2 which is exactly what you had right there so either way you do it you get the same answer the solution for your system is 2 Comma negative 3. now i want to eliminate one source of confusion

before we move on and look at another problem once we find the value for one of the variables i told you clearly that you can plug it in for either equation right either original equation and get the result for the other unknown so in other words if i figured out that y was negative 3 i could plug that in here or here doesn't matter and solve for the unknown x and That would be the x that's the solution to the system one thing that's a little different if when i originally take an equation and i

solve it for one of the variables i can't plug it back into that one right so if you do that you're going to get 0 equals 0. let me show you that real fast before we move on because i think it's important so let's say i solve this for y so y equals again i'd add x to both Sides of the equation so x minus 5. if i plug in x minus 5 for y there i'm just going to get 0 equals 0. so in the first step you always want to plug into the other

equation it's once you've figured out what one of the values is going to be then you can plug it into either equation okay so i want to make sure that you understand that so let's just plug that in real Quick so you can see that so you'd have negative x plus y is equal to x minus 5. so i'm going to plug in an x minus 5 and then equals negative 5. so negative x plus x is 0 then i have minus 5 equals negative 5. i could add 5 to both sides of the equation

that would cancel and that would cancel and i would get 0 equals 0. so if you end up with something like that and you're like what happened well You plugged back into the equation that you just started off with and you can't do that so if you originally solve this one for y or solve it for x you've got to plug into the other equation to get started once you've gotten a solution for one of the variables that's when you can plug it into either one so let's look at the official procedure real quick and

then we're just going to attack a bunch of problems So substitution method with two variables so solve one equation for either variable doesn't matter which one again look for a variable with a coefficient of one or negative one then you want to substitute for that variable in the other equation again very important in the other the other equation then we're going to solve for the other unknown right so how do we do that well once we substitute it In we get a value for one of the variables and then we can plug it back into

either of the original equations and that's how we're going to solve for the other unknown and then we just check our result remember the solution has to work in both all right so let's take a look at another one so we have negative 3x plus y equals 7 we have negative 2x plus 2y equals 10. so again i'm looking for a variable that has a coefficient of 1 or Negative 1 if that exists sometimes it doesn't but in this case it does right i have a coefficient of 1 here so that means that this equation

is really easy to solve for y i have negative 3x plus y equals 7. single step here just add 3x to both sides of the equation and i'm going to get that y is equal to 3x plus 7. now once i have that again i plug back in to the other equation so i use this equation to start So that means i'm going to plug in 4y in this equation because i've solved for y so i'm saying that y is equal to or the same as 3x plus 7. so i'm going to plug this in

right here that way i have a linear equation in one variable and i can get a result for x so i'd have negative two x plus two times y y is this quantity three x plus seven And this should be equal to ten let me just drag this down here okay so we'd have negative two x plus two times three x is six x and then two times seven is fourteen so plus fourteen this equals ten negative 2x plus 6x is 4x so i'd have 4x and then plus 14 equals 10. we would subtract 14

away from each side of the equation that's gone i'll have 4x is equal to 10 minus 14 is negative 4. Divide both sides of the equation by 4 and we're going to get x is equal to negative 1. all right so i know x equals negative 1 and now i can take that and i can plug it into either of the original equations i can plug it in here or here does not matter so i'm going to plug it into the second one so we have negative 2 times x is negative 1 so times negative

1 plus 2y equals 10. Negative 2 times negative one is two so that's positive two plus two y equals ten subtract two away from each side of the equation that's gone i'll have two y is equal to ten minus two is eight and then divide both sides of the equation by 2 and of course y is going to equal 4. so as an ordered pair this is negative 1 comma 4. and of course you always want to check Your work so make sure this ordered pair is a solution to this equation and this equation remember

it's got to work in both to be a solution for the system so we have negative 3 times x x is negative 1 so times negative 1 plus y which is 4 equals 7. you can eyeball that and see it's going to be true negative 3 times negative 1 is 3 3 plus 4 is 7. so that's going to Check out for the other one we have negative 2x so negative 2 times again negative 1. plus 2y so for y we're plugging in a 4 equals 10. so negative 2 times negative 1 is 2 plus

2 times 4 is 8 that should equal 10 and of course it does so this result the ordered pair negative 1 comma 4 is in fact the solution to each equation and so it's a solution for our system All right let's take a look at another one we have 4x minus 5y equals negative 8. we have 3x plus 5y equals negative 6. so again i want to solve one of the equations for one of the variables again doesn't matter which one i like to look for a variable that has a coefficient of 1 or negative

1. in this case we don't have that so we might be better off using a different method another method we're going to Learn in the next lesson is called elimination it would be perfect here because i have negative 5y and positive 5y and i know that probably doesn't make sense to you yet but you'll see in the next lesson for right now we're using substitution and it always works it's just a little bit more work so just solve one of these for one of the variables so let's go ahead and just solve this first one

for the variable x So we have 4x minus 5 y equals negative 8. so i would add 5y to both sides of the equation that's gone we'll have 4x is equal to 5y minus 8. then i'm going to divide both sides of the equation by 4 that's going to cancel i'll have x is equal to 5 4 y Minus 8 over 4 is 2. all right so now i'm going to plug in 4x in the other equation again it's got to be the other equation at this stage so here's my x and i'm saying x

is equal to or the same as 5 4 y minus 2. so i'm just going to plug this in right there so i'd have 3 times the quantity remember this is a quantity here 5 4y minus 2. make sure you use parentheses You don't want a silly mistake to cost you an answer on a test so then plus 5y equals negative 6. so 3 times 5 4 would be 15 4 times y minus 3 times 2 that's 6. then plus 5 y equals negative 6. let's go ahead and clear that denominator let's multiply both sides

of the equation by 4. so 4 times 15 4 is obviously 15 then times y minus 4 times 6 that's 24 then plus 4 times 5y that's 20y And this equals negative 6 times 4 that's negative 24. all right so 15y plus 20y is going to be 35y then minus 24 equals negative 24. and what i've taught you before is that if you see the same thing on both sides of the equation you can get rid of it because of the addition property of equality if i add 24 to both sides of the equation it's

just going to go away it's just Going to become 0. so i'll have 35y on this side and on the right side i'm just going to have 0. so this equals 0 and obviously for 35y to equal 0 y has to be 0. all right i divide both sides by 35 and y equals 0. okay so i know that y equals 0. now i just need to find a value for x and again i can use either equation and substitute a 0 in for y Solve for x so let's go ahead and use the second

equation so we have 3x plus 5 times 0 equals negative 6 and so 5 times 0 is 0 so that's gone i'd have 3x is equal to negative 6 divide both sides by 3 and you get x is equal to negative 2. so this ordered pair would be negative 2 comma 0. all right so again the last thing we want to do is just check our work make sure we got the right answer so we have 4 times for x i'm plugging in A negative 2 minus 5 times 0 equals negative 8 and that's true

4 times negative 2 is negative 8. so 5 times 0 is 0 and so we have minus 0 there so it's basically just negative 8 is going to be equal to negative 8 so this one is true so then the next one we have 3x so 3 times negative 2 plus 5y 5 times 0 equals negative 6. so 3 times negative 2 is negative 6 Then plus 5 times 0 is 0. so plus 0 i can just kind of get rid of that so negative 6 equals negative 6 so this one works out as well

so negative 2 comma 0 that ordered pair is a solution for our system so all in all not a very difficult procedure to execute just have to remember what to do in what order right i mean you're basically just going to end up solving a linear equation in one variable to get your value for x and to get your value For y so nothing that's too difficult all right let's talk a little bit about special case scenarios so you'll recall when we did our lesson on graphing that we looked at two special cases one special

case was that we had parallel lines right so we had a a system that would never have a solution because the two lines would never ever cross so there would be no ordered pair that would satisfy both then you also have a special case Scenario where you have the exact same line right so one line was manipulated to look like it was different but it was the same and so you had an infinite number of solutions now when you have a special case scenario with no solution you're going to end up with a nonsensical statement

like i don't know 4 equals 12 or 0 equals negative 36 the two sides won't be equal but you'll Have an equal sign there so let me show you with this example we have negative 4x minus 5 y equals 1. we have 12x plus 15y equals 3. so let me solve this first equation for y so we have negative 4x minus 5y equals 1. i'll add 4x to both sides of the equation and i'll have negative 5y is equal to 4x plus 1. Now i want to divide both sides of the equation by negative 5.

and so what i'd have is y equals 4 over negative five that's just negative four fifths then times x and then minus we have one over negative five so i put minus and then one fifth so that's solved for y and now we can plug in for y in the other equation so remember y is equal to or the same as Negative four fifths x minus one fifth so i'm going to plug in 4y here and again what i'm plugging in has only an x variable so i'd have a linear equation of one variable now

so 12x plus 15 times again you've got to use parentheses here because this is a quantity so negative 4x over 5 or negative 4 5x if you want to say it that way Minus 1 5th okay and this is going to be equal to 3. so unfortunately when we go through and solve this we're going to get some nonsense so we'll have 12x plus 15 times negative four-fifths the 15 would cancel with the 5 and give me 3. 3 times negative 4 is negative 12. so i might as well just put minus 12x And then

15 times negative one-fifth the 15 would cancel with the five and give me a three three times negative one is negative three so minus three equals three so you can see if you combine like terms here 12x minus 12x is zero so i'd end up with this cancelling and me having negative 3 equals 3 which is false so if you get something like this don't freak out you have a system that has no Solution you have a new solution and that's how you can tell you get a nonsensical statement like that negative 3 equals 3

right that's clearly false all right so to kind of prove that this has no solution i want to also solve this equation for y and show you that you have the same slope but a different y-intercept so we'd have 12x plus 15y equals 3. we're going to Subtract 12x away from each side of the equation we're going to have 15y is equal to negative 12 x plus 3. divide both sides of the equation by 15. and what are we going to have this cancels i'll have y is equal to negative 12 is divisible by 3

and so is 15. so negative four fifths Times x plus 3 15. each is divisible by 3 so i'll have one fifth so it's really easy to make a mistake here and think that you have the exact same equation but you don't and what's different here is a minus and a plus so you have y equals negative four-fifths x minus one-fifth so the y-intercept occurs at zero comma negative one-fifth For this one we have y equals negative four-fifths x plus okay see how that's different this is a minus this is a plus one-fifth so the

y-intercept occurs here at zero comma one-fifth so they're very similar but they're not the same equation and so this is not a scenario where we have an infinite number of solutions we don't have a solution at all so these two equations will never cross on the coordinate plane if you graph Each one they'll never touch each other you have two parallel lines and so again there's no solution all right so let's look at one that's an infinite number of solutions so some of you will get really good at this really quickly and you'll be able

to spot that one of the equations is just a manipulated version of the other so in other words if i look at x plus 6 y equals 8 and i multiplied both sides of the equation by 2 i would get 2x plus 12 Y equals 16. so it's the same thing i just took this equation and multiplied it by 2 to get to this equation right i'd have 2x plus 12y equals 16. so knowing that in advance i could just put that there's an infinite number of solutions because whatever works as a solution here is

also going to work as a solution here and a linear equation in two variables by itself has an infinite number of Solutions so in this particular case again the solution for the system would be an infinite number of solutions but let's go through and use substitution i see that i have a coefficient of 1 for my variable x so i'm going to go ahead and solve for that to start so i'd have x i'm going to go ahead and subtract 6y away from each side of the equation so equals negative 6y Plus 8. again all

i did was subtract 6y from here and here that's why i have a negative 6y on the right so negative 6y plus 8. now i'm going to plug this in 4x in the other equation so negative 6y plus 8. again that's a quantity so i've got to use parentheses so it's 2 times the quantity negative 6y plus 8 then plus 12 y equals 16. so 2 times negative 6y is negative 12y plus 2 times 8 that's 16 Plus 12y equals 16. so negative 12y plus 12y that's going to cancel and i'm left with 16 equals

16 or if i wanted to i could subtract 16 away from each side and get 0 equals 0. it doesn't really matter you get the same value on each side of the equation you're left with a true statement so yeah this is true 0 does equal 0 or 16 does equal 16 but it's not really what we're looking for We're looking for a variable equals some number so when the variables drop out and you're left with just a true statement you know that you have a system that has an infinite number of solutions right the

equations are what we call dependent so there's an infinite number of solutions [Music] Hello and welcome to algebra 1 lesson 23. in this video we're going to learn about solving systems of linear equations by elimination so in the previous two lessons we learned about solving systems of linear equations by graphing and then by substitution so kind of the final method that we're going to talk about in algebra 1 is known as elimination so another algebraic method for solving A linear system is again known as elimination so the method uses the addition property of equality to

eliminate one of the variables of the system so rather than start with kind of the textbook procedure i want to just look at two quick nice clean and easy examples and i think we can learn the basics from this then go through kind of the textbook Procedure and then you know move on to some harder problems so i'm going to start out with this system that has negative 4x plus 5y equals 11 and negative 3x minus 5y equals negative 18. so the first thing that i want to do when i'm using the elimination method is

i want to take the left side of one of the equations and add it to the left side of the other equation now i'm going to label these for Convenience i'm going to label this top equation as equation 1 just so i can refer to it and this bottom equation has equation 2. again just for reference sake so i'm going to take the left side of equation 1 which is negative 4x plus 5y and i'm going to add to that the left side of equation 2 which is negative 3x minus 5y Now i'm going to

set this equal to the sum of the two right sides so the right side of equation 1 is 11 and then the right side of equation 2 is negative 18. so what's going to happen here is that one of the variables is going to drop out and specifically the reason for that is because the coefficients are opposites okay so let's do our addition here negative 4x plus negative 3x is negative 7x and then 5y plus negative 5y is 0. right so this variable here y has been eliminated so we have negative 7x is equal to

we have 11 plus negative 18 that's negative 7. so just that quick we have a linear equation in one variable and we can solve this very quickly for x divide both sides by negative 7 the coefficient of x That will cancel and i will have x is equal to 1. all right so just from using some common sense now you can kind of think about what would we do to get y if 1 is the value for x for my solution for the system that means i can plug a 1 in for x in either

original equation and i can solve for y right i have a linear equation with just y in it very easy to solve for that so let's just plug it into the first Equation we'd have negative four times one just plugging in a one for x plus five y equals eleven negative four times one is negative four plus five y equals 11. add 4 to both sides of the equation that's going to cancel i'll have 5y is equal to 11 plus 4 is 15. divide both sides of the equation by 5 and you're going to get

that y is equal to 3. So this represents the ordered pair 1 comma 3. and before i move on and kind of explain why i did what i did and why it's legal i want to go in and check 1 comma 3 that ordered pair and each equation of the system so let's start out with equation 1 we'd have negative 4 times 1 plus 5 times three is equal to eleven negative four times One is negative four plus five times three is fifteen equals eleven negative four plus fifteen is eleven so you get eleven equals

eleven so yes it does work as a solution for this first one all right for the second one i have negative 3x so that's negative 3 times 1 minus 5y so that's minus 5 times 3. this equals negative 18. so negative 3 Times 1 is negative 3 and then we have minus 5 times 3 is 15 and this equals negative 18. and it does negative 3 minus 15 is negative 18. so you get negative 18 equals negative 18. so it works out here as well so now that we've verified that 1 comma 3 that ordered

pair is a solution for the system let's talk about what we did to get it and why this method is mathematically legal so the very first thing i want to call your Attention to is that these two equations are set up in standard form right it's a x plus b y equals c and i know some of you are going to scream and say well that's not standard form this isn't this isn't greater than or equal to zero okay i'm talking about standard form where we just think about a b and c as real numbers

okay just for this section i want you to think about it as a b and c just being real numbers and a and b Just can't both be zero okay that's your only restriction so once we have them both set up in this format when you put the two on top of each other it's going to be very easy to see right away if one of the variables would be eliminated through addition and how do you tell that well you look at the coefficients if one pair Have opposite coefficients so in other words this y

variable here has a coefficient of five this y variable here has a coefficient of negative 5. so obviously if i add 5 and negative 5 that's going to be 0. so i'd have 0 y or just 0. if i have negative 3 and positive 3 that's 0. if i have negative 212 and positive 212. that's going to be zero we know that if we add a number and it's opposite it's zero So if i see one pair of variable terms with opposite coefficients i know that this is going to be a very quick and easy

method to get a solution for the system so let me erase this real quick and let's talk about why this is legal now because this is probably the thing that trips students up the most they say well how are you able to add this side to that side and that that doesn't make any sense why are we able to do that Well mathematically if i took the first equation and said negative 4x plus 5y equals 11. you recall that the addition property of equality tells us that we can add the same quantity or the same

amount to both sides of an equation and we won't change the solution So in other words i could add 3 to both sides of this equation and anything that was a solution before is now still a solution so kind of expanding on that if i think about equation 2 for a moment and i look at this equal sign let's think about this for a second this equal sign means the same as the same as so that means that we are saying Negative 3x minus 5y is the same as negative 18. so they represent the same

quantity so what that means is that i can add this to this side of the equation so plus negative 3x let me kind of scooch this down a little bit minus 5y and add this to the other side of the equation And i've added the same quantity this is equal to this so i can add them to both sides of the equation legally it's the same thing as when i added 3 to both sides of the equation it represents the same amount or the same quantity so let me kind of draw that like that and

put same quantity okay so that's why we're legally able to do That so using this strategy if we have one pair of variable terms with opposite coefficients it's going to drop out and it's going to leave me with a linear equation of one variable that i can then solve in this case we solved and we found that x was equal to 1 then i can plug that back in for x in any of the original equations and i can figure out what the other unknown is all right let's take a look at another one and

again we're going to look at Another easy one and we have 4x minus 2y equals negative 6. we have negative x plus 2y equals 3. so again you see i wrote it in the format of ax plus b y equals c ax plus b y equals c okay so it's already set up for you like that now right away if you kind of look at things i have a coefficient of 4 here and a coefficient of negative 1 here so those aren't opposites 4 negative 1 are not opposites I have a coefficient of negative 2

here and a coefficient of positive 2 here those are opposites so negative two plus two is zero so i know when i do my first step add the two left sides together this variable is going to be eliminated now kind of a quicker way to do this and a lot of students prefer to use this method you can add vertically okay so you can kind of just add it the way it's set up you can put a plus sign out here if you Want you don't need to i never do i just add going down

so 4x plus negative x is 3x negative 2y plus 2y is 0y so that's eliminated and this is equal to negative 6 plus 3 is negative 3. so now i have a linear equation one variable and i can divide both sides by three the coefficient of x and get that x is equal to negative one so once i find that x equals negative One i can plug a negative 1 in for x in either of the original equations solve for y so let's go ahead and plug into this second equation looks like it's easier to

work with we have negative and then i'm plugging in a negative 1. so minus a negative 1 would be plus 1 then plus 2y equals 3. i'm going to subtract 1 away from each side that's Gone i'll have 2y is equal to 2. divide both sides of the equation by 2 and i'm going to get that y is equal to 1. so this is the ordered pair negative 1 comma 1. again let's check to make sure we got the right answer so i have 4x minus 2y equals negative 6 so i'd have 4 times negative

1. minus two times positive one equals negative six four times negative one is negative four so negative four Minus two times one is two equals negative six negative four minus two is negative six so you get negative six equals negative six so it checks out in the first one let's take a look at the second one so i have negative x so negative plug in a negative one so minus the negative one again is plus one plus two times plug in a one for y equals three So essentially this would be one plus two times

one is two equals three we know one plus two is three we get three equals three so it checks out there as well so now that we've had a chance to look at two very easy examples let's cover kind of the textbook elimination method procedure so the elimination method for two variables and obviously this would be with two equations we want to write both equations in Standard form so ax plus b y equals c and if you want to use the stricter definition that's fine you don't need to you can just have a b and

c as real numbers where as long as you don't have a and b both equal to zero you would be fine the next step and we really haven't gotten to this yet because we haven't looked at any difficult problems We want to transform one or both equations so one pair of variable terms are opposites and we're going to see this in a minute in the examples we've looked at we've come into easier problems where we've started out with one pair of variable terms our opposites so we didn't need to do this step so once this

is accomplished we want to add the left sides of the equations and set this equal to the sum of the right sides Then we're going to solve for one of the unknowns we're then going to plug in the result from step 4 into either original equation and find the remaining unknown and then we're going to check our solution in each original equation all right let's take a look at one that's a little bit harder so we have x minus 6y equals negative 1 and negative 3x minus 12y equals 3. so if we look here We

see that they're both in standard form already ax plus b y equals c the next thing we want to think about is looking at the coefficients for each variable if i look at the coefficients for x i have a coefficient here of 1 and a coefficient here of negative 3. so those aren't opposites if i look at the coefficients for y i have negative 6 and negative 12. those aren't opposites So i've got to transform one or possibly both of the equations into equivalent equations where one pair of variable terms would have opposite coefficients so

kind of the easiest thing i can think of right now if i look at negative 3 here well all i would have to do is multiply this equation here both sides by 3 and what that would do is that would give me a 3x here instead of 1x Perfectly legal because i can multiply both sides of an equation by the same number as long as it's not 0 and i preserve my solution so let me just do that down here i'm going to drag this guy down here and i'm going to multiply both sides by

3. 3 times x is 3x minus 3 times 6y that's 18y this equals negative 1 times 3 which is negative 3. So then here i'm just going to copy this equation negative 3x minus 12y equals 3. so once we've transformed this equation such that we have a pair of variable terms that have opposite coefficients we can go through and do our add the left sides together set that equal to the sum of the right sides so i'm just going to add vertically you can just add going down 3x plus negative 3x is 0. so This

is going to be eliminated then the next thing is we have negative 18y plus negative 12 y that's negative 30 y and this equals negative 3 plus positive 3 which is 0. divide both sides of the equation by negative 30 and you get that y is equal to 0. all right so we know that y is equal to 0 and again once i have that information i can plug that back in for Y in either original equation it doesn't matter which one so let's go ahead and plug it into the first one so i would

have x minus six times zero just plugged in a zero for y equals negative one so we know that six times zero is zero so i can just kind of cross this out and i would have x is equal to negative one so x is equal to negative 1 so that's the ordered pair negative 1 comma 0. so let's check our result in each Original equation make sure that we didn't make a mistake so i have negative 1 plugged in for x minus 6 times 0 plug that in for y equals negative 1. we'd have

negative 1 minus 6 times 0 is 0. so negative 1 minus 0 is just negative 1. so you get negative 1 is equal to negative 1. so it works out here let's take a look at the other one we have negative 3x so negative 3 Times negative 1 minus 12y so minus 12 times 0 equals 3. negative 3 times negative 1 is positive 3 and then negative 12 times 0 is 0. so plus 0 equals 3 you get 3 equals 3 so it works out here as well so we verify that the ordered pair negative

one comma zero is the solution for our system all right let's look at one that's a little bit more Complicated now i have negative four x plus five y equals negative twelve and i have negative four y plus two equals negative 9x so the first thing that you'll notice is that each equation is not in standard form so when you look at and you try to compare things it's not easy and it's a cluster so you want to put each equation as ax plus b y equals c so This one i'm going to leave alone

this one i'm just going to move the x variable over here and i'm going to move the constant over here i'm just going to do that with the addition property of equality so i'm going to add 9x to both sides of the equation and i'm going to subtract 2 away from each side of the equation so when the dust settles i'm going to have 9x 9x Minus 4y this will cancel so equals this will cancel negative 2. so now we have 9x minus 4y equals negative 2. all right so let's erase this real quick and

i'm just going to drag this other equation down here so they're on top of each other so i have negative 4x plus 5y equals negative 12. okay So now that the first step is completed and they're both in standard form the next thing that i want to do is i want to look at the coefficients on my variables so for the x variable i have a 9 and i have a negative 4. for the y variable i have a negative 4 and i have a 5. so i don't have anything with opposite coefficients so nothing

would cancel right now if i did the addition so i've got to transform one Or both equations to where one of the variables would drop out now it doesn't matter which one you pick i could pick the x variable and eliminate it or i could pick the y variable doesn't matter let's go ahead and work with the x variable so i have a coefficient here that's nine i have a coefficient here that's negative four so one of them is positive one of them is negative so kind of the easy thing to Do is just multiply

both sides of the first equation by four right that will give me 36 here or a 36 x and then multiply both sides of the kind of second equation by 9 that would give me a negative 36 x right so let's go ahead and make that happen we'll multiply both sides of this equation by four let me scooch this over a little bit and let's multiply both sides of this Equation by 9. so for the first equation 4 times 9x is 36x minus 4 times 4y which is 16y and this equals negative 2 times 4

which is negative 8. then we have 9 times negative 4x that's negative 36x 9 times 5y that's going to be plus 45y this equals 9 times negative 12 which is negative 108. all right So now that we have transformed both of the equations into equivalent equations where through the addition process this variable x would be eliminated right you have opposite coefficients here 36 and negative 36 we can go through and do that addition step now so we're going to add the two left sides together so 36 x plus negative 36 x that's zero so this

is eliminated Negative 16 y plus 45y is 29y this is equal to negative 8 plus negative 108 which is negative 116. now if i divide both sides of the equation by 29 the coefficient of y i'm going to get that y is equal to negative 4. negative 116 divided by 29 is negative 4. so y is equal to negative 4. so now i can plug a negative 4 in for y in either original equation solve for X so let's go ahead and plug it into the second equation so we'd have negative 4 times negative 4

plus 2 equals negative 9x negative 4 times negative 4 is 16. so you'd have 16 plus 2 equals negative 9x 16 plus 2 is 18. so you'd have 18 equals negative 9x divide both sides of the equation by negative 9. and you're going to get that negative 2 is equal to x or you could say x is Equal to negative 2. so this would be the ordered pair negative 2 comma negative 4. all right so let's plug in for x and y in the original equations and make sure that this is the solution for the

system so we'll have negative 4 times plug in a negative 2 for x plus 5 times plug in a negative 4 for y equals negative 12. negative 4 times negative 2 is 8 5 times negative 4 is negative 20. so minus 20 equals negative 12. and this is true 8 minus 20 is negative 12. so you get negative 12 equals negative 12. it checks out there all right for the next one we have negative 4y so negative 4 times for y i have a negative 4 plus 2 equals negative 9 times negative 2 plug in

the negative 2 for x Negative 4 times negative 4 is 16 so you'd have 16 plus 2 that's 18. so this equals negative 9 times negative 2 that's 18 also so it works out here as well so we verify that this ordered pair negative 2 comma negative 4 is the solution for the system let's take a look at one final problem again as you practice this it becomes very very easy we have negative two x minus seven y equals fourteen and we have one plus one half y plus seven six X equals zero so this

equation is in standard form this one's not so let's go ahead and fix that so because this is all addition here i can switch the order around i could rewrite this as 7 6 x plus one half y plus one equals zero and i could simply subtract one away from each side of the equation And i would just basically have equals negative one now a lot of us don't like to work with fractions so it's very easy to clear fractions from an equation you multiply both sides by the lcd the lcd here would be 6.

let me kind of scooch this over a little bit and scroll down so 6 times 7 6 the 6's would cancel i'd have 7x And then when i multiply 6 times 1 half the 6 would cancel with the 2 and give me a 3. 3 times 1 is 3. so i'd have plus 3y this is equal to negative 1 times 6 or negative 6. so i have 7x plus 3y equals negative 6 and then my other equation again is negative 2x minus 7y equals 14. all right so now that i have my two equations in

standard form again i want to check and see do i have Any coefficients for any of the variables that are opposites so in other words i look at the x variable in each case i have a 7 and a negative 2. so it's a no go there i look at the y variable i have a 3 and a negative 7. doesn't work there either so now i've got to do an additional step i've got to transform one or both equations into equivalent equations where the coefficients for one pair of Variable terms are going to be

opposites right that's crucial to the elimination method so let's just work with x i see that one coefficient is 7 the other is negative 2. so i just like to write them out there think about okay well if i multiply 7 by 2 i'd get 14 if i multiply negative 2 by 7 i get negative 14. so i'm going to multiply my top equation by 2. so 7x plus 3y Equals negative 6. again i'm just going to multiply this by 2. i'm going to multiply my bottom equation by 7. so we're going to have negative

2x minus 7y that's multiplied by 7. this equals we have 14 that's multiplied by 7. so 2 times 7x is 14x plus 2 times 3y that's 6y and this equals negative 6 times 2 that's negative 12. over here i have 7 Times negative 2x that's negative 14x 7 times negative 7y is negative 49y this equals 7 times 14 which is 98. so now we can see that we have a 14 that's the coefficient for x there and a negative 14 that's the coefficient for x there so we have successfully transformed these two equations such that

one variable is going to be eliminated when we do our next step so in the next step remember we add the Two left sides together set that equal to the sum of the two right sides so 14x plus negative 14x would be zero this is eliminated and then 6y plus negative 49y is negative 43 y over here negative 12 plus 98 is going to be 86 divide both sides of the equation by negative 43 To isolate y you're going to get that y is equal to negative 2. again y equals negative 2. and now i

can plug this back in for y in either original equation i'm going to go ahead and use this one because it actually looks like it's easier so we'd have 1 plus 1 half times negative two plus seven six x equals zero so this would cancel and be negative one so you'd have one plus Negative one that's zero so this whole thing is just gone so basically i have seven six x equals zero we know that can only be true if x is zero so i know x is zero here and again now we just wanna

check the ordered pair 0 comma negative 2 and each original equation make sure that it's a solution for the system so i would have negative 2 times x we have a 0 for x minus 7 times y we have a negative 2 for Y equals 14. this is going to go away 0 times anything is just 0. i have negative 7 times negative 2. that's going to give me 14. so i get 14 equals 14 so it does check out here and the second one i have 1 plus 1 half times negative 2 plus 7

6 times 0 equals 0. we know this is going to go away because 0 times anything is just 0. And i know this is going to cancel with this and become negative 1. so essentially what i'm going to have is 1 plus negative 1 which is 0 so i get 0 equals 0. so it checks out here as well so 0 comma negative 2 that ordered pair is a solution for the system hello and welcome to algebra 1 lesson 24. in this video we're going to learn about applications of a linear systems so we previously

learned how to set up and solve a word problem That involves a linear equation in one variable here we will apply a similar process to solve problems that involve a linear system and two variables all right so let's quickly cover this procedure and it's just a modified version of the original one i gave you so solving a word problem that involves a linear system and two variables so obviously the first thing you want to do is read the problem carefully Get a clear understanding of the objective then we want to assign a variable to represent

each unknown and because we're working with linear systems in two variables we're going to write two equations using both variables then we want to solve the linear system we're going to state the answer and generally speaking when we're working with word problems we want to make a nice little sentence and then lastly we're going to check and remember with Word problems we want to make sure that the answer is reasonable if you had a question like how many cars were produced at a factory and you get negative 37.2 it's probably not the right answer right

the factor would produce either zero or some positive integer amount all right let's go ahead and take a look at the first problem we have enough experience with word problems to just kind of jump right in So kayla and kristen are selling wrapping paper for a school fundraiser customers can buy rolls of plain wrapping paper and rolls of holiday wrapping paper so kayla sold six rolls of plain wrapping paper and five rolls of holiday wrapping paper for a total of 149 kristen sold six rolls of plain wrapping paper and two rolls of holiday wrapping paper

for a total of 92 dollars so we want to Find the cost of one roll of plain wrapping paper and one roll of holiday wrapping paper so remember our first step was to read the problem and get a clear understanding of the objective well the objective is right here we want to find the cost of one roll of plain wrapping paper and one roll of holiday wrapping paper so i should be able to make a nice little sentence in the end that says the cost of one roll of plain wrapping Paper is this much and

the cost of one roll of holiday wrapping paper is this much all right so for the next step we want to assign a variable to represent each unknown we don't know how much is one roll of holiday wrapping paper i'm just going to put w period p period that's for wrapping paper and then we also don't know how much is one roll of plain wrapping paper so plain And then w period p period just to abbreviate a little bit so let's assign a variable to each one and let's just use x and y because those

are the most common so let's let x let's let x equal the cost or you could say sale price of one roll of plain wrapping paper so i'm going to put wp for that And i'm going to put some periods there just for abbreviation w period p period for wrapping paper so then for our other unknown i'm going to say y is equal to the cost of one roll of holiday wrapping paper so again w period p period and don't get confused here when i say cost i mean sale price so a lot of Students

will see that and say well is that the cost that they pay and it's a little confusing so i want to just clarify and say that the cost here when we say that we're saying the sale price or in other words the cost for the customer the person who's buying it all right so for step three we want to write two equations using both variables so in order to do that we need to read back through the problem so i'm going to Go to this part where all the information is given so reading back through

here we see that kayla sold six rolls of plain wrapping paper and five rolls of holiday wrapping paper for a total of a hundred forty nine dollars so we can use this information to make our first equation so six rolls of plane wrapping paper five rolls of holiday wrapping paper again for a total of 149 dollars So putting that into an equation just using our variables for the unknowns we would have 6 times x right x is the cost of one single roll of plane wrapping paper plus five times y why is the cost of

one roll of holiday wrapping paper should be 149 the total cost when these two are added together so again this is going to be 6x plus 5y is equal to 149. and let's label this so We don't get confused so 6 represents the number of rolls sold and so does five so let me draw a little arrow over to five for x this is the cost of one roll of plain wrapping paper so the cost of one roll of plain wrapping let me kind of get rid of this Paper then for y it's the cost

of one roll of holiday wrapping paper so again 149 is the total the total money she brings in so i can put the total from the sale of both so she sells six rolls of plain wrapping paper so that's modeled with six the number of rolls sold times x which Is the cost per roll again of plain wrapping paper then we add this to five which again is the number of rolls sold times y which is the cost of one roll of holiday wrapping paper so we sum these two amounts together and we get the

total amount that she gets from the sale of both that's 149 so there's your first equation now for the second one let's read about kristin so Kristen sold six rolls of plain wrapping paper and two rolls of holiday wrapping paper for a total of ninety two dollars i can go through and highlight that so six rolls of plain wrapping paper two rolls of holiday wrapping paper again this is for 92 dollars so we're going to come down here and just make a similar equation so we'd have 6 times x plus 2 times y is equal

to 92. If we went through and labeled it would be the same right you'd have six which is the number of roles sold times x which is the cost of a single roll of plain wrapping paper plus and then you have two which again is the number of rolls sold times y which is the cost of one roll of holiday wrapping paper and the sum here is going to equal 92 right that's the total from the sale of both all right so we've correctly modeled Each situation for the two girls so let's look at these

two equations now and see if we can solve this linear system so i'm going to rewrite them down here so i have 6x plus 5y equals 149 and then i also have 6x plus 2y Equals 92. now it'll probably be easiest to use elimination here now none of my variables would drop out if i just went through and started adding what i need to do is make a slight modification you can see that the coefficient of x is 6 here and 6 here so if i multiplied one of these equations Both sides by negative 1

one of these would turn into negative six and one would be positive six and so when we did our addition the x variable would drop out so i'm gonna multiply this second equation here by negative one and so i would have 6x plus 5y equals 149 still for my first equation for my second equation negative 1 times 6x is negative 6x Then negative 1 times 2y is minus 2y and this equals negative 1 times 92 that's negative 92. all right so let's add the two left sides together so i'm going to do this vertically so

6x plus negative 6x is 0. so that's been eliminated 5y minus 2y is 3y and this equals now we're going to sum the right sides 149 plus negative 92 is 57. so the final step we divide both sides Of the equation by 3 and we get that y is equal to 19. so once you have this you're not done you need to figure out a value for x and we look at our two original equations again y is equal to 19. i can plug a 19 in for y in either equation looks like it'll be

a little bit easier here so i'd have 6x plus 2 times 19 is equal to 92 so then this is 6x plus 2 times 19 is 38 So 38 equals 92 subtract 38 away from each side of the equation that's going to cancel i'll have 6x is equal to 92 minus 38 is 54. and then we divide both sides of the equation by 6 and we get that x is equal to 9. so this is the ordered pair 9 comma 19. and just for the sake of completeness i Like to go through and check make

sure that that ordered pair works as a solution to both equations and then we go back up and then we state our answer so 9 is my value for x so 6 times 9 plus 5 times 19 is my value for y and this should equal 149 6 times 9 is 54 plus 5 times 19 is 95 so this should equal 149 and it does you get 149 equals 149 So we're good to go here let's look at the second equation all right for the second one we're going to again plug in a 9 for

x so we'd have 6 times 9 plus 2 times 4 y we're plugging in a 19 and this should equal 92. 6 times 9 again is 54 plus 2 times 19 that's 38 this should equal 92 and 54 plus 38 is 92 so you get 92 equals 92. so yeah checks out here as well now Remember with a word problem you've got to make sense of everything you can't just stop and say my answer is the ordered pair 9 comma 19. that doesn't relate to what you got your teacher would only probably give you partial

credit for that so let's go back up to where we wrote what the variables were representing so remember x was the cost or again the sale price of one roll of plain wrapping paper and we found that x was equal to nine so X was equal to nine so then one roll of plain wrapping paper is nine dollars then why is the cost of one roll of holiday wrapping paper we found that y was nineteen dollars so to answer this we can say that each roll of plain wrapping paper which again i'll abbreviate Sells for

nine dollars while each roll of holiday wrapping paper [Music] sells for 19 and although we've checked the equations go back into your actual problem and make sure that that makes sense so again kayla sold six rolls of plain wrapping paper and five Rolls of holiday wrapping paper for a total of 149 dollars so you're checking this just like you checked your equation it's basically the same thing so six rolls of plain wrapping paper again each roll is nine dollars so that would be fifty four dollars and five rolls of holiday wrapping paper again each roll

is nineteen dollars five times 19 is 95. so you sum these two amounts together you would get a total of 149. so that Part's good to go then again the second part kristen sold six rolls of plain wrapping paper six times again nine is fifty four so that's fifty four dollars two rolls of holiday wrapping paper two times nineteen right nineteen dollars a piece for the holiday wrapping paper that's thirty eight dollars some of these two amounts together and you would in fact get ninety two dollars So our answer here is correct each roll of

plain wrapping paper sells for nine dollars and each roll of holiday wrapping paper is going to sell for 19 all right let's take a look at another one so a plane traveled 700 miles to far land and back the trip there was with the wind and took only 7 hours the trip back was into the wind and took 14 hours find the speed of the plane in still air and the Speed of the wind so when you get these type of problems you're always thinking about okay although it might not be a hundred percent accurate

in real life if you're going with the wind the wind is pushing on the back of the plane or a lot of times you'll see these with a kind of a current problem where the current is pushing the boat so you have a current or you have a wind that's pushing at your back that's making you Go faster when you fly into the wind or when you are in a boat and you go against the current you're fighting it so you're going slower so the main thing you're trying to figure out here your main objective

is to find the speed of the plane in still air so meaning the air is not moving at all and the speed of the wind so the speed of the wind so we're going to use two variables to Represent each right those are our two unknowns so let's let x equal the speed of the plane in still air and then y can be the speed of the wind now again for this type of problem because it involves motion we're going to be using our distance formula so you should remember this from when we're Solving word

problems with linear equations of one variable the distance formula looks like this you have your distance is equal to the rate of speed multiplied by the amount of time traveled and again this is very very intuitive you jump in a car you go 70 miles per hour for four hours forget about the units just multiply the Numbers together 70 times 4 is 280. so that would be 280 if you think about the units it would be 280 miles very very easy to calculate your distance if you know a rate and a time just multiply the

two together you could manipulate this in many ways you could solve for rate and figure that out if you have the time and the distance where you could solve for time and figure that out if you had the rate and the distance right so on and so forth but what we're Going to do in this particular problem is set up a nice little table so i have that distance is equal to rate times time so i'm going to put this up top draw a little line down here all right so we have two scenarios we

have the trip there we have the trip there And we have the trip back let me get a little bit more room let me kind of scoot this over all right so let me put this down here equals equals so let's go and get some information so for the trip there we know the distance is 700 miles in each case so the trip there and the trip back the plane traveled 700 miles too far and back so the distances will be the same so let's fill that out so the trip there The distance is 700

the trip back the distance is also 700. now let's talk about the rate the trip there was with the wind so that means that whatever the plane would do in terms of its speed you have to add to that the fact that the wind is pushing at your back so you take the plane speed in still air which we represented with x and you would add to that the speed of the wind which were represented with y So for your rate for the trip there it's going to be x plus y now in terms of

time it says it took only seven hours so we have a rate we have a time and we have a distance so the time is seven now for the trip back it says it was into the wind and it took 14 hours so into the wind that means that i'm taking the speed of The plane in still air which is x and i'm subtracting away the air speed or the speed of the wind so that's minus y and then the time is 14 right for 14 hours so then we'll have our two equations essentially we

have an equation for the trip there which is a rate times the time we have seven which is our time times our rate which is x again this is the speed of the plane in still air Plus y which again is the speed of the wind or the wind speed and when we multiply these two together we're going to get the distance right this is a time this is a rate rate times time gives me a distance so this should be equal to 700 then for the trip back i set up another equation again i

have a time that's 14 and this is multiplied by the rate which is x again the speed of the plane in Still air now i'm subtracting away the wind speed because if i'm going against the wind i'm fighting it so i got to subtract that away so then minus y and this equals 700. so i have a time times a rate or a rate times a time gives me a distance same thing up here i have a time times a rate that gives me a distance so i have two equations and two unknowns so let's

go ahead and just solve this System so 7 times x is 7x plus 7 times y that's 7y and this equals 700 and then i have 14 times x that's 14x and then minus 14 times y that's 14y and this also equals 700. now again it would probably be pretty easy to use the elimination method specifically because i have a negative here and a positive here and all i need to do is multiply this top equation both sides by 2 and i would have opposite Coefficients for my variable y so let's go ahead and do

that if i multiply this side by 2 and this side by 2 i would have 14x plus 14y is equal to 1400 and i'm going to go ahead and erase this for now and these are going to be the two equations i work with so i'm going to Use my elimination method 14x plus 14x is 28x when i add here negative 14y plus 14y that's going to be eliminated and then this equals 700 plus 1400 which is 2100 or 2100. so now to get x by itself i divide both sides of the equation by 28

and that's going to give me that x is equal to 75 but again i'm not done i've got to find out the value for y so let's go back And look at one of these original equations again we found that x was equal to 75 so we had 7x plus 7y equals 700 and we had 14x minus 14y equals 700. so plug in a 75 for x in either one so let's just take the first one so 7 times 75 plus 7y equals 700. what's 7 times 75 that's going to be 525 so we'd have

525 plus 7y equals 700 and then we'd subtract 525 away From each side of the equation that's going to cancel and we'd have 7y is equal to 700 minus 525 is 175 then if i divide both sides of the equation by 7 i'm going to end up with y is equal to 25. all right so now we can write this as an ordered pair we'll say it's 75 comma 25 and let's just check this guy in both equations so we'd have 7 times 75 Plus 7 times 25 equals 700. so 7 times 75 is 525

plus 7 times 25 that's 175 this should equal 700 and it does 525 plus 175 is 700 so these two sides are equal so this one checks out then for the next one we have 14 times 75 minus 14 times 25 and this should equal 700. So 14 times 75 is 1050 minus 14 times 25 which is 350 and this should equal 700 and again it does 1050 minus 350 is 700 so you get 700 equals 700 so it works out there as well all right so remember with a word problem you always want to

make a nice little sentence so we're just going to say that the speed the speed of the plane and still air Was 75 miles per hour and the wind speed the wind speed was 25 miles per hour let's see if this just makes sense in terms of what the problem said so a plane traveled 700 miles to far land and back the trip there was with the wind and took only seven hours so we know that if the plane was going 75 miles per hour And there was a 25 mile per hour wind then each

hour the plane is traveling 100 miles so then it does this for seven hours it's going to get 700 miles worth of distance so that checks out now the trip back was into the wind so if it's going 75 miles per hour and the wind is 25 miles per hour you subtract you would get 50 miles per hour and then it does that for 14 hours well 50 times 14 is also 700 so that checks out as well so Again the speed of the plane in still air is 75 miles per hour and the wind

speed is 25 miles per hour hello and welcome to algebra 1 lesson 25 in this video we're going to learn about solving systems of linear inequalities so so far we've seen how to graph a linear inequality in two variables such as and i have a quick example for us we have 4x minus y is greater than 2. now there's a few different methods to Kind of graph this but the quickest one for me is to solve this for y so in other words i would subtract 4x away from both sides of the inequality i would

have negative y is greater than negative 4x plus 2 and then i would divide both sides of the inequality by negative 1 and remember if i'm dividing by a negative or multiplying by a negative i've got to flip the direction of the Inequality so this symbol right here tells me i have a greater than i want to make this into a less than okay so the direction of the inequality is going to change now negative y over negative 1 is y so now it's y is less than negative 4 over negative 1 is positive 4

then times x and then 2 over negative 1 is negative 2. so i get y is less than 4x minus 2. Now the reason i solve for y is it's easier for me to figure out where to shade i don't have to go through and make a test point or anything like that for a less than or a less than or equal to i just shade below the line now i'm getting ahead of myself remember this line that i'm talking about is called the boundary line and we create the boundary line by replacing the inequality

symbol with an equal symbol so i would graph y equals 4x Minus two to begin so let's go down to the coordinate plane so again we have y is less than four x minus two and then our boundary line we're gonna graph y equals four x minus two so one other piece of information before we kind of draw this boundary line remember there's two different situations the boundary line will be a dashed line or a broken up line if you have a strict Inequality like we have here in this case the points that lie on

the boundary line are not part of your solution the other scenario that you run across is a non-strict inequality so let's say i had a less than or equal to or i had a greater than or equal to in this case we want a solid boundary line because the points on the line are part of the solution for this specific case though we have a strict inequality and so when we draw our boundary line we want it to Be dashed or broken up again all we need to do is graph the line y equals 4x

minus 2 that's going to be our boundary line so the y-intercept occurs at 0 comma negative 2 so that's right here and my slope is positive 4. so i'm going to go up 1 2 3 4 to the right 1. and i can do that one more time go up one two three four to the right one so let's draw a solid line and then We'll use our eraser to kind of break it up [Applause] okay and i'm just going to take my eraser and just kind of break it up so that you know it's

a dashed line all right so that's my boundary line for y is less than 4x minus 2. now one of the benefits to solving your inequality for y if it's a less than or less than or Equal to you just shade below the line if it's a greater than or greater than or equal to you shade above the line i know in your textbook you probably read something about test points and all this other stuff if you solved it for y you don't need to do any of that stuff again less than or less than

or equal to shade below the line greater than or greater than or equal to shade above the line so this is a less than so we shade Below the line nice and simple so basically anything in that region would be a solution to our inequality which was 4x minus y is greater than 2. all right so suppose we add another inequality into the mix and create a system so this is the inequality we just graphed 4x minus y is greater than 2. what about this one x minus y is greater than or equal to negative

1. well let's Graph this one real quick i would solve it for y so i'm just going to subtract x away from each side of the inequality and i would have negative y is greater than or equal to negative x minus one i would divide both sides of the inequality by negative one scroll down get a little room going so i would have y this would flip all right i got to flip the direction of the inequality right now it's a greater Than or equal to so i'm going to flip that direction and now it's

going to be a less than or equal to and then negative over negative is positive so this is just positive x negative over negative is positive so plus one so y is less than or equal to x plus one so i'm looking at y is less than or equal to x plus one now again i'm going to graph my boundary line to start the boundary line is found By just replacing this inequality symbol with an equal symbol so i'm going to graph y equals x plus 1. and in this situation we have a non-strict inequality

so we want our boundary line to be a solid line so let's go ahead and graph this so the y intercept would occur at zero comma one so that's right there and then our slope is one so we go up one into the right one so we share a point here up one into The right one up one into the right one and so we have a few points going so let's go ahead and graph that guy okay remember this is a solid line so we're going to keep it that way and then it's a

less than or equal to so that means i want to shade below the line so let me use this kind of color i'm going to use an orange so i would shade everything that goes below this line Now i know there's a lot of stuff going on in the graph it's kind of busy but what is the solution for the system well recall that the solution for a system of linear equations is the single point that lies on both lines so in this case since we're dealing with inequalities it's going to be the region of

the coordinate plane where the two graphs are going to overlap so if i take my highlighter here Where is that going to be well it's got to be on this line starting there and i know i might have went over a little bit and then kind of following down this path here kind of all this area all this area and it continues forever and ever and ever but i'm just going to highlight the best that i can to kind of give you an idea here let me just kind of fill this in So you can

see that let me kind of figure out a different color here you can see that this part right here this area right here it satisfies one of them so let me highlight this in blue this satisfies one of them but not both of them so that's the why it's not part of the solution so let me go ahead and just erase all that so this wouldn't be part of the solution and this area up here is the same as Well this would satisfy one of them but not both of them so that's not part of

the solution either so i know it went over the line in some of the cases but you can essentially see where the two graphs overlap that's the part of the coordinate plane where y is less than 4x minus 2 and also y is less than or equal to x plus 1. so that specific area satisfies the system alright so pretty easy overall let's Take a look at another one we have x minus 2 y is less than or equal to negative 2 and we have y is less than or equal to 3 halves x minus

1. so this guy right here is already solved for y so i don't need to do anything to that i can just take it down and graph it for this one i want to solve it for y first so let's pull this one down and do a little extra work to begin We have x minus 2y is less than or equal to negative 2. so i would subtract x away from each side of the inequality and i'll have negative 2y is less than or equal to negative x minus 2. now to get y by itself

i'm going to divide both sides of the inequality by negative 2. so this guy is going to flip instead of it being a less than or equal to it's going to become a greater than or equal To so this is going to cancel and become y is greater than or equal to negative over negative is positive so this would be 1 half x and then negative 2 over negative 2 would be plus 1. so i've got y is greater than or equal to 1 half x plus one so y is greater than or equal to

one half x plus one so my boundary line for this again remember all you need to do to make a boundary line is replace this guy with an equal sign so i would just graph the Line y equals one half x plus one the only thing i need to know is am i dealing with a strict inequality or a non-strict inequality well this is a greater than or equal to so it's a non-strict inequality so we want a solid line so to graph y equals one half x plus one go to zero comma one which

is right here and the slope is one half so up one over to the right two up one over the right two Or i can go down one to the left two okay remember this is going to stay a solid line i don't need to break it up at all and then since y is greater than right it's a greater than or equal to but it's a greater than i would shade here above the line so let me shade above the line okay so that's y is greater than or equal to one half x plus

one now we had another one we had y is less than or equal to three halves x minus One so y is less than or equal to again three halves x minus one so to graph the boundary line for that again i just replace this with equals so i have y equals three halves x minus one and again this is a non-strict inequality so we're going to have a solid line so my y-intercept occurs at 0 comma negative 1. so it's going to be right Here and my slope is 3 half so i'm going to

go up 1 2 3 and then to the right 1 2. up one two three to the right one two or i can go down one two three to the left one two so this is a less than or equal to so i'm going to shade below the line so i'm going to shade below the line here so where do these two graphs overlap Well for this one compared to our last example is kind of a small area remember for this first one this purple line it's above right it's above that's the solution for this

blue line it's below the stuff in orange it's going down so the overlap would be this area right here and i know i'm going over the line a bit but it would be this so that's where the overlap is anywhere outside of that you're not going to be In the solution range for the system all right let's take a look at one final problem we have y is greater than or equal to 2 and then we have 3x minus 2y is greater than or equal to 2. so this one's easy enough to graph this one

right here i just want to solve it for y so i'd have negative 2y is greater than or equal to i would subtract 3x away from both sides of the inequality so i'd put negative 3x Over here then plus 2. divide both sides of the inequality by negative 2. and remember if i do that i've got to flip the direction of the inequality so right now it's a greater than or equal to we're going to change this to a less than or equal to so negative 2 over negative 2 is 1 so this would be

y this is less than or equal to negative over negative is positive so 3 halves x And then 2 over negative two is minus one so y is less than or equal to three halves x minus one okay so for the boundary line i just replace this with equals so y equals three halves x minus one and it's a non-strict inequality so my boundary line will just be a solid line so i'm going to go to zero comma negative one so that's right here that's my y intercept And my slope is three halves so up

one two three to the right one two up one two three to the right one two or down one two three to the left one two again this is a solid line so no need to take the eraser out and it's y is less than or equal to so with a less than or less than or equal to i shade below the line so i would shade everything going this Way now the other one we had was y is greater than or equal to two so y is greater than or equal to two so the

boundary line there is that y equals 2 and again non-strict inequality so it's a solid line so to graph y equals 2 i just find 2 on the y axis and i draw a horizontal line okay and then this is a greater than or equal To so i shade above the line so i'm going to shade everything going above the line and so where's my area of intersection where do the graphs overlap well it's this area going like this everywhere i'm highlighting here would be the area that's a solution for both so in real life

this is an infinite area but just to get a picture of it we can see what we have here and i went a little bit over the line here it's got To be exactly on that green line and on that purple line those would be part of the solution for the system but nothing above that okay so there's your solution for the system that area that is highlighted hello and welcome to algebra 1 lesson 26 in this video we're going to learn about the product and power rules for exponents so back in pre-algebra we learned

that whole number exponents One or larger were used as notation to display repeated multiplication of the same number i want you to recall that the number being multiplied is known as the base whereas the number of factors of the base is known as the exponent so for example let's say we had something like three to the fifth power three The bigger number is known as the base this again is the number that's going to be multiplied by itself then this little guy up here this 5 is known as the exponent this again tells us how

many factors of the base that we have so three to the fifth power is equal to three times three times three times three times 3. you have 1 2 3 4 5 factors of 3. as another kind of Refresher let's say i had 7 to the fourth power again 7 the bigger number is the base and then 4 the little number up here is the exponent and again this is telling me i have 4 factors of 7. so this is 7 times 7 times seven times seven one two three four factors of seven now we

can also do this with variables and in pre-algebra we didn't talk about Variables at all but it would be the same thing if i said okay i have x to the sixth power x would be the base and six would be the exponent so what would this be equal to well i would just have six factors of x so x times x times x times x times x times x one two three four five six factors of x so in a short time we're going to start Working with polynomials and before we get to this

topic we have to have a complete understanding of how to work with exponents it is very very critical to your success with polynomials so the first thing i'm going to teach you is called the product rule for exponents so it's very very simple and it's very very intuitive we have a to the power of m times a to the power of n Is equal to a to the power of m plus n i know some of you will have a glazed overlook on your eyes right now but essentially something you'd see in your textbook to

kind of translate into something you can understand when we multiply two numbers or expressions in exponent form with the same base so that's the key here with the same base here i have a and then i have the same a here so what we do is we keep the Base the same so i don't change a at all and then i just add the exponents so you notice how the exponent here is m and the exponent here is n and we have m plus n right there so even if that doesn't make sense right away

let's take a look at some examples and kind of let it set in a little bit so we have 2 to the fourth power times 2 squared so again we're looking for the same base so i have a base of 2 And i have a base of 2. so it's the same base it's got to be the same number if one is 2 and 1 is let's say 8 that's not going to work the base has to be the same so what i'm going to do is i'm going to keep that base the same so

i'm not going to change it so it's going to stay as a 2. and then i'm just going to add the exponents so i have an exponent here 4 and i have an exponent here of 2. so i would do 4 plus 2 Which is 6. so this is 2 to the 6th power now where does this come from let's think about this for a second if i have 2 to the fourth power that's 2 times 2 times 2 times 2. it's 4 factors of 2. now if i multiply that by 2 squared i'm essentially

just multiplying it by Another two factors so two times two so this is two factors two factors of 2. so to figure out how many total factors i have i just add the number of factors i have here which was 4 given to me by this exponent so the number of factors i have here which is 2 given to me by this exponent so doesn't it make more sense just to go through and just keep the base the same the number that's being Multiplied by itself and just add the exponents right because that's going to

tell me how many total factors of this base that i'm going to have in the end this is a lot easier to do than writing all this out every time and saying okay this is 2 to the sixth power and you can go through and evaluate this if you want you don't have to usually at this stage of the game they just want you to write 2 to the 6th power But 2 to the 6th power is 2 times 2 which is 4 times another 2 which is 8 times 2 again which is 16 times

2 once more which is 32 and times 2 one more time which is 64. and by the way by the time you get out of algebra 1 you're going to know up to 2 to the 8th power by heart because you're going to use it so often all right let's take a look at another one we have 3 to the fifth power times 3 squared So again i'm looking to see that i have the same base so this is the same as this so there's my base and there's my base here's my exponent and here's

my other exponent now the rule tells me if i have the same base which i do here i'm going to keep the base the same so just write a 3 and then i'm going to add the exponents so i'm going to add five plus two so It'll be three raised to the power of five plus two which is three to the seventh power and again one more time to get a complete understanding of this if i had three to the fifth power i'd have three times three times three times three times three again five factors

five factors of three now if i multiply by three squared i'm just multiplying by Three times three so this would be two factors two factors of three so how many total factors of three do i have well i have seven right i have five factors here plus i have two factors here which is exactly what i did up here so i have 3 raised to the power of 5 plus 2 or 3 raised to the 7th power and again you can go through and evaluate that if you want but that's not Really what we're trying

to do we're just trying to kind of simplify it and put it in this form usually in your book they're not going to say hey evaluate 3 to the 7th power they're just going to say what is 3 to the fifth power times 3 squared write that using an exponent right so that's what we've done we have 3 to the 7th all right let's look at one with a variable so we have x squared times x to the fourth power So is there anything that's going to be different here no so i'm just going to

do the same thing i'm looking at the base the base is the same i have an x and i have an x so when i do this multiplication the base stays the same i have x and then i'm just going to add the exponents i have a 2 here and a 4 here so 2 plus four is six so this is x to The sixth power all right let's take a look at another one we have z to the eighth power times z times z cubed so can we extend this to multiplication where now we

have three of these guys involved well yes we can i know the original definition said a to the m times a to the n is equal to a to the power of m plus n but it doesn't matter i could go through And put z to the fourth times z to the ninth times z to the twelfth i can keep going and going and going as long as the base is the same okay as long as the base is the same there's really no limit so base is the same i keep it the same and

then i just add the exponents so i have an exponent here of 8 then plus here i don't have a visible exponent so what am i going to put and this is very Very important if you don't have a visible exponent the exponent is 1. because i have z to the first power there or one single z and it's very important that you put that because some students will skip that over and just put 0 there well no you have to account for that factor of z in the end result right so If you put

a 0 there you will be one short now this one is z cubed so i'm going to put plus 3. so then i would have z to the power of 8 plus 1 is 9 then 9 plus 3 is 12. and again if you wanted to write that out and see that you would be short if you left this one off go ahead and write out 8 factors of z you'd have z times z times z times z times z times z times z times z that's Eight one two three four five six seven eight

then you have one factor of z so times z and then times you have three factors of z z times z times z so as a total you have eight here you have one here and then you have three here so that's eight plus 1 plus 3 or 9 plus 3 which is 12. if i had not counted this one If i had said okay well there's there's no exponent there so i just i'll leave it off well then you're not going to count this and you're going to have 8 plus 3 which is 11

it's going to give you the wrong answer right so this is an understood exponent of 1. make sure that if you come across that situation you account for it okay so the product rule for exponents very very simple very very easy to use Just practice it and you'll memorize it very very quickly so the next thing we want to look at are some power rules and kind of the first one would be the power to power rule so that's where you have something like a to the power of m so a is already raised to

something and then you raise that whole thing to another power so a to the power of m is now raised to the power of n you can see this is equal to a Raised to the power of m times n so again i don't like to give these kind of textbook definitions but essentially you're just keeping the base the same so this original base stays the same so that's the same and then you're just multiplying the exponents so something like 3 squared raised to the fourth power again keep this original base 3 the same so

keep this the same and then just multiply the exponents so I would have 2 times 4. so 3 raised to the power of 2 times 4 which is 3 raised to the power of 8. it's really really that simple you kind of think about why this works 3 squared 3 squared is being raised to the 4th power so that means i'm going to have 4 of these so times 3 squared times 3 squared times 3 squared so i already know from the product rule for exponents that we just learned i would Keep three the same

and i would have two plus two which is four plus two which is six plus two which is eight so three to the eighth power but another way that you could think about this is that this is going to happen four times so if i have this many factors that's gonna occur four times i can just multiply two times four would give me eight that's kind of the shortcut because in the end if i was to Completely simplify this i would end up with what two factors of three times two factors of three times two

factors of three times two factors of three or again three to the eighth power all right let's take a look at seven to the fifth power and then this is raised to the sixth power so again i'm just going to keep the base 7 the same And multiply the exponents so i'd multiply 5 times 6 and that's 30. so we get 7 to the 30th power what about if we had a variable involved is anything going to be different well no if i have x to the fourth power and then this is raised to the

third power or cubed i'm going to keep x the same that's my base and then i'm going to multiply 4 times 3 is 12. so i would get x to the 12th power what about z raised to the 9th power and then this is raised to the 8th power again z is going to stay the same multiply the exponents 9 times 8 is 72. what about y to the fifth power and then this whole thing is raised to the third power again keep the base the same so y is going to stay the same multiply

the exponents 5 times 3 is 15. All right so continuing with the power rules the next thing we want to talk about is raising a product to a power and this one can go back and forth so sometimes you'll need to simplify you'll start out in this format and you need to go to this format sometimes you start out in this format and you want to go to this format you can go back and forth between the two it doesn't really matter so if i have this product here which is a times b so That

product and it's raised to the power of m i can split it up and say this is equal to a to the power of m times b to the power of m so in other words if i have 2x and this is raised to the power of 2 or it's squared i can break this up and say that i have 2 squared times x squared Right all i did was take this and make it an exponent on each one of those individually now i could further simplify this if i wanted to because i know 2

squared is 4. so i could write this as 2 squared times x squared like that or i could write it as 4 x squared and there's all kinds of manipulation that we're going to have to go through when we get to polynomials it's just Very important to know the rules so you know what's legal and what's not so here we have an example of something that's presented a little differently we have 4 squared times 3 squared so to simplify this if you get a problem remember i have the same exponent in each case so what

i can do is multiply the numbers or the bases together so four times three and then it's raised to that exponent That they each have which is two so i have four times 3 that's going to be squared and we know 4 times 3 is 12 so we could write this as 12 squared so again just different ways to kind of simplify things all right what about 5xy this is going to be raised to the 9th power so we could simplify this and say we have five to the ninth power times x to the ninth

power times y to the ninth power so kind of Separating this each one is going to be raised to the ninth power or i could go the other way if i got this one a test i could write it this way where i have 5xy raised to the ninth power because the exponent is the same in each case what about 3zq raised to the second power or 3zq squared well again i can write this as 3 squared z squared q squared what about negative 8 x to the Fourth power well one of the interesting things

here is you have this negative out in front so you can write this as you can write this as negative 8 to the fourth power times x to the fourth power now notice how i use parentheses around that negative that's highly highly highly important when you're working with exponents Particularly when you have an exponent that's even remember that from pre-algebra if i have something like negative 2 squared it's not going to be equal to negative 2 inside of parentheses squared these two are not going to be equal this right here is basically saying i have

the opposite of 2 squared so it's kind of like negative 1 times 2 squared which is negative 1 times 4 or negative 4. If i have negative 2 inside of parentheses squared this is saying i have negative 2 times negative 2 which equals positive 4. so pay close attention to that especially when we get into algebra because you're going to be expected to know this and most teachers are not really going to cover that again they expect you to kind of remember that from pre-algebra all right so let's take a look at our last power

rule So we have a over b so you can think of this as a fraction or division remember those are the same this is a divided by b however you want to think about it and this is raised to the power of m what we're saying is that this is equal to a that numerator raised to the power of m over b that denominator raised to the power of m and this again can go back and forth sometimes you'll need to write It this way sometimes you'll need to write it this way one thing to

note is that b cannot be zero and why is that the case b is in the denominator b is in the denominator and we cannot divide by zero a is undefined all right so we have eight over four and this is raised to the second power so we could rewrite this as 8 squared your numerator raised to that power Over 4 squared your denominator raised to that power and there's all kinds of different things you can do to simplify this you know if you wanted to you could have cancelled this with this to start and

had two so this is two squared or four but for the purposes of you know kind of learning this this is how we're going to simplify for right now and you can kind of go through and prove this to yourself We know 8 over 4 would be 2 2 squared is 4. so that's what this is equal to it's equal to 4. over here 8 squared is 64. 4 squared is what that's 16. well if i divide 64 by 16 i do get 4. so it's the same either way and of course it should be

this should be mathematically equal to this and when we go through and do the simplification in this matter we get four we go through and do it the kind of Original way where we divide eight by four to start then raise it to the second power we also get four so again it's the same either way what about five halves raised to the fourth power well again i would just take 5 the number in the numerator raise that to the fourth power and put this over 2 the number in the denominator Raise that to the

fourth power what about x over y raised to the 7th power again all i'm going to do is take x that's what's in the numerator and raise it to the seventh power put this over y which is in the denominator and raise that in the seventh power now generally speaking when we're working with variables and we have a variable in our denominator it's assumed that that variable in the denominator is not Allowed to be zero because that would be undefined but you can specifically specify that y is not equal to zero okay but again if

you don't put that it's logical to assume that that variable cannot be 0. all right what about z over 2q and this is raised to the 14th power so again whatever's in the numerator which is z that's raised to the 14th power and this is over What's in the denominator i have 2q so the 2 would be raised to the power of 14 and so would the q that would be raised to the power of 14 as well you can kind of start out by thinking about it as okay this in the denominator the 2q

is raised to the 14th power we've already learned that if we have that we can kind of separate the 2 and the q and raise each to the 14th power so again we End up writing this as 2 to the 14th power times q to the 14th power hello and welcome to algebra 1 lesson 27 in this video we're going to learn about integer exponents and the quotient rule so in our last lesson we learned about the power rules for exponents and we also talked about the product rule for exponents now we're going to expand

on this and start to learn about negative exponents The power of 0 and then the quotient rule for exponents so i want to start today's lesson out by just asking a simple question what can we do if we see something like 2 to the power of 0 or 2 to the power of negative 3. if i have something like 2 to the 4th power 2 to the 4th power i know that this is 4 factors of 2 2 times 2 is 4 4 times 2 is 8 times 2 is 16. Then if i decrease the

exponent by one and i have two to the third power now i have three factors of two that's eight if i have two squared again if i decrease the exponent by one i have only two factors of two or 4 and then i know that if i just have the number 2 that has an exponent that's understood to be 1 so that's just the number 2. so now i'm going to put 2 to the power Of 0 here and i'm just going to put question mark for right now i'm also going to put 2 to

the power of negative 1 and i'm going to put 2 to the power of negative 2. and let's see if we can use some basic reasoning to get some answers here so the first question i would ask you is do you notice any pattern here well Every time we increase our exponent by one you will notice that we take the result from the previous part and we multiply it by two so in other words to get the answer for 2 squared if i didn't know it i could just take the result from 2 to the

first power and i could multiply it by another factor of 2 and i would get the result for 2 squared so that would give me 4. What i could do again since i'm increasing my exponent by 1 i could just take this result which is 4 and multiply it by another factor of 2. that would give me 8. again if i go up 1 in terms of my exponent i go to 2 to the fourth power if i don't know what that is i can just take 8 which is the result from the previous one

and again multiply it by 2. 8 times 2 is 16. Now that's obvious because each time i increase my exponent by 1 i'm basically taking the previous answer and i'm multiplying by another factor of 2. what happens if i go down though now i'm going to be dividing right the opposite of multiplication is division so in other words let me kind of erase this real quick if i start out at two to the fourth power the exponent is four If i go down to two to the third power i've decreased my exponent by one so

if i didn't know what any of these were i could just start with 16 and say okay well i have 16 here i would just divide by 2 right i'm just going to end up with the number 8 because essentially what i'm doing is i'm saying okay i had 2 to the fourth power which is two times two times two times two And now i'm going to remove a factor of two because i'm decreasing my exponent by one so essentially i'm dividing by two so that i can cancel one factor out and now i have

2 cubed which is this right here so i'm dividing that previous result which was 16 by 2 and now i'm going to get to 8. now i'm going to get to 8. and i can keep doing this going all the way so down other words if i want 2 squared i take 8 I divide it by 2 and i get to 4. if i want 2 to the first power i take 4 i divide it by 2 and i get 2. if i want 2 to the power of 0 i take 2 and i divide

it by 2 and i get 1 okay i get 1. this is 2 over 2 and this equals 1. if i want 2 to the power of negative 1 i take one and i divide it by two so that's one over two or one half If i want two to the power of negative two i take one half and i divide it by two so one half divided by two or you can think of it as one half divided by two over one and we know this equals one-half times one-half which equals one-fourth let me

show you another example real quick and then i'm going to give you a few rules that you want to write down Let's start out with a base of 3 and do the same thing if i have 3 cubed or you could say 3 to the third power we know this is 3 factors of three or twenty-seven three times three is nine nine times three is twenty-seven what about three squared three times three is nine what about three to the first power we know that's three then 3 to the power of 0 Let's do 3 to

the power of negative 1 and 3 to the power of negative 2. again as i go from here to here as i decrease the exponent by 1 i just divide this answer by the base so in other words if i erase all this and i don't know any of it if i want to go from a known result of 3 cubed which is 27 to an unknown result of 3 squared all i do is take the result here and divide by 3. 27 divided by 3 is 9. To get to here 9 divided by 3

is 3. to get to here 3 divided by 3 is 1. again 3 to the power of 0 is one to get to here three to the power of negative one okay one divided by three is one-third to get to here one-third divided by three is one-ninth so in each example we saw 2 to the power of 0 was 1 and 3 to the power of 0 was 1 also so here's the rule that you want to Write down and memorize any nonzero non-zero real number that is raised to the power of 0 is 1.

so a raised to the power of 0 is 1 and a is any real number but a is not allowed to be 0. okay so a is not going to be 0. so any number you can pick doesn't matter what it is let's say i have 1032 if i raise it to the power of zero it is equal to one Because essentially what a to the power of zero is is it's a over a it's the same number divided by itself and we know that any non-zero number divided by itself is 1. so here are

some very very easy examples and i'm sure you're going to see stuff like this in your textbook so you have the number 168 raised to the power of 0 this is equal to 1. You have the amount negative 412 and this is inside of parentheses so everything here is raised to the power of 0 and so this is equal to 1 and then you have the opposite of 8 raised to the power of 0. again remember this rule so this is equal to negative 1 times 8 to the power of 0 8 to the power

of 0 is 1 so this is negative 1 times 1 which equals negative 1. pay close attention to that because for sure your teachers are going to try to trip you up with these and then of course the same rules are going to apply when we work with variables but again you cannot raise 0 to the power of 0. so we have xy this is raised to the power of 0. it's inside of parentheses so this is equal to x to the power of 0 times y to the power of 0 and each is going

to be equal to 1 So you get 1 times 1 or just 1. but kind of the shortcut is to realize that okay both are raised to the power of 0. so this whole thing is raised to the power of 0 so that's just going to be equal to 1. and again you're not going to see this at first but recall that you can't divide by zero and so x and y both have a limitation here neither one can be allowed to be equal to zero now what about something like z to the power of

0 y squared and Then this is all raised to the power of 0. well you don't need to go through anything you could do it individually if you wanted to you could do power to power rule and go okay well z to the power of 0 times 0 and then this is multiplied by y to the power of 2 times 0. in the end anything times 0 is just 0. so you can kind of skip this step because it ends up being z to the power of 0 Times y to the power of 0 which

is equals 1 times 1 or 1. but again the shortcut is to realize that everything here inside the parentheses is raised to the power of 0 so the result is just going to be 1. so i could have just as easily said okay z to the power of 0 y squared this everything's raised to the power of zero so this just equals one all right so for raising something to the power of zero very very simple Remember you raise any non-zero number to the power of zero it's just one okay just memorize that you're going

to use that a lot moving forward so let's talk about negative exponents so for negative exponents we think about these using reciprocals so i want you to recall that we had 2 to the power of negative 1 and this was equal to 1 over 2 or one-half We had 2 to the power of negative 2 and this was equal to 1 4. we saw 3 to the power of negative 1. this was equal to one-third we saw three to the power of negative two this was equal to one over three squared or one-ninth now i'm

gonna give you the rule here for any non-zero real number a okay a just stands for any non-zero real number And any integer n we have the following rule so a raised to the power of negative n is going to be equal to 1 over a to the power of n so you can just kind of substitute into this formula here and you'll never get the wrong answer so in other words a here In this specific example is going to represent 2 right there so then all i would do is plug a 2 in where

that is so kind of to do this over here i would have 1 over plug a 2 in for a and then here i see that i have negative n and then here i have positive n so in Other words this starts out as negative 1 it becomes positive 1 and 2 to the power of positive 1 is just 2. what we're doing here is we're taking the reciprocal of the base so let's say i had something like 2 to the power of negative 3. i take the reciprocal of that guy 2 and that's going

to be 1 over 2 or 1 half and then i'm raising 2 to the same exponent the only difference is instead Of it being negative we make it positive so 2 to the power of negative 3 is equal to 1 over 2 cubed or 2 to the power of negative 4 is equal to 1 over 2 to the fourth or 2 to the power of negative 12 is equal to 1 over 2 to the power of 12. it's not very complicated once you kind of memorize and do enough practice very very straightforward rule all right

so 4 to the power of negative 3. what do we Do here flip the base so this becomes 1 4 and then this exponent is now going to be positive it's still on 4 but it's just now just going to be positive so 4 to the power of negative 3 is equal to 1 over 4 to the power of 3 or 1 over 4 cubed 6 to the power of negative 2. again flip the base so 6 will now be 1 6 and then make that exponent positive so This is going to be 2. so

1 over 6 squared 4 to the power of negative 7. so again flip that base 4. so this is going to be 1 over 4 and then your exponent here instead of being negative 7 is positive 7. so this is 1 over 4 to the seventh power what about x to the power of negative 3 y squared well you can kind of separate these think about this as x to the power of negative 3 times y squared like that when you separate stuff with a Multiplication sign although you don't need it to imply multiplication it

makes it easier for you to kind of separate these two in your mind so this one right here is really 1 over x cubed right i have the base here which is x i take the reciprocal of that so i get 1 over x and then x is still going to be raised to a power i just make that power Positive so instead of being raised to negative 3 it's positive 3. so really what this becomes is 1 over x cubed times y squared which i can write as y squared over x cubed right because

y squared times one would just be y squared so the next situation you're going to encounter is a negative exponent in the denominator so what happens when your negative Exponent is in the denominator well all you really need to do is bring your denominator into the numerator and make the exponent positive so i'm just going to take this guy up here and the exponent is going to be positive so 5 comes up here the exponent negative 2 becomes positive 2 and this is times 1 which is still going to be 5 squared so 1 over

5 to the power of negative 2 is just equal to 5 squared and if you want To think about this using the rule that i just gave you let's suppose we wrote this out 1 over 5 to the power of negative 2. so let's say this is equal to 1 over 1 divided by 5 to the power of negative 2 is 1 over 5 to the power of 2 or 5 squared right because this guy we take the reciprocal so that's 1 over 5 make this guy positive so raised to the power of 2. now

i wrote this original 1 here as 1 Over 1 so we could divide with fractions so how do i divide fractions again this part would stay the same so i'd have 1 over 1 times the reciprocal of this guy which would be 5 squared over 1. at the end of the day it's 1 times 5 squared over 5 squared over 1 which is just 5 squared so again the shortcut to that process is you're going to take the number from the denominator that's raised to a negative power bring it into the numerator and then you're

going to Take that exponent that's negative and you're going to make it positive let's say i came across a problem like 3 to the power of negative 3 over 5 to the power of negative 3. so what is this equal to so the shortcut is to take what's in the numerator here bring it into the denominator so 3 goes into the denominator instead of the exponent being negative it's now going to be positive And then this 5 here is in the denominator bring it into the numerator instead of the exponent being negative make it positive

so this becomes 5 cubed over 3 cubed or i could write it as 5 over 3 that amount cubed now if you want to go about it the slow way you can certainly do that so 3 to the power of negative 3 is 1 over 3 cubed and this is over we have 5 to the power Of negative 3 is which is 1 over 5 cubed we know how to divide fractions the first one stays the same so one over three cubed we're going to multiply this by the reciprocal of one over five cubed which

is five cubed over one of course we get five cubed over three cubed so that leads us to this little simple rule here if we have a negative exponent move the base across the fraction bar and change the sign of the Exponent so in more mathy terms for any non-zero real numbers a and b and any integers m and n we can set up this a to the power of negative m over b to the power of negative n is going to be equal to i would take this b and bring it into the numerator

and raise it to a positive n put it over a which is going to go into The denominator and raise it to a positive m so as an example let's say i had something like i don't know 3 to the power of negative 6 over 2 to the power of negative 7. all i need to do is take 2 bring it into the numerator and make the exponent positive so this is 2 to the 7th up here Take 3 bring it into the denominator make the exponent positive so this would be two to the seventh

over three to the sixth so again if we have a negative exponent move the base across the fraction bar so that's why i move this 2 across the fraction bar so from the denominator into the numerator see how it crosses above the fraction bar and then you just change the sign of the exponent so it was negative and now it's Positive i did the same thing here i drag this below the fraction bar it was negative was negative and now it's positive so let's look at some quick examples here we have 4 7 and this

is raised to negative 1. so using our rules from our last lesson we know we can write this as 4 to the power of negative 1 over 7 to the power of negative 1. and then what we just learned is we can Take this 4 and drag it down here raises the power of positive one we can take this seven and drag it up here raise this to the power of positive one so this is seven over four just seven fourths and the same goes when you're working with some variables so for y to the

power of negative 3 over x to the power of negative 2 this variable x is going to come up into The numerator and i'm going to now raise it to positive 2 so that's x squared now this y is going to go into the denominator and i'm going to raise it to positive 3. so i get x squared over y cubed okay so now let's talk about the last rule that we're going to learn and it's known as the quotient rule for exponents so i want you to suppose that you need To simplify something like

eight to the seventh power over eight to the fourth power so with the knowledge we already have learned we could take eight to the seventh power and write it as eight times eight times eight times eight times eight times eight times eight this over eight times eight times eight times Eight all right we can cancel common factors between numerator and denominator we cancel this with this this with this this with this and this with this and i'm left with three factors of eight or eight to the third power but how can we do this more

quickly like i realize that the exponent that i have right here in the top minus the exponent that i have Right here on the bottom gave me the ending exponent so i could have started out with saying okay i have 7 factors of 8 and i'm going to be cancelling away four factors of eight so that's going to leave me with seven minus four or three factors of eight so if i saw something like seven to the ninth power over seven to the fifth power you guessed it i can take this number right here this

9 and i could subtract Away this number right here this 5 and i would get 7 to the 4th power right and to see that once more again if i write out 9 factors of 7 1 2 3 4 five six seven eight nine over five factors one two three four five what's gonna happen is i'm gonna end up canceling five factors of seven here in other Words if i think about this using the exponents i'm starting out with 9 and i'm subtracting away 5 i'm going to end up with 4. right cancel this cancel

this cancel this cancel this and cancel this this is what i'm left with i'm left with four factors of seven or seven to the fourth power so this leads us to the following rule for any non-zero real number a and any integers m and n we want to keep the Base the same and subtract m minus n so in other words if i have the same base here the same non-zero real number and they're involved in a division that looks like this all i need to do is keep the base the same so i would

keep a the same and i would subtract the exponents i would take the 1 in the numerator which is m and i would subtract away the 1 in the denominator which is n So for 5 9 over 5 cubed again this is the same base so i write 5 so i take my exponent the numerator which is 9 and i subtract away my exponent the denominator which is 3 and 9 minus 3 is 6 so this is 5 to the sixth power again you can prove that to yourself you can write out nine factors of

five and then underneath you can write out three factors of five and you can cancel three factors of five away from The problem you create there and you're end up with six factors of five which again is our result all right here we have x to the fourth power y cubed over x squared y so for x to the fourth power over x squared i would keep the base x the same then i would subtract exponents so i'll take the exponent in the numerator which is 4 subtract away the exponent of the denominator which is

2 and then i have y cubed over y so i'm going to put y down Keep the base the same then i'll have 3 minus this exponent will be 1. so 3 minus 1. so this equals x squared 4 minus 2 is 2 times y squared also 3 minus 1 is 2. so you could write x squared y squared or you could write x y this inside of parentheses squared so even when negative exponents are involved we follow the same rules so we have 7 to the power of negative 4 over 7 cubed so What

i'm going to do is i'm going to keep that base 7 the same and then i'm going to subtract the exponent in the numerator which is negative 4 minus the exponent the denominator which is 3. so this gives me 7 to the power of negative 4 minus 3 which is 7 to the power of negative 7 which we write as 1 over 7 to the seventh power all right let's take a Look at some practice problems to kind of wrap up the lesson so we have x y squared this is in parentheses and it's raised

to the power of negative two then this is over we have negative two times x y cubed times negative two times x to the power of negative one so let's start out in the numerator let's use our power to power rule to simplify So we have x raised to the power of negative two so that's x to the power of negative two and then we have y squared that's raised to the power of negative two so y stays the same we multiply the exponents 2 times negative 2 is negative 4. in the denominator i have

a negative 2 i have an x i have a y cubed and i'm multiplying this by another negative 2 And x raised to the power of negative 1. now before i simplify any further between numerator and denominator let me simplify this denominator here let me just erase this and make a new line so negative 2 times negative 2 is 4 or if you want to kind of keep with the spirit of exponents i could write negative 2 inside of parentheses squared like that then i have x Y cubed and we're multiplying this by x raised

to the power of negative one now this can be combined with this x to the power of one times x to the power of negative one is what well remember x to the power of negative 1 is 1 over x to the first power so really what this becomes is x over x which is 1. And you can verify this using your product rule for exponents this is x to the first power times x to the power of negative 1 well this is x to the power of 1 plus negative 1 which equals x to

the power of 0 which by definition is 1. so essentially these two would cancel out you could just erase them So i'm just left with this negative 2 inside of parentheses squared so i'm just going to go ahead and write as 4 times y cubed now to simplify this further in the numerator i have x to the power of negative 2 and then i have y to the power of negative 4. now to simplify i have a y cubed down here so again the quotient for exponents tells me that same base I can take that

base keep it the same which is y and i could subtract take the exponent the numerator which is negative four subtract away the exponent of the denominator which is three so then this is over four and this equals x to the power of negative two times y raised to the power of negative four minus three which is negative seven this is over 4 so this is going to be Equal to these are both going to get moved down into the denominator so i'm going to write the numerator as a 1. this guy right here again

like i just said is going to come down here and so is this guy so i have my 4 that's going to stay the same so when i drag this below the fraction bar the x stays the same but then the exponent's going to go from negative to positive then the y is going to go down There the exponent goes from negative to positive and because these two both went down there i'm going to write a 1 as my numerator about something like x squared y cubed times 2x squared y to the power of negative

2 over y to the power of negative 2 and then everything's inside of brackets and it's raised to the power of 0. well the shortcut here is to remember that everything is raised to the power Of 0 here anything raised to the power of 0 is 1. so you see that on a test you just put 1. but you can go through it and do the power to power rule for everything what's going to happen is every single thing in here is going to end up being raised to the power of 0 so you'll just

have 1 times 1 times 1 times 1 to a bunch of ones being multiplied by itself and then down here would be over one so you'd end up with just the number one Hello and welcome to algebra 1 lesson 28 in this video we're going to learn about scientific notation so one immediate application of integer exponents is scientific notation this will make it more convenient to write really large or really small numbers so before we get into scientific notation i want to review some things from pre-algebra so the first thing is that when Multiplying by

10 or 10 raised to a whole number one or larger the decimal point moves right one place for each zero in the power of 10. so for example if i take let's say the number five and i was to multiply five by let's say one hundred well this is equal to i could just write five as five point zero and i can move this decimal point one place to the right for every zero in this power of 10. so there's two zeros Here two zeros so this would go one put another zero in there two

places to the right and we'd end up with the number 500 and we all know 5 times 100 is 500 but that's kind of another way you can do it and you could also think about this as let's say 5 times i don't know let's say we had 10 000. so there's one two three four zeros so four zeros So again i could put equals 5.0 and i can move this decimal point one two three four places to the right and i would end up with fifty thousand right so five times ten thousand is fifty

thousand okay keep doing this let's do one more let's say we had five times one million so we have one two three four five six Zeros so i could put a five here point zero and i'm going to move this six places to the right so let me go ahead and add one two three four five zeros so this can go one two three four five six places to the right and i'm going to end up with the number 5 million all right so additionally i want you to recall that we have a shortcut to

evaluate 10 raised to a whole number that's one or larger So 10 cubed is what it's a 1 followed by 3 zeros 1 2 3. so if i multiply by 10 cubed or if i multiply by a thousand i'm moving the decimal point three places to the right because i have one two three zeros there all right so the exponent is three that gives me the number of zeros in that power of ten so ten to the fourth that's a one followed by one two three four zeros that's ten thousand ten to the ninth is

a one followed by one two three Four five six seven eight nine zeros so that is one billion so i have here basically if we multiply a number by 10 raised to a positive integer we move the decimal point to the right by the number of places equal to the exponent so for example if we have 7.3 times 10 to the fifth power this number right here i can just copy it 7.3 i'm multiplying by 10 to the fifth power So i'm multiplying by a power of 10 with what five zeros so the decimal point

just goes five places to the right so let me go ahead and add four zeros behind that three and i'm going to go one two three four five places to the right and my decimal point would be there so i'd end up with 730 thousand about 6.829 times 10 to the 4th again just copy this number right here 6.829 and i'm multiplying by 10 to the fourth So the exponent is 4. if i was to evaluate 10 to the 4th power it'd be a 1 followed by 4 zeros that power of 10 would tell me

to move this decimal point four places to the right if i'm multiplying by that so this would go one two three put a zero behind that nine four places to the right and so i end up with 68 290. so additionally we have a similar trick when dividing by 10 Or a positive integer power of 10. so dividing by 10 or a 10 raised to a whole number 1 or larger so we're going to move the decimal point one place to the left for each zero and the power of 10. now if i'm multiplying by

10 or a 10 raised to a whole number one or larger i'm just moving the decimal point one place to the right for each zero and the power of ten so it's left for division right for multiplication so for example i have 64 divided by 100 i'll take 64. Put a visible decimal point here i've got two zeros so this is going to go one two places to the left and i'll end up with point six four or a lot of people would like to write zero point six four what about one hundred twenty eight

thousand over one thousand one trick is to cancel zeros so in other words i can take this and cancel with this every time i'm doing that i'm removing a factor of 10. So i'm canceling this with this this with this and i'll end up with 128 over 1. but another way i could have done that another way i could have done that is just written 128 000. put my decimal point here i've got one two three zeros so i would move this one two three places to the left and again i'd end up with the

number 128 either way what about 64 times 10 to the Power of negative one we learned about negative exponents in our last lesson so this would be 64 times 10 to the power of negative 1 is 1 10 right it's 1 over 10 to the power of 1 so 1 10. and so essentially here you have 64 divided by 10 and so i can do this as saying okay i have 64. put the decimal point at the end and then move this one place to the left Because i have one zero here so this would

end up being 6.4 so let's revisit a problem real quickly we have 128 000 times 10 to the power of negative three we've done this a few examples ago but it just looked different we have a hundred twenty eight thousand times ten to the power of negative three is one over ten cubed or one over one thousand right so i can really write this as 128 000 Over 1000 or 10 cubed so really again i'm taking 128 000 and i'm taking my decimal point i'm moving it three places to the left one two three that's

going to give me the number 128. so for the next one i have 5.476 times 10 to the power of negative 4. so do i need to keep writing this as 5.476 times 1 over 10 to the fourth power i don't Because what i need to realize here is that i'm just going to take if i have a negative exponent on 10 okay something like negative 4 all that's going to end up happening is i'm going to take this value here and i'm going to move my decimal point 4 places to the left right if

it's a negative 4 if it was a negative 7 i'd move it 7 places to the left it was a negative 12 i'd go 12 places to the left so really all i need to do is write this number 5.476 and then look at that and say okay it's a negative so that means i'm going to go to the left and then buy four places right because the exponent is negative four so this would go 1 2 3 4 places to the left right there you'd have 0.0005476 as your answer and if you wanted to

you Could do that though the long way you could have written this as 5.476 times 1 over 10 to the fourth power i said okay i have 5.476 over 10 to the fourth power and said okay well i have four zeros in this power of 10 that i'm dividing by so yeah this is going to be 5.476 this is going to move 1 2 3 four places to the left but again the way i showed you a minute ago was much much quicker i don't have all this tedious stuff i've got to go through all

right now that we've reviewed everything let's talk a little bit about scientific notation so it's kind of easy to do this the first thing you want to do is place the decimal point to the right of the first non-zero digit in the number so if the number was let's say 672 the first non-zero digit means coming from the left and going to the right this is non-zero so it's going to the right of that first digit so that's where my decimal point is going to go if i had something with a bunch of zeros let's

say i had point zero zero zero zero zero two seven five six this is going to get moved to right There ignore the zeros you're coming from the left and going to the right so it's to the right of the first non-zero digit it's got to go right there okay so that's kind of your first step with scientific notation now the second thing you're going to do is count the number of places the decimal point was moved this will be the absolute value for the exponent on 10. So again count the number of places you

moved that decimal point so let's say i had something like 682 well my decimal point is here and i'm going to move it here right because i'm moving it immediately after the first non-zero digit so it's going to go there so it went one two places to the left so i'm gonna say okay the number that i'm looking for is a two that's going to be the absolute value for the exponent on ten or something like point zero zero Zero zero zero eight one two this is going to go here and i'm gonna move it

one two three four five six places to the right so the absolute value for the exponent on 10 will be a 6 in this case now step 3 the exponent is positive if the original number is larger than the new number formed then the exponent is negative if the original number is smaller than the new number formed And that's going to make sense once we start doing some examples and then for step 4 we just set up the scientific notation as a times 10 to the power of n where a is the number formed in

step one right where we had put a decimal point after the first non-zero digit then n is going to be the exponent we got that in step two right we counted the number of places that we had to move Our decimal point and in step 3 we figured out what the sign was going to be now i know these steps when you kind of just read them out loud and you don't really have an example it seems like man it's going to be really difficult you practice 10 or 15 of these you're going to really

have this down pack super super easy all right let's take a look at an example so we have 96 million and we want to write this in scientific Notation so the first thing is again you want to place the decimal point to the right of the first non-zero digit in the number so coming from the left going to the right the first non-zero digit is a nine so i'd put the decimal point right there so it would be 9.6 you'd have all these zeros after the six so that's one two three four five six so

one 2 3 4 5 6. but The question is do i want to write them well they don't add any value to the number so you don't have to you might be asking why don't they add any value to the number remember we have a decimal point here now so after your decimal point and the final non-zero digit the zeros don't matter they don't add any value so you can just erase them you don't need them you just have that number 9.6 now these two are not equal yet but they're going to be now the

next Thing i want to do is count the number of places the decimal point was moved well it was moved 1 2 3 4 5 6 7 places so my exponent on 10 the absolute value of it is 7. absolute value is 7. So i'm going to multiply this by 10 i know the absolute value of the exponent will be 7. but will it be positive or will be negative knowing the absolute value is not enough well again in step 3 i told you the exponent is positive if the original number is larger than the

new number here's the original number 96 million here's the number that we formed 9.6 so clearly this is bigger so the exponent would need to be positive so this is Going to be positive 7 and we're done right we've written it in the form described in step 4 which is a times 10 to the nth power now a couple of pointers for you here the first thing is that you don't need to go through this elaborate process every time to get your sign for your exponent just think about okay if i'm here 9.6 and i

want to get to here 96 million what i need to multiply by a power of 10 that's going to tell me to move my Decimal point to the right or the left remember if this is positive i'm going to be moving my decimal point to the right if it's negative i'm going to be moving my decimal point to the left i want to move this 7 places to the right i want to go 1 2 3 4 5 6 7 places to the right so i can get back this original number of 96 million so

96 million written in scientific Notation is 9.6 times 10 to the seventh power all right let's take a look at this one we have point zero zero zero zero zero zero zero so i'm going to move this guy to immediately after the first non-zero digit so after that eight so it's going to end up right there this is going to be equal to i'm going to have eight 8.5016 all the zeros that preceded that 8 i'm going to delete i don't add any value to The number anymore then i'm going to multiply this by 10

raised to what power well how far did i move the decimal point one two three four five six seven eight now do i want ten to the eighth power or ten to the power of negative eight well i clearly want ten to the power of negative eight because if i to go from this right here which is a larger Number to this which is a smaller number this decimal point has to move to the left so this needs to be a negative 8 right because i want that decimal point to go to the left so

if i have this number with this decimal point here 8.5016 and i multiply by 10 to the power of negative 8 this is going to go 1 2 3 4 5 6 7 eight places to the left put me right back where i started with this number that's point zero zero zero zero zero zero zero eight five zero one six what about seventy eight trillion nine hundred ninety four billion well again i want to put my decimal point immediately after the first non-zero digit so that's after that seven And then i'm gonna write the eight

the nine the nine and then the four these zeros i can just delete i don't need them they don't add any value to the number now because my decimal point is here now how far did i move the decimal point well it went from here to here so that's 1 2 3 4 5 6 7 8 9 10 11 12 13 places so times 10 To the power of do i want positive 13 or negative 13 well again this original number is bigger so in order to get from the smaller value to the bigger value

i need that exponent to be positive this is going to be positive 13. so this will be 7.8994 times 10 to the 13th power what about something like 0.321 well i'm going to move this decimal Point to right after the first non-zero digit first non-zero digit is a three so it's going to basically just go one place to the right so this is going to be 3.21 and then times 10 my exponent here is going to be negative 1. it went one place to the right so i need to move it one place to the

left to get back right so this is a bigger number The 3.21 than the original number the 0.321 so we need a negative exponent here again negative 1 tells me to take this decimal point and move this one place to the left so we end up with 3.21 times 10 to the power of a negative one all right what about 15 million 65 000 again put the decimal point immediately after the first non-zero digit so 1.5 and i can delete those zeros i don't need them And then times 10 raised to what power so we

went one two three four five six seven places to the left so in order to get back i need to go to the right the original number is bigger my exponent's gonna need to be positive so this would be 10 to the seventh power right positive seven so you'd end up with one point six times 10 to the seventh power all right Let's take a look at 0.0001095 so again move the decimal point to immediately after the first non-zero digit it's going to go right there so this would be 1.095 then times 10 raised to

what power well this went one two three four places to the right so then to get back there i'd have to go four places to the left so i want a negative 4 as my exponent again if the original number is smaller Than this number that i'm forming here the exponent is going to be negative right so i end up with negative 4. if i took my decimal point there and i moved it four places to the left one two three four and ended up back with what i started with so you get one point

zero nine five times ten to the power of negative four all right so let's try writing a few without exponents you want to go back and forth get as much Practice as you can and so this would be take this number here 1.9051 i'm multiplying by 10 to the fifth power so i'm moving five places to the right so let me add a 0 here this is going to go 1 2 3 4 5 places to the right and i'm going to end up with 190 510 about 3.56 times 10 to the power of negative

9. so 3.56 and i'm going to move this decimal point Nine places to the left again that negative is telling me to go to the left and the nine says nine spaces so let me go ahead and put in eight zeros here one two three four five six seven eight this is going to go one two three four five six seven eight finally nine and so i'm going to end up with .0000356 As my answer what about 1.205 times 10 to the third power well again just take this number 1.205 and we're going to move

the exponent 3 places to the right so 1 2 3. it's going to end up right there you're getting up with 1 205 as your answer what about 6 times 10 to the power of 0 well we know anything to the power of 0 is 1 so this is basically 6 times 1 or 6. and just for future reference this is How you're going to write a single digit number in scientific notation your teacher might say how do you write 4 in scientific notation well you put 4 times 10 to the power of 0 or

how do you write 13 in scientific notation 13 times 10 to the power of 0 because in each case you're just multiplying by 1 right anything to the power of 0 other than 0 is 1. so we can also perform operations with numbers in scientific Notation there are different ways to kind of go about it you have 1.17 times 10 cubed times 1.3 times 10 squared so kind of the simplest or the easiest way to do this is to say okay i have 1.17 times 1.3 kind of do that first that would give you 1.521

now 10 cubed times 10 squared you would use the product rule for exponents so you'd have 10 as your base that would Stay the same 3 plus 2 is 5. so this would end up being 1.521 times 10 to the fifth power we know this would be 1.521 move the decimal point five places to the right so let me put in two zeros here this would go one two three four five places to the right and so you're going to end up with 152 100. all right what about 2.7 times 10 to the power of

negative 2 is multiplied by 2 times 10 To the power of 2 or 2 times 10 squared well again you would multiply the numbers first 2 times 2.7 that's going to give you 5.4 and then with this one you have 10 to the power of negative 2 times 10 squared what is that really if you look at it well that's 10 squared over 10 squared or it's just 1. you can see that using your product rule for exponents you'd keep the base the same which is 10 and you'd raise it to the power of Negative

2 plus 2 which is 0. 10 to the power of 0 is 1. so this is really 5.4 times 1 which is 5.4 all right what about 1.8 times 10 to the power of negative 3 over 2 times 10 to the power of 1. so you can do 1.8 divided by 2 first that will give you 0.9 and then you would multiply this by use your quotient rule for exponents you have 10 to the power of negative 3 over 10 to the power of 1. So times 10 to the power of negative 3 minus 1.

so this would be equal to 0.9 times you would have 10 raised to the power of negative three minus one is negative four so then to completely finish this off you'd have point nine if i have ten to the power of negative four i need to move this four places to the left so let me put one two three four zeros in This would go one two three four places to the left so i'd end up with point zero zero zero zero nine as my answer hello and welcome to algebra one lesson 29 in this

video we're going to learn about adding and subtracting polynomials all right so we're going to start off today by just talking about some basic vocabulary so we previously learned the definition for a term so recall that a term is a number A variable or the product of a number and one or more variables so as you may recall in an algebraic expression terms are separated by plus and minus signs so for example we have 2x squared minus 7x plus 4. so this is a term this 2x squared and then we have a minus sign this

is a term this 7x And we have a plus sign and then this 4 is a term so notice how this is a barrier right between your terms so you have the term 2x squared the term 7x and then the term 4. all right so i also want you to recall that the number that is multiplying the variable is called a coefficient so for example if i have something like 3x to the 5th power The 3 that's multiplying this x to the fifth power is known as the coefficient or as another example you could have

let's say 7x squared 7 is the coefficient and again i just give you these definitions because i'm going to be referring to things later on i might say hey just add the coefficients together so you need to know what i'm talking about all right so now let's talk again about like terms So when we say like terms we mean terms with the same variable or variables raised to the same power or powers so the coefficients can be different but that's the only thing that can differ so let me give you some examples here we have

5x squared and we have negative 7x squared so these are like terms these are like terms again look at the variable this is an x and this is an x and then Look at the power that it's raised to this is raised to the second power and so is this the coefficients are different right if i highlight that in a different color this one is 5 and this one is negative 7 but that's ok again same variable or variables if if you have more than one raised to the same power or could be powers again

if you're dealing with one of those situations all right let's look at this one right Here we have 3x to the fourth power y squared then we have 9x to the fourth power y squared so again these are like terms if you look at the variables again we have 2 here now we have x and we have x we have y and we have y now you're looking at the powers we have x to the 4th power x to the 4th power we have y squared and y squared so these are like terms and let's

say you got a situation where You had something like 3 x to the fourth power y squared and then this differed by just a little bit let's say it was 9 x cubed y squared these are not like terms not like terms because this right here this is x cubed and this is x to the fourth power so everything in terms of your variable and the power that it's raised to has to be exactly the same it can't differ even by A little bit so in this example here where we have negative 2z and 4x

obviously those aren't like terms right they don't even have the same variable this one has a variable of z this one has a variable of x so these are not like terms all right so let me give you the textbook definition for a polynomial now and you're going to read this it's going To be a little bit confusing at first and hopefully i can give you a little insight into kind of how you can look at the examples and say this is a polynomial and this is not a polynomial so a polynomial in x or

could be another variable like y or z whatever you want is a single term or the sum of a finite right finite means it has an end amount of terms ax to the power of n Where a is any real number and n is any whole number so what they're saying is that we have a x to the power of n a is any real number so anything you can think of can be plugged in for a could be a negative could be a positive could be a fraction could be anything you can think of

can be plugged in there so any real number any Real number x is the variable so that doesn't have to be x as i just said up here could be y or z or w whatever you want to use for that and then n is the exponent on x and it says that it has to be a whole number so it can be 0 it can be 1 it can be 2 something like that so what you want to pay attention for you want to look for Negatives and you want to look for fractions and

things that are not whole numbers if you see that if you see that in place here you don't have a polynomial okay you don't have a polynomial so this any whole number so let me give you an example we have something here 3x squared okay and remember we're looking for ax to the Power of n so a can be any real number and so 3 here is representing a so that's fine then we have our variable x and it's squared remember n has to be a whole number 2 is a whole number so this is

fine this is a polynomial this is a polynomial so if somebody asked you on a test it's 3x squared a polynomial you would say yes But there's a further definition for this type of polynomial if you have a polynomial with a single term meaning it's just one term here 3x squared it's called a monomial okay so this has a special name monomial so let's say we saw something similar to this on a test let's say we saw something like 3x to the power of negative 2. Would this be a polynomial well the answer would be

no the exponent on x cannot be negative it has to be a whole number so 0 1 2 3 you know et cetera et cetera if you saw something like x raised to the power of one half that's not a polynomial okay that doesn't work so we'd have 3x to the power of one-half not a polynomial so this one yes this one no all right what about this 5x To the fourth power minus 7x squared just look at the variables and look at your exponents so 5 is okay anything can be the coefficient for the

variable remember that's any real number and we think about this as ax to the power of n so this can be anything the exponent on the variable has to be a whole number and in this case it is so this is okay so this one checks out and Then this checks out as well this is a 2. so you have a 4 and then a 2. so everything checks out so this is a polynomial now we have a special name for a polynomial with two terms only okay it's called a binomial okay so this is

a polynomial with two terms only again a monomial which we saw in the previous example is a polynomial with one single term a binomial polynomial with two terms all right what About a variation of that so we have 5x to the power of negative 4 minus 7x squared well no this is not going to work if i look at my exponent on x it's a negative 4. you are not allowed to have a negative as an exponent when you're thinking about a polynomial right it's got to be a times x to the nth power where

n is a whole number whole number This is an integer negative four so that's not going to work and later on we're going to see that if you have something like this really i could write this as 1. i could write it as 5 over x to the fourth power right using our rules for exponents that we've previously learned then minus 7 x squared and we're going to talk about these types of expressions they're known as rational expressions we'll see those here shortly but for Right now you just need to know that when you see

something like this it's not a polynomial all right so you might also have polynomials that have more than one variable involved same rule is going to apply you're basically looking at the variable parts and making sure that you don't have any negative exponents and any fractional exponents basically you want the exponents to be whole numbers only so if i look at negative x to the fifth power Y cubed plus 3x squared y minus 4 just look at the exponents so i have 5 a 3 a 2. this is a really a phantom 1 here so

i can put that as a 1. and then if we wanted to we could write this as 4 times x to the power of 0 y to the power of 0. we could do that remember something to the power of 0 is 1. so really this would simplify to 4 times 1. but this doesn't violate our definition of a polynomial because the exponents for the variables have to Be whole numbers 0 is a whole number so this would still be a polynomial so we have negative x to the fifth power y cubed plus 3x squared

y minus 4 this is a polynomial now if i see something like negative x to the fifth power y to the power of negative 3 plus 3x squared y minus 4. the difference here is that i have a negative 3 and so now because i don't have a whole number as my exponent this is not a Polynomial and again you might see a variation of this you might see something like negative x to the fifth power let's say y to the power of one third plus three x to the one-half power y minus four you

can't have fractional exponents either again everything has to be a whole number in terms of the exponent for you to have a polynomial all right so kind of the next thing we want to talk about is the degree of a Term okay so you're going to have a degree for your term and then you're going to have a degree of the polynomial okay they could be the same or they could be different just depends on the situation so the degree of a term okay the degree of a term is the sum of the exponents on

all variables of the term so for example if i want to look at 7x to the fourth power What's the degree of this term well if i look at the definition again the degree of the term is the sum of all exponents on all variables of the term so i just have an exponent of 4 here and that's it so the degree is 4 and that's for the term it's also the degree of the polynomial but again those could be different it just depends on the situation so the degree let me be more specific and

say The degree of the term is 4. now for this one we have 3x squared y to the sixth power so the degree here is going to be eight the degree of the term is eight and that's because i looked at the exponent here that's two and i looked at the exponent here that's 6. i just sum them 2 plus 6 is 8 so that's how i got Them so if i look at the next one i have negative 2xy cubed z to the ninth again i'm just going to sum my exponents now a common

mistake here you don't see anything written as an exponent for x but remember it's a one x to the first power is x so don't forget that because a lot of students will just go okay 3 plus 9 is 12 and they're done they forget about That x raised to the first power so you really have a 1 here a 3 here and a 9 here so 1 plus 3 is 4 4 plus 9 is 13. so the degree the degree of the term is 13. and what if i see a single number like 5

and someone asks you what's the degree of this term well remember as i just showed you you could write this as five times some variable raised to the power of zero So if i sum the exponents here i just have an exponent of zero and that's my degree so the degree of the term is 0. all right so now let's talk about the degree of a polynomial so the degree of a polynomial is the largest degree of any non-zero term of the polynomial so essentially what you're going to do is look at each term of

the polynomial find out which one has the largest degree and then That's the degree of the polynomial so for example if we have 8x to the fourth power minus 7x cubed plus 5 i look here i have a degree of 4 a degree of 3 and then a degree of basically 0 right i can put this as x to the power of 0. so i know that the largest degree of any term is 4 so the degree of the polynomial is 4. so this polynomial has a degree of four well again when you start getting

stuff with more than one Variable in it and you're going to get a lot of that in algebra two not so much in algebra one you really have to take your time and add all of them up and try to see what the largest degree is for any term sometimes you'll get like seven or eight of these in a row they're trying to trip you up and you know you'll have multiple multiple ones in each one so you really have to go through and take your time So for this one i just count the exponents

so i have a 7 and a 7. so 7 plus 7 is 14. so my degree for this term is 14. for this one i have a 5 and remember i have a 1. so you have to remember that so important that you don't forget so this is a 1 there so 5 plus 1 is 6. and then here if i see a constant the degree is always going to be 0 right because i could write this as x to the power of 0. so i don't really Need to put that the degree of 0.

obviously that's not going to be the winner unless i just had you know a single term polynomial that was just a constant so in this case 14 is the largest degree of any term so the degree for the polynomial here is 14. so this polynomial has a degree of 14. so what about something like a constant Like negative 10 well the first thing is how do we make this match the definition of a polynomial remember a polynomial you get something like ax to the power of n where a is a real number and n is

a whole number we could write negative 10 times again x to the power of 0 and this fits your definition a is a real number that's negative 10 x is there and then n has to be a whole Number 0 is a whole number and then if we were to simplify this it's just negative 10 times 1 which is what it's just negative 10. right so that matches up with our definition now for the degree for this thing again you're just looking at the exponent which in this case is a zero so the polynomial here

just has a degree of zero so this polynomial has a degree of 0. so just look out for that on a Test because you might see a single number like that and say well this doesn't have a degree there's no there's no variable to look at yes it does the variable is whatever you want it to be it could be x or y or z or whatever you choose raised to the power of 0. all right so kind of the last thing we're going to talk about before we get into actually adding and subtracting polynomials

when we write polynomials it is custom To write them in what's known as standard form so this means the term with the highest degree is first then followed by the next highest and so on and so forth so something like this 3x plus 7x squared minus 5x to the fifth so the term with the highest degree would be this one negative 5x to the fifth power right your exponent there is 5 this is 2 and this is basically 1. so This would be first so you'd have negative 5 x to the fifth power the next

highest you have a 2 here and you have a 1 here so this is the next highest so 7x squared and then this is the last term so plus 3x so this is that polynomial written in standard form again it's just placing it where the term with the highest degree or the largest degree is first then followed by the next largest and so on And so forth so we have this term has a degree of five then this term has a degree of two then this term has a degree of one right so on and

so forth all right let's take a look at one more of these so we have 9x plus 12x to the ninth power minus 4x squared plus 3x to the seventh power plus 5x to the fifth power so the highest degree is going to go first so if i look here i have a degree of one I have a degree of nine i have a degree of 2 a degree of 7 and a degree of 5. so 9 is the largest so i start by writing 12x to the ninth power and i'm just going to line

this out just so i know i've used it already and if i look at the numbers that are remaining 7 is the next largest so then plus 3x to the 7th power line that out then 5 is the next largest so plus 5x to The fifth power line that out then i have this negative 4x squared right that 2 is the next largest so minus 4x squared find that out and then this is what's left so plus 9x so i can just erase all this now and so 9x plus 12x to the ninth power minus

4x squared plus 3x to the seventh power plus 5x to the fifth power written in standard form is going to be Equal to 12x to the ninth power that's first plus 3x to the seventh power that's second plus 5x to the fifth power that's third minus 4x squared that's fourth and then finally in fifth place plus nine x so you can see how it goes in descending order you have an x to the ninth an x to the seventh and x to the fifth an x squared and then an x or an x to the

first power all right so now let's get to the meat of the lesson and this is actually the easiest part of This lesson once you get past all the kind of vocabulary stuff that you have to learn it's really really easy to add and subtract polynomials so to add polynomials we simply combine like terms we've been doing this since the beginning of algebra super super simple at this point so we have 2x plus 3 that polynomial plus 4 minus x that polynomial so you can add this vertically if you Want a lot of students like

to do that they'll go okay i have 2x plus 3 and i have you could rearrange this and say i have negative x plus 4 and you can kind of add like this because you're adding like terms so 3 plus 4 is 7 2x plus negative x or 2x minus x is x so your result would be x plus 7. or kind of more traditionally and what i just like to do you could just rearrange The terms so i could just put like terms next to each other so i'd have 2x plus negative x and

then plus 3 plus 4 and then just combine like terms 2x and negative x are like terms again you keep the variable part the same so the x stays the same and i just add the coefficients so here my coefficient because i just have a negative there is Negative 1. so 2 plus negative 1 or 2 minus 1 is 1 so i would have x or just x from combining like terms here so this would just be x and then obviously three plus four is seven so plus seven again super super easy all right so

we have five n squared minus n cubed and we're adding to this n cubed minus n squared so again just put your like terms next to each other so i have 5n squared and i have negative n squared so i'd have 5 N squared you can put minus n squared or you put plus negative n squared doesn't matter and then i have minus n cubed and i have plus n cubed okay so everything is written next to each other so these are like terms same variable same power and then these are like terms same variable

same power so again i'm just going to say 5n squared minus n squared is what that's 4n squared The variable part n squared stays the same and i'm just subtracting coefficients so the coefficient here is a five the coefficient here is basically a negative one five minus one is four so that's how i got that four there and then the variable part n squared stays the same then we have a negative n cubed plus a positive n cubed this is basically like saying i have negative one plus positive one that's zero right the variable part

would stay the same Zero times anything though is zero so it would basically cancel itself out so we'll just have four n squared as our answer all right now we're looking at negative 14x to the fourth power minus 5x cubed minus 5 plus 6x cubed minus 12x to the fourth power minus 1. so i'm just going to start out again by just writing my like terms next to each other so i have negative 14x to the fourth power i have minus 12x to the fourth power i have negative 5x cubed i have plus 6x cubed

then i have minus 5 and then minus 1. so these are like terms these are like terms and these are like terms so that's what i can combine so negative 14x to the fourth power minus 12x to the fourth power again x to the fourth power stays the same i just add the coefficients so negative 14 minus 12 is going to be negative 26 So this would be negative 26 x to the fourth power and then if i look at negative 5 x cubed plus 6 x cubed x cubed is going to stay the same

i just add the coefficients so what is negative 5 plus 6 that's 1. so this is going to be plus 1 x cubed or just x cubed and then lastly i have negative five minus one and we know that's going to be negative six so then minus six and this is already written in standard form kind of when i think about doing This process a lot of times i'll just set it up to where my results are going to be in descending order sometimes i forget and don't do that but if you notice that it's

out of order you can quickly just rearrange it right it's easy to tell okay this term is of the highest degree then this is the second highest degree you know so on and so forth so your teacher might not explicitly say it But you know for the terms of moving forward you always want to try and write your polynomials in standard form and that just means that okay i have x to the fourth power here all by x to the third power followed by you know x to the power of 0 right so on and

so forth all right now i have 11x minus 5x cubed minus 11 plus i have negative 7x cubed plus 3x plus 5. so if i wanted to i know that this x Cubed here is going to have the highest degree so i could start out with that one if i wanted to or i could just rearrange it at the end it doesn't really matter so i can just say okay i have 11x and then plus 3x and then i have negative 5x cubed and then plus negative seven x cubed or minus seven x cubed and

i have minus eleven and then plus five So i put the like terms next to each other and because i wrote it this way i'm going to have to rearrange my terms in the end so i put it in standard form so if i wanted to save myself a little bit of work i could have shifted this kind of down here right i could have started out with that okay so moving forward we have 11x plus 3x 11 plus 3 is 14 x stays the same so we get 14x And then we have negative 5x

cubed minus 7x cubed negative 5 minus 7 is negative 12. and then x cubed stays the same and then we have negative 11 plus five that's going to be negative six so minus six and this right now is again not in standard form because this is x to the first power this is x cubed so this one has to go there right this has to go in the leading position because it's The term that has the highest degree so we'd have negative 12x cubed out in front and then plus 14x in second place and then

minus six right highest degree second highest degree smallest degree all right let's look at one that's a little bit harder and you might not see this in your textbook but i wanted to throw this in there you're gonna get a lot of these in algebra two where you have multiple multiple variables you Might have five six variables in there with all kinds of coefficients and you know it's not any more difficult it's just the fact that it's more tedious and you have to pay close attention to what you're doing so we have six x to

the fourth power y squared plus eight x y squared minus two x y plus negative four x to the fourth power y squared plus six x y minus x cubed y squared plus negative x to the fourth power y squared Minus six x cubed y squared minus four x y so we're gonna pay close attention to what the like terms are so i'm gonna start out with we have six x to the fourth power y squared and every time i use something and put it over here i'm going to go ahead and line it out

just so i know i've used it we can go back and erase that later just to keep track of what's going on so then plus I have this term here in the other polynomial so that's a negative 4x to the fourth power y squared these are like terms we have x to the fourth x to the fourth we have y squared and y squared so those are going to be next to each other and then this one will be a like term also we have minus x to the fourth power y squared so i've used

those three now the next term in the first polynomial i have 8xy Squared so 8 x y squared any like terms with that i don't have an x y squared here i don't have an x y squared here no x y squared anywhere else so moving on now i have minus 2 x y so minus let me kind of write this down here minus 2xy so i've used everything from the first polynomial in the second polynomial i have plus 6xy Line that out in the third polynomial i have minus 4xy so i've used all those

okay now everything from the first polynomial is in here the second polynomial i still have a minus x cubed y squared and then i have that in the final polynomial as well so minus 6 x cubed y squared and so i've represented everything from all three polynomials in this where all the like terms are next to each other And once you've done that you can go through and erase your little you know lines that you put in there if you don't like to see that or you can leave it it doesn't really matter okay so

now i'm going to combine like terms so these three are like terms so just think about leaving the variable parts the same just add the coefficients so i have 6 negative 4 and negative 1 basically So six plus negative four or six minus four is two and then two minus one is one so this would be one or just don't have to put anything x to the fourth power y squared so i've taken care of these and then next i have this 8xy squared plus 8 x y squared nothing to match with that right no

like terms for that and then here i have negative 2xy i have positive 6xy and i have negative 4xy so negative 2 plus 6 is 4 4 minus 4 is 0. so what happened there those are going to be eliminated right they're going to cancel each other out because 0 times anything is just 0. so these you could line them out right they cancel then i'd have negative x cubed y squared minus 6 x cubed y squared these are like terms so i have basically negative 1 minus 6 which is negative 7. so minus 7

times x Cubed y squared and so this is our answer but is it in standard form let's take a look the degree for this term is 6. the degree for this term remember you have a 1 here is 3 and the degree for this term is 5. so these are not in descending order right this needs to get switched with this this needs to go here so to write this in standard form I would have x to the fourth power y squared then i would have minus seven x cubed y squared and then i would

have plus eight x y squared and once you get a little bit of practice with this it becomes very very easy all right so the next thing we're going to talk about is subtracting polynomials so this is just as easy there's just one additional step So to subtract polynomials change the minus sign to a plus sign and change the sign of each term in the polynomial that is being subtracted away very very simple and goes back to this fundamental rule if i have 2 minus 3 right we know that's negative 1 this is the same

thing as if i said it is two plus right change this to plus and then change this to its opposite it's the Same thing as two plus negative three it's still negative one so that's all we're going to be doing here using that same rule so if i have 5p squared minus 4p minus p minus p squared again you know this from earlier if i see a minus sign out in front of a set of parentheses in order to remove the parentheses again you think about this as just multiplying by like a phantom negative 1.

a lot of your teachers are Going to say that you change the sign of each term so instead of p that's going to be negative p this is going to be negative p instead of negative p squared you're going to change the sign this is going to be plus p squared so i would have 5p squared minus 4p and then minus p plus p squared okay so if you're removing parentheses you change the sign of each term inside The parentheses that's just the rule whether you're working with polynomials or or anything really so now i

can just set this up just like we did in the previous problems so i have 5p squared plus p squared put my like terms next to each other minus 4p minus p so these are like terms and these are like terms so 5p squared plus p squared that's going to be 6p squared and then negative 4p minus p that's Minus 5p so there's your answer 6p squared minus 5p again whenever we see a negative outside of parentheses a lot of teachers are just going to say make this a plus negative 1 and distribute the negative

1. that's going to remind you to change the sign of everything so this becomes negative 1 times 6x or negative 6x negative 1 times 5 or minus 5. so that's just a rule to remember just Moving forward again not just when you're working with polynomials but when you're just working in math in general negative outside of parentheses change the sign of each term inside the parentheses to drop the parentheses all right so 3x minus 6x plus 5. so again i'm going to change this to a plus and then i can think about this as a

plus negative 1 right each sign inside the parentheses is going to change so I'm going to have 3x plus negative 1 times 6x is minus 6x where i could drop the plus sign and just put minus 6x it doesn't matter and then negative 1 times 5 is minus 5. 3x minus 6x is negative 3x and then minus 5. and there's your answer negative 3x minus 5. what about something like negative 1 plus v cubed plus 12v and i'm subtracting away 14v plus 7 plus 8v cubed so again this part right here this first part stays

the same so we have negative 1 plus v cubed plus 12v but again if you want to save yourself a little bit of time you can put the polynomial in standard form to start so i could reorder this just kind of put the negative 1 behind everything and now it's in standard form and then i have this minus here so what i'm going to do is i'm going to Distribute i'm going to put plus negative 1 and i'm just going to distribute the negative to each term inside the parentheses so negative 1 times 14 v

is minus 14v negative 1 times 7 is minus 7 negative 1 times 8 v cubed is minus 8 v cubed okay so now again this first polynomial i put it in standard form so i'm just going to Follow that format so i have v cubed minus 8 v cubed i have plus 12v minus 14v i have minus 1 and then minus 7. and now i just follow this v cubed minus 8 v cubed that's going to give me negative 7 v cubed right 1 minus 8 is negative 7. so negative 7 v cubed and then

12v minus 14v and that kind of looks like a u let me Make that look a little better so 12v minus 14v is negative 2v and then negative 1 minus 7 is negative 8 or minus 8. and notice how this is already in standard form for me i have v cubed v to the first power and you can think of this as v to the power of 0. so it's in standard form so negative seven v cubed minus two v minus eight that's your answer all right now we're looking at four m to the fifth

power minus three m cubed Minus ten m and we're subtracting away 13 m to the fourth power plus 4 m to the fifth power minus 14 m cubed so again this first part stays the same so i have 4 m to the fifth power minus 3 m cubed minus 10 m and then look i have this minus sign that's out in front the parenthesis so just put plus negative one and i know this is kind of off Down to the next line so let me put the negative one over here actually make it a little

easier to see so the negative one we multiply this by 13 m to the fourth power so minus 13 m to the fourth power and we multiply it by 4 m to the fifth power so minus 4 m to the fifth power and lastly we multiply it by negative 14 m cubed so plus 14 m cubed all right so if i look at this i can Kind of combine like terms here i don't need to write everything next to each other that's going to save you a little time so 4m to the fifth power and

negative 4m to the fifth power those are going to cancel right those are opposites next i'm going to look at this negative 13 m to the fourth power negative 13 m to the fourth power there's no like terms to combine with this so we're just going to leave this out in front that's Going to be my term that has the highest degree so it's going to go first and i can cross it out as i use it and then i have negative 3m cubed and 14 m cubed negative 3 plus 14 is 11 so this

is going to be plus 11 m cubed cross it out as you use it and then minus 10 m and again these little tricks that i give you for you know crossing stuff out and using that these are things that a lot of people use just so that they don't get confused a lot of these Problems start getting longer and longer and more tedious and you need some kind of way to just keep track of what you've worked on and what you haven't right so that you get the right answer because the last thing you

want to do is completely understand something and just make some silly error and you know not get 100 on your test right you want to get the highest grade possible so we're going to end up with a result here of negative 13 m to the fourth Power plus 11 m cubed minus 10 m hello and welcome to algebra 1 lesson 30. in this video we're going to learn about multiplying polynomials so before we jump in and start talking about multiplying polynomials want everybody to recall the distributive property so have something like 6 times the quantity

4 minus 6 remember i can remove the parentheses here and write this as 6 times 4 6 times 4 minus a minus 6 times 6. so 6 times 4 is 24 minus 6 times 6 which is 36 and this gives me a result of negative 12. and i can see that by just working inside the parentheses first if i had done 4 minus 6 that would give me negative 2 and then i would have 6 times Negative 2 which would also give me negative 12. as another example if i have negative 7 times the quantity

five plus one again i can remove the parentheses by multiplying negative seven times five negative seven times five plus negative seven times one negative seven times one so negative seven times 5 is negative 35 plus negative 7 times 1 that's negative 7 and this gives me negative 42. again i could have found the same thing by just saying okay 5 plus 1 is 6 negative 7 times 6 is negative 42. so at this point we should be very familiar with the distributive property since we've been using it all throughout algebra 1 and most of pre-algebra

so now we're going to use the same technique to multiply a monomial right that's a polynomial that has just one single term Times a binomial right this is a polynomial with two terms so we're going to have 4x times the quantity 3x minus 4. so how are we going to do this we're going to use the distributive property remember i'm going to take this guy out here and i'm going to multiply it by each term inside the parentheses so 4x times 3x and then i have that minus there and then we have 4x times 4.

so for something like 4x times 3x how do We do that well you'd multiply the number of parts together the coefficients right what is 4 times 3 well that's going to be 12. then you'd multiply the variables together so i have x times x so i know from my rules of exponents that if i'm multiplying and i have the same base i would just add the exponents right i keep the base the same add the exponents so this has an exponent of one and so does this so x times x is x Raised to the

second power or x squared right keep the base the same so x stays the same and then add 1 plus 1 that's 2. all right then we're subtracting away now we have 4x times 4. so 4 times 4 is 16 and then x just comes along for the right so that's 16x so we end up with 12x squared minus 16x and when you're working with polynomials you always want to make sure that you don't have any more like terms that you Can combine i have an x squared here and i have an x here those

are not like terms so this is the best that we can do 12x squared minus 16 x all right for the next one i have 5x times the quantity 5x plus 1. and again i'm just going to use my distributive property so 5x times 5x 5x times 5x and then plus next i'm going to have 5x times 1. 5x times 1. So 5x times 5x 5 times 5 is 25 and then x times x again is x squared and then plus 5x times 1 is 5x so these two are not like terms right you have

x squared and you have x so that's going to be our answer 25x squared plus 5x what about something like 3p times the quantity 4p plus 2. again i'm going to use my distributive property so 3p times 4p 3p times 4p plus right this guy's going to go over There 3p times 2. 3p times 2 so 3p times 4p you do 3 times 4 that's 12 and then p times p is p squared then plus next we have 3p times 2 2 times 3 is 6 and then p just comes along for the right so

that's 6p so i end up with 12p squared plus 6p as my answer all right now i'm going to Make it a little bit more tedious we have a monomial that's being multiplied by a trinomial and a trinomial is just a polynomial with three terms you have one two three terms same principle here i'm going to take this guy right here multiply by each term inside the parentheses so we have negative 8n times this quantity negative 6n squared plus 10n minus 13. and i'm just going to say okay negative 8n times negative 6n squared So

negative 8n times negative 6n squared then plus i'll have negative 8n times 10n so negative 8n times 10n and then minus minus i have negative 8n times 13. and one of the things you can do here you can think about this as negative 13 right so you don't make a sign mistake Or you can have a minus you'd end up with minus a negative we know that would be plus a positive but just to make this easy on us i'm just going to put plus right here and i'm going to write this as a negative

13. okay so let's go through and crank out the multiplication so negative 8n times negative 6n squared negative 8 times negative 6 is first off a positive right negative times negative Is positive 8 times 6 is 48 and then if we have n times n squared well the base n stays the same and then we just add the exponents remember this has an exponent of one so one plus two would give me three so this is 48 n cubed then next i have negative eight n times 10 n well negative eight times ten is negative

80. so minus 80. and then n times n is n squared Again keep the base and the same and you add your exponents right each case has an exponent of 1 1 plus 1 is 2 so you get n squared okay then we have negative 8n times negative 13. so negative times negative is positive and 8 times 13 is 104. so this would be 104 and then don't forget the n so there's your answer you get 48 n cubed minus 80 n squared plus 104 n All right so now that we've seen how to multiply

a monomial which again is a polynomial with one single term times a polynomial and the specific examples we looked at we looked at multiplying by a binomial which is a polynomial with two terms and a trinomial which is a polynomial with three terms but that really doesn't matter because if i had a polynomial with four terms or five terms or whatever it would be the Same process i would take the monomial and i would multiply it by each term in that polynomial okay that's all we need to do we're using our distributive property now we

multiply two polynomials and neither is a monomial okay neither one is a monomial we multiply each term of the first polynomial each term of the first polynomial by each term of the second polynomial so let me show you that So let's say we have something like negative 4x minus 4 and we're multiplying this by 6x plus 8. so this is the product of two binomials and this is going to happen so often we have like a little shortcut for this called foil and i'm going to teach you that in the next lesson but for right

now i'm going to kind of show you a general method to do this so i'm just going to take each term of the first polynomial and multiply it by each term Of the second polynomial so i'm going to break this up into two different problems so i'm going to have negative 4x multiplied by this 6x plus 8. so i'll have negative 4x multiplied by 6x plus 8. and then i'm going to add to that i'm going to take negative 4 and i'm going to multiply it again by 6x plus 8. So this is each term

of the first polynomial being multiplied by each term of the second polynomial so again now i'm just going to use my distributive property just like i used in the previous example so for the first one i would do negative 4x times 6x so i'd have negative 4 times 6 that's negative 24 and then x times x is x squared then i'd have negative 4x times 8. so i know negative times positive is negative So i'm going to put a minus there 4 times 8 is 32 and then x will come along for the ride so

the result from this is negative 24x squared minus 32x all right now next we're looking over here i have plus and then negative 4 times 6x so negative times positive is negative 4 times 6 is 24 so i'd have 24 and then x comes along for the ride so Minus 24x and then i have negative 4 times 8. so negative times positive is negative 4 times 8 is 32. so i'm not done in the previous examples i never had any like terms or anything i would not report this as my answer because i have negative

32x and i have negative 24x those can be combined right they're like terms so what is negative 32 minus 24 well that's negative 56. so this is Going to end up being negative 24x squared minus 56x minus 32. that would be my answer all right let's take a look at another one we have 8x plus 8 times 7x plus 4. so again i'm going to take each term of the first polynomial and multiply it by each term of the second polynomial so i'm going to do 8x times x plus four So eight x times seven

x plus four and then plus next i'm gonna do eight times again seven x plus four all right so i'm going to do 8x times 7x 8 times 7 is 56 x times x is x squared the next i'd have 8x times 4. 8 times 4 is 32 so i'm going to put plus 32 and then x comes along for the ride so plus 32x all right the next i have 8 times 7x 8 times 7 is 56 and then x would come along for the ride so this is plus 56x and lastly i have

8 times 4 and that's 32 so plus 32 and again i have like terms that i can combine here so i have to do 56x squared plus 32x plus 56x is 88x and then plus 32 so that's my answer 56x squared plus 88x plus 32. all right for the next one i have a binomial multiplied by a Trinomial so nothing's going to change here i'm still going to multiply each term of the first polynomial by each term of the second polynomial so i have negative b minus 12 times 8b squared minus 9b minus 5. all

i'm going to do is i'm going to take this guy right here negative b and i'm going to multiply it by this whole thing here so negative b times inside of parentheses 8b squared minus 9b Minus 5. and then i'm going to put plus next i have this negative 12 okay negative 12 times this whole quantity so negative 12 outside of the parentheses and then inside of parentheses we'll have 8b squared minus 9b minus 5. let's go ahead and crank this out so i'll have negative b times 8b squared negative times positive is negative and

Then you basically have b times b squared that's b cubed so you would have 8 that comes along for the ride and then b cubed all right so negative 8b cubed and then the next one we'd have a negative b times a negative 9b so negative times negative is positive b times b is b squared so this will be positive 9 b Squared then i'd have negative b times negative 5. so negative times negative is positive you have b times 5 that's just 5b okay now moving on to this over here i'll have negative 12

times 8 b squared negative times positive is negative 12 times 8 is 96 so this would be minus 96 b squared then next i have negative 12 times negative 9b negative times Negative is positive 12 times 9 is 108. so this would be 108 b and lastly i have negative 12 times negative 5 negative is negative is positive 12 times 5 is 60. so we have negative 8b cubed plus 9b squared plus 5b minus 96b squared plus 108b plus 60. so where am my like terms here nothing to combine with this one But i have

9b squared and i have negative 96 b squared those are like terms and then i have 5b and i have 108b those are like terms so we want to simplify by combining like terms here before we report our answer so negative 8b cubed that's going to stay the same and then 9b squared minus 96b squared that's going to be negative 87 b squared And then i have 5b plus 108b that's going to be plus 113b and then finally plus 60. so there's your answer negative 8b cubed minus 87b squared plus 113b plus 60. all right

for the next one i have negative 11v minus 1 and we're multiplying this by negative 7v squared minus 8v plus 4. so again i'm going to take each term of the first polynomial and multiply by each and every term of The second polynomial so i'm going to take this negative 11v multiply it by this guy so negative 11v times you have negative seven v squared minus eight v plus four and then plus next i'm going to take negative one and multiply it by this guy as well so plus negative one times negative seven v squared

minus eight v Plus four so negative eleven v times negative seven v squared negative eleven times negative seven is seventy seven v times v square is v cubed negative 11v times negative 8v negative times negative is positive 11 times 8 is 88. v times v is v squared negative 11v times 4 would be negative And then 11 times 4 is 44 and then v all right then moving on over here i have negative 1 times negative 7 v squared it's basically just a sign change right you multiply by negative 1 you're just changing the sign

so this would be plus 7v squared negative one times negative eight v that would be plus eight v again just changing the sign negative one times four would be minus four again lastly just changing the sign All right so now let's look for a like term so 77v cubed nothing i can combine with that 88 v squared and 7v squared those are like terms these are like terms so what is 88 plus 7 well that's 95 so then plus 95 v squared and then for other like terms we have negative 44v and 8v those are

like terms so if we can combine those you'd end up with negative 36v So minus 36v and then lastly minus four so we end up with 77 v cubed plus 95 v squared minus 36 v minus 4. all right let's take a look at the next one so now we have a trinomial times another trinomial so this isn't getting more difficult it's just becoming more and more tedious right as you expand these things out and get you know polynomials with more and more terms this just Becomes more tedious and harder to keep track of you

know kind of where everything is all right so again i'm just going to multiply every term of the first polynomial by every term of the second polynomial that's all i'm doing so i'm going to take 4x squared and i'm going to multiply it by this polynomial here 6x squared plus 6x minus 6. so without going through and kind of creating all this you know kind of extra Work let's just do it and write our answer here so 4x squared times 6x squared 4 times 6 is 24. x squared times x squared is x to the

fourth power right x is the base that's common so that stays the same you add your exponents 2 plus 2 is 4. next i'd have 4x squared times 6x so 4 times 6 is 24 so plus 24 x squared times x remember this is x to the first power so x squared times x is x cubed Then next i have 4x squared times negative 6. so that's going to be minus and then 4 times 6 is 24 and then x squared all right so we're done with that one now we're going to do 5x times

this guy so 5x times 6x squared is going to be what 5 times 6 is 30. so plus 30 x times x squared is x cubed then 5x times 6x is 30x squared so plus 30x squared And 5x times negative 6 would be minus 30 x all right lastly i'm going to do 2 times this polynomial so 2 times 6x squared is 12x squared so plus 12x squared 2 times 6x is plus 12x and then 2 times negative 6 is minus 12. so again it starts to become just more and more tedious right it's not

necessarily harder it's just now i got a Lot of stuff that i got to keep track of so 24x to the fourth power nothing i combined with that i'm just going to rewrite that here 24x to the fourth power then i have some like terms here so i have 24x cubed and 30x cubed nothing else to combine with that so 24 plus 30 is 54 so this would be plus 54 x cubed and then next i have negative 24x squared i have 30x squared and i have 12x squared so these are all like terms these

are all like terms so what is negative 24 plus 12 well i know that's negative 12. and then if i had negative 12 and i had 30 that's going to give me positive 18. so this would be plus x squared now i've used all of these so now i just have let me mark this in a Different color negative 30x and 12x as like terms so negative 30x plus 12x is negative 18x and then lastly i have minus 12. nothing combined with that so as my answer here i get 24x to the fourth power plus

54x cubed plus 18x squared minus 18x minus 12. so what happens when we want to multiply three or more polynomials well it's just like when you're multiplying three numbers if i have something like three Times five times two i can multiply in any order as long as i multiply everything together so in other words i can do 3 times 5 that's 15 then 15 times 2 is 30. or i could switch this up and say okay i can do 3 times 2 times 5. so 3 times 2 is 6 6 times 5 is 30 or

i could do it as five times two times three that's another way to do it five times two is ten ten times three is Thirty you know as long as i multiply everything together it doesn't really matter the order right when we do multiplication it's commutative right so the order is unimportant so when i look at eight times this negative eight p plus five times eight p plus four i could multiply eight times this guy first or i can multiply eight times this guy first or i could multiply these two guys together first it really

doesn't Matter so kind of the easiest way for me to do it i just kind of go in order so i would start out by just saying okay i have 8 times negative 8 p that's negative 64 p then i have 8 times 5 that's 40. so plus 40. so now i have this guy times this guy and now it's a problem that's pretty easy to deal with so negative 64p would be multiplied by this guy this is going to multiply by this guy so Negative 64 p times 8 p what is negative 64 times

8 we know that's negative and then 64 times 8 is 512. so negative 512 p times p is p squared and then what is negative 64 p times 4 we know that's again negative and then 64 times 4 is 256 so negative 256 p and then we have 40 times this guy So 40 times 8p 4 times 8 is 32 and then put a 0 at the end that'd be 320. so plus 320 don't forget about the p and then 40 times 4 that's easy 4 times 4 is 16 put a zero at the end

it's 160. so now i can combine like terms here so i could put negative 512 p squared negative 256p plus 320p is 64p so plus 64p and then plus 160. so there's my answer Negative 512p squared plus 64p plus 160. now if you wanted to i'm not going to do in this video in the interest of time you can multiply 8 times this polynomial here 8p plus 4 first then multiply that product by this guy you're going to get the same answer either way so the order is not important again when you're multiplying and we

know that from basic pre-algebra all right for the last problem we have something that's going to be a little tedious we have negative 5x minus 2 times negative 3x minus 6 times negative 2x minus 1. so again if you're multiplying the order isn't important so i can go left to right i can multiply these two together first and then multiply it doesn't matter so what i like to do is just multiply these two together first get a product and then take that and multiply it by this so i'm going to start out by just finding

the product of these two So if i had that problem here i would have negative 5x times this right here this negative 3x minus 6 and then plus i'd have negative 2 times negative 3x minus 6. let's figure out what that is first so negative 5x times negative 3x that would be plus 15x squared negative 5x times negative 6 that would be plus 30x Negative 2 times negative 3x would be plus 6x negative 2 times negative 6 would be plus 12. so essentially what i'm transforming this problem into is this guy right here i'd have

15x squared combine like terms here 30x plus 6x's plus 36x and then plus 12 times this guy this negative 2x minus 1. so an easy way to think about the step That we're at now if i had 3 times 2 times 5 well i would do these two first if i wanted to 3 times 2 is 6 i'd have 6 times 5 now that's exactly what i did here i multiply these two together first and i have that product here times this kind of last one so now i have this product here times this kind

of last one here so the end result is third and again you can switch the order Around if you want it doesn't matter you're going to put the same result so now i've got to take each term of this first polynomial multiplied by each term of the second polynomial so i'm going to do 15x squared times this negative 2x minus 1 and then plus i'm going to have 36x times this negative 2x minus 1 and then plus i'm going to have 12 times this negative 2x minus 1. all right so 15x squared times negative 2x

that's going to give me negative 30 x cubed then 15x squared times negative 1 is negative 15x squared then 36x times negative 2x is going to be negative 72 x squared then 36x times negative 1 is going to be negative 36x and then lastly i have 12 times negative 2x that's minus 24x and then 12 times negative 1 is negative 12. so combine like terms here i have Negative 30x cubed i have negative 15x squared and negative 72x squared that's going to give me negative 87 x squared and then i have negative 36x minus 24x

that's going to be negative 60x and then lastly minus 12. so we end up with negative 30x cubed minus 87x squared minus 60x minus 12. hello and welcome to algebra 1 lesson 31 in this video we're going to learn about foil alright so in the last lesson we Learned all about how to multiply two or more polynomials together now in this lesson we're going to focus specifically on kind of an easy method to get the product of two binomials so that's what foil is foil is an easy method to find the product of again two

binomials so it's only going to work if i have something like x plus five times let's say two x minus one right those are each binomials we Multiply them together we can use this foil or something like x minus three times x plus two right two binomials being multiplied together all right so foil is an acronym and basically each letter stands for something so the f in foil stands for first terms so i'm going to start by multiplying my first terms together the o stands for outer terms so kind of my second step would be

to multiply the Outer terms together the i stands for inner terms so my third step is to multiply the inner terms together and then the l means the last terms or the final terms so my final step is to multiply the final terms together now when you first hear about foil it seems a little confusing you're like what are the first terms what are the outer terms once you do it let's say 10 or 15 times this becomes a method that You use for the rest of your life i go through every time i multiply

two binomials together i use foil because it's just really really easy to keep track of what's going on so let's start out with this first problem and the first way i'm going to do it is the way i showed you in the last lesson so i have 2r plus 7 times 2r minus 3. so in the last lesson i kind of took each term from the first polynomial and just multiplied it by the second so I did 2r times 2r minus 3 and then plus i would do 7 times 2r minus 3. and this is

not wrong right we can still use this so 2r times 2r is 4r squared 2r times negative 3 is minus 6r and then plus 7 times 2r is 14r and then 7 times negative 3 is minus 21. so i would combine like terms here we would have 4r squared and then negative 6 plus 14 would be positive 8 So this would be plus 8r and then minus 21. so you get 4r squared plus 8r minus 21. now if i had used foil i would get the same thing and let me kind of drag this answer

up here i'm going to erase everything else if i used foil f stands for first terms so you multiply the first terms together so this is the first one in this polynomial this is the first one in this One so 2r times 2r is 4r squared so 2r times 2r gives me 4r squared then i would do my outer terms my outer terms so what's on the outside this is on the outside this is on the outside so for my outer terms i'm going to do 2r times negative 3. so 2 times negative 3 is

negative 6 then times r i get negative 6r then i would do my inner terms so what's on the inside so this is on the inside And this is on the inside so i get my inner terms i get 7 times 2r 7 times 2r that gives me 14r and then lastly i do my last terms my l or my final terms so this is the final term here this is the final term here so i get 7 times negative 3 and that's negative 21. so i have 4r squared i have negative 6r i have

14r and i have negative 21. so if i combine like terms You would see that you get the exact same thing as you got here you would end up with 4r squared negative 6r plus 14r again would be plus 8r and then you'd have minus 21. so this foil method is not producing a different result it's just a convenient way to remember how to get the product of two binomials and again once you start using this you're going to love it something you're going to use for the Rest of your life all right let's take

a look at this one we have 5n plus 5 times 6n minus 2. so let's use foil so i want to start out with the f first terms so the first terms we have 5n times 6n 5 times 6 is 30. n times n is n squared now i want to use my outer terms so here's the outer term here's the outer term so my outside terms 5n times negative 2 would be negative 10 n Now i want to do my inner terms so i want to do 5 times 6n so these are on the

inside 5 times 6 is 30 then times n so 30 n then i want to do my last or final terms so i got 5 here is the final term negative 2 here's the final term 5 times negative 2 is negative 10 and so i've got 30 n squared minus 10 n plus 30 n minus 10. so all i need to do now is just combine like terms here negative 10 n plus 30 n Is 20 n so this gives me a final answer of 30 n squared plus 20 n minus 10. all right for

the next one i have 4x minus 5 times 5x plus 7. so again i'm going to use foil so first terms i have 4x and i have 5x or x times 5x is 20x squared then i'm going to move to the outer terms the o so that's 4x times 7. 4x times 7 is 28x So plus 28x then i'm going to move to my inside terms so i have negative 5 and i have 5x so negative 5 times 5x is negative 25x then i'm going to move to my last or final terms so that's negative

5 times 7 and that's going to give me negative 35. so all i need to do now is just combine like terms i have 28x minus 25x that's going to give me 3x so this would be 20x Squared plus 3x minus 35 as my final answer all right what about 7x minus 7 times 4x plus 2. again i'm going to use foil so my first terms i have 7x and i have 4x 7x times 4x is 28x squared my outer term 7x times 2 would be plus 14x my inside terms i have negative 7 times

4x that's negative 28x And then my last terms i have negative 7 times 2 that's negative 14. so we end up with 28x squared 14x minus 28x is negative 14x and then minus 14. so 28x squared minus 14x minus 14 is my answer all right let's take a look at another one i have negative x plus 2y times 2x minus y so i'm going to do my first terms negative x times 2x is Negative 2x squared i'm going to do my outer terms negative x times negative y is plus xy i'm going to do my

inside terms 2y times 2x would be plus 4xy and then i'm going to do my last terms 2y times negative y is going to be negative 2y squared so what can i do here to combine like terms i have xy and i have 4xy so those Are like terms so basically you think of this again as 1xy plus 4xy 1 plus 4 is 5. so this ends up being negative 2x squared plus this would be 5xy and then minus 2y squared let me kind of make this a little better again our answer is negative 2x

squared plus 5xy minus 2y squared all right let's take a look at negative 5x minus 3y Times 2x plus 4y so my first terms negative 5x times 2x negative 5 times 2 is negative 10 x times x is x squared my outer terms negative 5x times 4y that's negative 20xy my inner terms negative 3y times 2x that's negative 6xy and then my last or final terms negative 3y times 4y is negative 12 y squared so here are my like terms in the middle here negative 20 minus 6 would be Negative 26 so you get negative

26 x y so you'd have negative 10 x squared and then minus 26 x y and then minus 12 y squared so again my answer negative 10 x squared minus 26 x y minus 12 y squared all right for the next one we have 3x plus 5y times negative 8x minus 7y so 3x times negative 8x is negative 24x squared then 3x times negative 7y is minus 21xy then 5y times negative 8x is minus 40xy then 5y times negative 7y is minus 35y squared all right so now we want to combine the like terms in

the middle here so i'd have negative 24 x squared i'm going to make that a little better and then negative 21xy minus 40xy would be negative 61xy And then we have minus y squared so again the answer here would be negative 24x squared minus 61xy minus 35y squared all right so for this last problem here we can't use foil to get the product of three binomials but what we can do is we can use foil for the product of any two so if i wanted to start out by multiplying these two together i could use

foil here Then once i got that product i can multiply this guy whatever that product is by this guy again you can only use foil for the product of two binomials very important that you understand that so let's go ahead and use foil to multiply these two together so 8x times negative 5x that would be my first terms and that would give me negative 40x squared My outside terms 8x times negative 8y would be minus 64 xy my inside terms here negative 2y times negative 5x would be plus 10xy then my last or final terms

would be negative 2y times negative 8y which would be plus 16y squared now that i have this i want to combine like terms in the middle negative 64xy plus 10xy is going to be negative 54xy so i have this guy now which is the result of the multiplication of these Two binomials and then times this guy so i can't use foil here because i have a binomial times a trinomial oil does not work please don't try that you've got to use the original method that we talked about so you're going to take each term of

this and multiply it by each term of this so i'll do negative 40x squared times negative 5x that's going to be X cubed then negative 40 x squared times negative 5y that's going to be plus 200 x squared y then we'll go here so negative 54xy times negative 5 x that's going to be positive 270 x squared y then negative 54 x y times negative 5 y that's going to be again positive 270 x Y squared i'm going to take this guy and multiply it by this polynomial so 16y squared times negative 5x 16 times

negative 5 is negative 80. so that's negative 80. and then x y squared and then 16y squared times negative 5y that's negative 80 y cubed all right then this equals so 200x cubed and then let's take a look at this we Have x squared y x squared y so no other like terms remember for you to have like terms it's got to be the same variable raised to the same power so i've got to have an x squared and then a y to the first power so that occurs here and here it doesn't occur here

doesn't occur here or here so just these two are like terms so 200 plus 270 is 470 and then x squared y Now i have 270xy squared and negative 80xy squared so these are like terms these and only these can be combined together so 270 minus 80 is 190 so this would be plus 190 x y squared and lastly i have minus minus 80 y cubed nothing i can combine with that so my answer here is 200x cubed plus 470x squared y Plus 190xy squared minus 80y cubed hello and welcome to algebra 1 lesson 32.

in this video we're going to learn about special products so you might see the title of this lesson and say what what do we mean by special products well essentially in this section i'm going to give you some shortcuts for common binomial products this is going to be a huge time saver for you moving forward So i'm going to start out by just defining again what a binomial is a binomial is a polynomial with two terms so something generic like let's say x plus y and let's say we squared this guy okay let's say we

square this guy what is this equal to if i said give me an answer for this give me a value for this what would you say Well a lot of students make the mistake of just trying to distribute this exponent 2 to each term inside the parentheses they say okay well this is x squared plus y squared that's a very very common mistake in algebra and this is wrong this is not the right answer okay and to understand why this is wrong why it's not the right answer let's go back to The definition of squaring

something let's say i had the number 4 and i want to square 4. what does this mean it means i take this base 4 and i multiply it by itself or i take the base four and i have two factors of four so this is four times four or sixteen well by that same logic if i have this quantity here x plus y and it's squared well the base x plus y Needs to be multiplied by itself so this is x plus y times x plus y right i have two factors my exponent is two

of x plus y if i changed that exponent up and i said i have x plus y to the third power or cubed well then i'd have three factors of x plus y right you've got to expand this out very very important that you understand that moving forward because again trying to say that X plus y that quantity squared is equal to x squared plus y squared again this is a very very common mistake common mistake so now that we understand the basic logic behind squaring a binomial let's talk about how we can kind of

get a generic formula going and speed our work up so if we know that x plus y that Quantity squared again is equal to x plus y times x plus y let's go through and use foil remember we can use that to get the product of two binomials to get our answer we would do the first terms so x times x that's x squared then we would do the outside terms so x times y that's plus xy then we would do the inside terms y times x which is y x or x y doesn't matter

so plus x y Then we do the last terms y times y is y squared so plus y squared so if i combine like terms in the middle here i'm going to end up with x squared plus 2xy plus y squared now this result is not unique i can follow that same kind of format to get an answer when i have something that looks like this if i have a binomial that's squared i'm always going to end up with the First term squared okay so here i have x is the first term that ends up

squared plus 2 times the first term times the second term so we have two times the first term x times the second term y and then plus the final or last term squared right in this case i have y so y is squared you can kind of see i've written this out here the square of a binomial we have x plus y that quantity squared again it follows this format first term squared plus two times the First term times the second term plus the final term or the last term squared and you might not remember

that right away but you need to write that on a flash card and kind of write some notes on that just keep looking at it over and over and over again and this is going to be one of the things that saves you so much time let me give you an example let's say i just throw something at you like x plus 7 that's squared Without doing foil can you give me an answer well i can get an answer just by again using the general formula so i know it's the first term squared so x

squared plus two times the first term times the second term or times the last term so two times x times seven two times seven would be fourteen so this would be 14x and then plus the final term or the last term squared 7 squared is 49. so without doing foil i'm able to very quickly produce the result for the quantity x plus 7 squared and if you don't believe that that works let's let's expand this and i'll show you that i got the correct answer so x times x is x squared plus x times 7

that's 7x Plus 7 times x that's 7x again plus 7 times 7 that's 49 and this equals x squared plus combine like terms 7x plus 7x is 14x plus 49 that's exactly what i got up here by just using that general formula now there's another one that we didn't look at yet and the only difference between the one we looked at already and this one is that there's a minus sign so there's a slight modification it's The same thing to remember it's just that there's a minus sign in front of the 2xy so for example

if i had something like x minus 3 that's squared well following this formula i would have x squared minus two times the first term times the second term so two times x times three would be six x And then plus your second or your last term squared three squared is nine so i get x squared minus six x plus nine now again i could expand this and show you kind of the slow way that i got the correct answer this is x minus 3 times x minus 3. so i get x squared right my first

terms and then x times negative 3 is negative 3x then negative 3 times x is negative 3x and then negative 3 times negative 3 Is plus 9. if i combine like terms here negative 3x minus 3x is negative 6x so i get x squared minus 6x plus 9. now one thing i want to call your attention to i get a lot of students that use this formula and at first they're making sign mistakes so what i want you to understand here when you use this formula the signs are already kind of calculated for you so

don't worry about putting it in as Negative three just think about what number is here and what number is here or it could be a variable or whatever you have so whatever's in this position is here squared whatever's in this position and this position don't worry about the sign it's going to be 2 times this times this the sign comes into play here okay so if this is a negative you put a negative There if it's a positive you put a positive right if this was a plus i would just put a plus there but

don't go through and start thinking of this as negative three and jam up your multiplication and say okay well i have you know two times x times negative three you get negative six x and then you put minus there from the formula put minus the negative 6x and you get plus 6x please don't do that the sign is already Figured out for you you just need to be very very simple here and say okay i have this is my first value and this is my second value don't think about the sign just follow the general

formula all right so let's do some quick practice we have this quantity a plus b that's squared so again following the general formula i know i'm going to take the first term and i'm going to square it so a squared Then i have plus i'm going to take 2 times the first term times the second term so 2 times a b then plus i'm going to have the last term squared so plus b squared and again you can check all of these you just take this a plus b this quantity squared and you expand it

a plus b times a plus b this equals a times a is a squared plus a times b is a b Plus b times a is a b then lastly b times b is b squared if i combine like terms here i get a squared plus 2 a b plus b squared which is exactly what i had right there what about a minus b square again only difference is going to be that sign there so i'm going to have the first term which is a squared then minus right because of that sign two times The

first term which is a times the second term which is b and then plus the second term squared so b squared so a squared minus two a b plus b squared and again if you don't believe me take this guy and go ahead and use foil to expand this right it has a minus b times a minus b and verify that this is exactly what you get all right here's something a little bit more typical and This is where it's really going to save you the time so i have 2p minus 7 that quantity squared

so using my formula whatever's in the first position that's this whole thing right here okay the 2p is in the first position that's my first term so that's going to be squared so the 2 p is squared and don't make the mistake of just writing 2p squared the Whole thing is squared so the 2 is squared and so is the p so if i evaluate this i would end up with what 2 squared which is 4 times p squared another common mistake would be to say okay i just have 2p squared no the whole thing

has to be squared then i have the minus here and then times 2 times whatever's in this position so that's times two p Times whatever's in this position which is seven two times two is four four times seven is twenty eight so this would be minus twenty eight p and then lastly i'm gonna add whatever's in this position squared 7 times 7 is 49. so end up with 4p squared minus 28p plus 49. now again this is one that i feel that we should check so i could show you we got the right answer so

We've got 2p minus 7 that's squared so this is going to expand out to 2p minus 7 times 2p minus 7. 2p times 2p first terms that's 4 2 times 2 is 4 and then p times p is p squared okay exactly what we have here then the next thing we'd have is the outer terms 2p times negative 7 that's minus 14p Then we'd have your inner terms negative 7 times 2p that's minus 14p then the last terms negative 7 times negative 7 is plus 49. when we combine like terms here we're going to end

up with 4p squared minus 28p plus 49 which again is exactly what we got by using our shortcut all right let's take a look at another one we have 6k plus 1 this quantity is squared So again what i want to do is take the first term in this case that's 6k and i want to square it so don't just write 6k squared again i'm squaring that whole thing so if i square the whole thing you know a good practice is just put everything in parentheses the 6 and the k and square it that way

you know that the 6 is squared and so is the k so 6 squared is 36 And k squared is k squared okay then plus we have two times whatever whatever's in this position which is 6k 2 times 6 is 12 times k is just 12k and then times whatever's in this position which is 1. so 1 times anything is itself so this is just 12k and then lastly plus whatever's in this position squared so 1 squared is just 1. so we get 36k squared plus 12k plus 1. And again we can prove this we

have 6k plus 1 squared so this is six k plus one times six k plus one six k times six k is thirty six k squared six k times one is plus six k one times 6k is again plus 6k then lastly 1 times 1 is 1. so we get 36 k squared Plus 6k plus 6k is 12k plus 1. again exactly what you got there by using the shortcut let's take a look at 3x plus 2y squared and again all i'm going to do is whatever i have in the first position i'm going to

square that so the 3 has to be squared and so does the x again this is 3x that term is squared the whole thing please don't just write 3x squared that's wrong so 3 squared is 9 And x squared is x squared then the next thing i want to do is have plus two times the first term times the second term so two times three is six six times two is twelve so plus twelve then of course the variables just come along for the right so x times y so 12xy then lastly i have the

final term which is 2y squared again i've got to square the whole thing Whole thing is squared so 2 squared is 4 y squared is just y squared right please again don't just write 2 y squared a lot of students they try to do this as fast as they can and they make little small errors like that this formula is only good if you practice it correctly right if you don't do it correctly might as well just go through and do it the long way so we're going to end up with 9x squared Plus 12xy

plus 4y squared and again let me do this the long way and prove this to you so if i expand this i get 3x plus 2y times 3x plus 2y 3x times 3x is 9x squared 3x times 2y is going to be plus 6xy and then 2y times 3x again is plus 6xy and then finally plus 2y times 2y is 4y squared so if i combine like terms in the middle here i get exactly what i got from my Shortcut which is 9x squared plus 12xy plus 4y squared let's look at one final problem here

and we're going to move on to some other common formulas and we can use this with fractions as well we have 5 7 x minus 10 7 y this quantity is squared so again i'm going to take whatever's in the first position which in this case is 5 7 x and i'm going to square it so to square this guy If i square 5 7 5 times 5 is 25 7 times 7 is 49 so that would be 25 49 that's the square of 5 7 and then we have x squared and remember i have

this minus here so that's going to go right there and next i'm going to have 2 times whatever's in this position which is 5 7 and then times x i'm going to put these at the end for the variables and then times 10 7. whatever's in this position and then times y I'm trying to just see if there's anything i can cancel and there's not so you're going to have 2 times 5 which is 10 10 times 10 is 100 so this is going to be 100 over 7 times 7 which is 49 and then

times x y all right then lastly i'm going to have plus this guy right here squared so 10 7 y this is squared so 10 squared is a hundred Seven squared again is 49 and y squared is of course just y squared so there's your answer you get 25 49 x squared minus 100 over 49 x y plus 100 over 49 y squared and again if you want to you can take this guy and you can expand it out use foil and you'll see that you get exactly this as your answer all right so let's

move on now and talk about something else so another common scenario occurs when We multiply two binomials and one is the sum of two terms while the other is the difference of the same two terms now the result of this we're going to learn later on is called the difference of two squares once you start talking about things like this you're going to know what that means so the difference of two squares is something you're going to come across all the time kind of once You hit it in factoring moving forward you're just going to

keep talking about it over and over and over again so it kind of looks like this you have x plus let's say a times x minus a so the same terms are in the same position of each binomial so i have an x in the first position here and an x and a first position here i have an a in the second position here and then the second position here The only difference here is that i have a plus here and a minus here so when you do this what happens is the middle terms from

the foil cancel themselves out so let me show you that so x times x is x squared the outer terms x times negative a is minus ax the inner term a times x is plus ax so you can see those would cancel and then a times negative a is minus Okay minus a squared so what happens is the middle part is going to cancel itself out and you're left with x squared x squared minus a squared so in other words the first term here is going to be squared minus the second term which is squared

okay whenever you come across this scenario again this is referred to as the difference Of two squares right because you have subtraction of all that's where difference comes in and then each term is squared so the difference of two squares so here's kind of the generic formula that you probably see in your textbook again it's x plus y times x minus y and again the first position of each one is the same i have x here and x here the second position of each is the same i have y here and y here you've got

to Have that you can't have something like you know x plus three times x minus seven that that's not going to work you gotta have x plus three times x minus three or x plus five times x minus five or it could be you know z plus five times z minus five but the first term in the first binomial has to be the same as the first term and the second binomial the second term in the first binomial Has to be the same as the second term in the second binomial has to be the same

the only thing that's going to be different is the plus here and the minus here so you get this first term squared so x squared minus your second term squared remember the middle terms are going to drop out when you do foil because they're opposites so if i was to show you this one more time If we have something like x plus y times x minus y if we use foil x times x is x squared the outer x times negative y is negative xy the inner y times x is plus xy right you see

those are going to cancel and then the last y times negative y is minus y squared so this is going to go away in the middle here that negative xy plus xy is going to cancel and what i'm left with is just x squared Minus y squared all right let's take a look at some examples so i have x plus 3 times x minus 3. so again same thing here and here same thing here and here so i'm good to go so all i need to do is take the first term and square it and

subtract away the second term and square it so 3 squared we know would be 9. so this ends up being x squared minus 9. and again you can verify this just by using foil x times x is x squared Let me kind of write this out real quick then the outer x times negative 3 is minus 3x the inner 3 times x is plus 3x so those are going to cancel and then the last 3 times negative 3 is minus 9. right this cancels again you end up with x squared minus 9. so now we

have 6x plus 1 times 6x minus 1. so again the first term is the same in each case the second term is the same in each case Just the plus and the minus is what's differing here so then the first term is squared again don't just write 6x squared and move on very very common mistake you've got to square the whole thing the 6 and the x are both squared 6 squared is 36 x squared is x squared then minus i'm going to take that second term which is 1 and i'm going to square that

We all know 1 squared is just 1. so this ends up being 36x squared minus 1. and again go through and use foil and verify that i got the right answer right what about something like 3n squared plus 1 times 3n squared minus 1. so 3n squared 3n squared and then 1 and 1. so that works out so we're going to take our first term 3n squared and we're going to square it so think about this 3n squared and we're going to square it So 3 squared is easy that's 9. what is n squared squared

right if i have n squared that means i have n times n and let's say i square that guy well that means that i'm going to have this twice so i have n times n times n times n or n to the fourth power again this is found using the power to power rule so i just take the base and leave it the same and i multiply the exponents so 2 times 2 would be 4 so this is n to the fourth power then i'm subtracting away i have this term here the second term and

that's going to be squared 1 squared is 1. so we end up with 9 n to the fourth power minus one all right what about something like x plus one half y times x minus one half y well again i'm looking at what's in the first position here i have x and i have x What's in the second position i have one half y and one half y so we're good to go there so i'm going to take what's in the first position which is x and i'm going to square it and then i'm subtracting

away what's in the second position i'm going to square that so if i square one half y if i square this whole thing one half squared one squared is one two squared is four so this would be a Four here and we know that y squared is just y squared so we end up with x squared minus 1 4 y squared all right what about 9m squared minus 4n times 9m squared plus 4n so again 9m squared 9m squared and we're looking at 4n and 4n so we're good to go there so we'll just start

out we have this first term and that's going to be squared so 9 m squared is going to be squared so what's that Going to give me well 9 squared is 81 right 9 times 9 is 81. m squared raised to the power of 2 again use the power to power rule base m is going to stay the same multiply exponents 2 times 2 is 4. now i'm going to put a minus there my second term is going to be squared so i'm going to have 4n and that's going to be squared so the 4

and the n are both squared again very Important you don't make that mistake 4 squared is 16 and then n squared is just n squared so i'm going to end up with 81m to the fourth power minus 16n squared all right so lastly let's look at a binomial that is cubed and this formula is a lot more difficult to memorize the formulas i gave you up to this point are very easy you can memorize them after about a day's worth of practice This one you kind of have to work on but i'm going to tell

you that it's more rewarding and it saves you much more time to memorize these because when this comes up on an exam if you've got this down packed this is going to save you a solid let's say minute and a half of you cranking stuff out at minimum right so if you know these off the top of your head and you're confident in your ability to execute the formula this is The biggest time saver that you're going to have for your kind of basic algebra one algebra ii test that involves cubing a binomial so i

have that a plus b that quantity is cubed you get a cubed plus three times a squared b plus three a b squared plus b cubed so essentially you have the first term cubed then you have three times the first term squared then times the second term Then plus three times the first term times the second term squared plus the final term cubed now here's kind of the trick to remembering this you know that you're going to have your first and your last term that's going to be cubed okay so what i do is if

i can't remember i write those out first so let's say i had something like i don't know x plus 2 and that's cubed So i know i'm going to have this first one cubed so x cubed and i know i'm going to have the last one the last one which is 2 cubed so plus 2 cubed is going to be 8. let me write that way down there now in between i can remember the number parts here because the cube part is a 3. so i know i have 3 and space space And then plus

three again space space now all you need to remember is that in each instance one of them is going to be squared and one of them is not so in other words in the first instance this first term is squared so i get x squared and the second term is not so times 2 and 3 times 2 is 6. so i'm going to erase this and put 6 for now so 6x squared now in the last one i put x squared the first term so in this one I'm going to make the second term squared

so 2 squared is 4. so this is going to be 4 and then the other one is going to be to the first power so 3 times 4 is 12 so this is going to be plus 12x so we get x cubed plus 6x squared plus 12x plus 8. and that might not be the way you remember it but this is always if i get stuck for a minute i remember that okay i know these first two i know i'm going To have first term cubed second term cubed and i know that i'm going to

have two spaces each with a three and i know that each one of these is going to be used in those spaces one's going to be squared in one instance the other is going to be squared in the other instance so once i remember that i just kind of go through a process of you know writing it out and then do that three or four times and then i'm good to go Now for the other one it's a little bit more difficult to remember because of the minus sign here you gotta remember that you have

a minus sign here and then here so it's right before the second term and right before the final term so that's just something you have to remember so for example if i had something like x minus 2 that cubed Everything would be the same so i would have x cubed but then the signs would be different so this would be minus 6x squared plus 12x and then minus 8. all right so let's take a look at an example we have x plus 4 this quantity cubed so again i know that i'm going to have this

first one cubed and i'm going to have the last one cubed so let's write that way over here let's put Plus 4 cubed we'll go through and do the math in a little while we know that 4 cubed is 4 times 4 which is 16 times 4 again which is 64. we'll just leave it as four cubed for now now i know that i'm going to have plus something plus something else right let me kind of scoot this down a little bit now just think about this right here this is going to be my number

so 3 And 3. then just think about these two one of them is going to be squared in one case and the other is going to be squared in the other case so in other words in the first case it doesn't matter the order that you do this in so i could say in the first case i'll just square the first one so x squared and the second one won't be squared so times four Then in the second case the four is going to be squared so times four squared times x now this one's not

squared so again this is kind of how i remember it it's it's my thought process you can use it or you know you can come up with something on your own or maybe there's another tutorial out there that makes more sense for you but this is how i remember personally so then i go through and just simplify so then x cubed Plus 3 times 4 is 12 so 12 x squared plus 3 times we have 4 squared 4 squared 16 3 times 16 is 48 so that's 48 x and then plus 4 cubed which is

64. so that's your answer and you know if i didn't sit there and kind of write things out if i was on a test and kind of did that in my head i would probably have that answer in under 10 seconds right whereas if i went through and had to expand this right here this X plus 4 cubed let's just think about how long this takes you have x plus 4 times x plus 4 times x plus 4. well i could save myself a little bit of time because remember this right here is the same

thing as x plus 4 squared right so i already know what that is that's x squared plus 2 times 4 which is 8 times x so 8x Plus 4 squared which is 16. so i could save myself a little bit of time with that you know just using that formula so this is x squared plus 8x plus 16 but i've still got to go through and multiply this guy by this guy so i've got to do x times x squared so x times x squared is x cubed i've got to do x times 8x that's

going To be plus 8x squared i've got to do x times 16 that's plus 16x and then i've got to do 4 times this so 4 times x squared is plus 4x squared 4 times 8x is plus 32x 4 times 16 is plus 64. so now i would combine like terms so i have x cubed plus 8x squared plus 4x squared is going to give me 12x squared then plus 16x plus 32x is 48x And then plus 64. so i got the same answer here but it took me a lot longer and remember i still

use the shortcut i still went through and said okay i know that part of this x plus 4 times x plus 4 i already know the formula for that so if i had went through and did foil for that first to get this first kind of answer that would have taken even longer so you can see already what a tremendous time saver this is going to be for you all Right so now let's take a look at z squared y minus 5y and this guy is cubed so again just trying to get a process for

you to remember this you'll always remember that the first guy the first guy is going to be cubed now the second term you have a space then plus you have a space then minus remember your signs before the second term and before the final Term so the final term is this guy right here the 5y cubed now what's going to be in the middle positions well again it's the same thing it's 3 remember that exponent is 3. so it's 3 here and 3 here and then in the first instance you square the first one so

that's z squared y that guy squared times this guy 5y in the second instance you square the Second one so z squared y times we're going to do 5y squared i'm kind of running into stuff so i know it's let me kind of scoot this down a little bit and now i can just simplify this so z squared cubed z stays the same 2 times 3 is 6. y cubed is just y cubed then minus I have 3 times 5 that's 15. and then let's think about this we have z squared squared that's z to

the fourth power then we have y squared then we have times y out here so y squared times y is y cubed then plus i have 3 z squared y and then we have 5y that's squared so 5 squared is 25 25 times 3 is 75 y squared is y squared so you'd have z Squared y times y squared y times y squared is y cubed then minus lastly we have 5y cubed 5 cubed is 125 and then y cubed is just y cubed so we get z to the sixth power y cubed minus 15

z to the fourth power y cubed plus 75 z squared y cubed minus 125 y cubed so this is a perfect example of how much time you would save using this Generic formula versus you know going through and expanding this into z squared y minus 5y times z squared y minus 5y times z squared y minus 5y go ahead and do that and see how long it takes you versus doing it this way and verify that my answer here is correct again z to the sixth power y cubed minus 15z to the fourth power y

cubed plus 75 z squared y cubed minus 125 y cubed Hello and welcome to algebra 1 lesson 33 in this video we're going to learn about dividing polynomials by monomials so far we've learned how to add and subtract with polynomials and we've learned how to multiply polynomials now the next logical step is for us to learn how to divide polynomials but before we get into the more serious division we want to start out with something that's very basic very simple and that's dividing a Polynomial by a monomial and before we kind of jump into that

we need to go over a few things from pre-algebra first so i want you to recall if i have something like a plus b over c legally i can write this as a over c plus b over c these two are the same now i know in this kind of general form it might not look familiar to you but all we're saying is that if we have a common denominator in this case we have A common denominator of c i can write the sum of the numerators which is a plus b right this numerator is

a this numerator is b so the sum of the numerators over that common denominator of c as an example with whole numbers so let's say you saw something like 20 plus 4 and this is over 2. well i can legally split this up into 20 over 2 plus 4 over 2. And we can see that we get the same answer either way if i just evaluated this 20 plus 4 is 24 if i divide it by 2 i get 12. if i do it this way 20 over 2 is 10 4 over 2 is 2

so 10 plus 2 is 12 also and what you're used to seeing is you're used to seeing this and going to this and all i'm saying is that it's the same Either way i can start out kind of in this format and go to this format and that's what we're going to need to do in this lesson so as another example let's say you saw something like negative 9 minus 6 over 3. again i can split this up and say i have negative 9 over 3 plus negative 6 over 3. this is just a common

denominator here of 3 and i can just go back and forth between these two right these are the same and again to kind of prove that if I take negative 9 and i subtract away 6 i get negative 15 then negative 15 divided by 3 would be negative 5. if i kind of do it this way negative 9 over 3 is negative 3 and then negative 6 over 3 is negative 2 so you would have negative 3 minus 2 which again is negative 5. right so either way it's the same result all right so now

that we have that little fun fact under our belt we're going to cover how to divide a Polynomial by a monomial so remember a monomial is a polynomial that is one single term so we're dividing by one single term so in other words the divisor in each case will be a monomial so to divide a polynomial by a monomial we want to divide each term of the polynomial by the monomial we're going to use the technique we just saw when we looked at those examples with whole numbers So starting out with this first one we

have 2k cubed plus 18k squared plus 18k divided by 6k squared so the first thing i want you to do when you get this problem is set it up as a fraction so what i'm going to do is i'm going to write this as 2k cubed plus 18k squared plus 18 k and then i'm gonna put a fraction bar and then my divisor is going to go as The denominator so i have my numerator or my dividend which is 2k cubed plus 18k squared plus 18k and then it's divided by we have our divisor or

the denominator 6k squared so in order to set this problem up it's very very simple again i'm just going to split it up into each term of this polynomial up here in the numerator divided by This monomial down here in the denominator it's just that simple so i would say i have 2 k cubed over 6k squared plus next i have 18k squared over 6k squared then next i have plus 18k over 6k squared and again this is perfectly mathematically legal i have a common denominator here of 6k squared so i could go right back

and write it this With the sum of the numerators 2k cubed plus 18k squared plus 18k over that single common denominator right so it's perfectly legal to go back and forth so once we have set it up in this format all we want to do is simplify each part separately so in other words i'm going to go through starting here and say okay if i have 2 over 6 i know i can cancel a common factor of 2. so this is going to cancel and be 1 this is going to cancel And be 3 right

6 divided by 2 is 3. and then k cubed over k squared we know that that would be k raised to the power of 3 minus 2. so this is going to cancel and this is going to be a 1 right so that's k to the first power so let's go ahead and put equals and then we'll write the results as we go so this is k to the first power or just k over just a 3 in the denominator And then plus if i look at this right here i have 18 k squared over

6 6k squared well 18 divided by 6 is 3 k squared over k squared is 1. so this is going to cancel completely that's going to be 1. so then plus i'm going to put a 3 here and then lastly i have plus 18 k over 6 k squared so 18 divided by 6 again is 3 k over k squared so this is k to the first power so remember if you use your quotient Rule for exponents this is going to be k to the power of 1 minus 2 or k to the power of

negative 1. k to the power of negative 1 is what it's 1 over k so essentially just eyeballing that you can see that this cancels out with one of the factors of k here you just have a k to the first power in the denominator so this would be plus 3 over k to the first power or just 3 over k So this is your answer you get k over 3 plus 3 plus 3 over k now i want you to remember something else from pre-algebra let me scroll down and get a little room going

for this let's say that i had something like 20 divided by 4 and we all know this is 5. so i have my dividend this is my dividend This is my divisor and then this is my quotient so how can i check the result remember i can go backwards i can take the quotient and multiply it by the divisor and i should get my dividend back so in this particular case this is our quotient so i should be able to multiply this quotient by our divisor remember the divisor is 6k squared and then i should

get the dividend back the 2k cubed plus 18k Squared plus 18k so let's check that out okay so i'm going to multiply this guy by 6 k squared and again all i'm going to use is the distributive property so 6k squared times k over 3. let's just write that out like this so 6 k squared times k over 3. well we know that the 6 would cancel with the 3 and give me a 2 and then k squared times k is k cubed So for the first part i'm going to put 2 k cubed let

me erase this then we have 6 k squared 6 k squared times 3. well that's easy it's just 3 times 6 which is 18. and then k squared comes along for the ride and then lastly i have 6 k squared times 3 over k so 6 k squared times 3 over k well we can see that one of the factors Of k here would cancel with this k and i'd have 6 times 3 that's 18 so 18 k so plus 18 k so let's see if this was our original dividend so we scroll back up

here and we see that it was 2k cubed plus 18k squared plus 18k 2k cubed plus 18k squared plus 18k so because we multiplied our divisor by the quotient and it gave us the dividend back we know we have the correct answer So it's very important especially when you first start to go back and check your answer make sure that you got the correct one that way if you didn't you can correct the mistake right away all right let's take a look at another one so we have eight r to the fifth power plus 12

r to the fourth power plus eight r cubed divided by four r cubed so again i'm going to write this as a fraction so i'm gonna write this as eight r to the fifth Power plus 12r to the fourth power plus 8r cubed that's the numerator right that's the dividend so then this is divided by 4r 4r cubed and then i'm just going to split this up right this is just a common denominator so i'm going to say this is 8r to the fifth power over 4r cubed Plus 12r to the fourth power over r

cubed plus 8 r cubed over 4 r cubed and then once i have it in this format it's very easy to go through and simplify each part so let's take a look at that we know that 8r to the fifth power over 4 cubed the 8 would cancel with the 4 give me a 2 r to the fifth power over r cubed is r squared So then let's go ahead and write these as we get them so this would be 2r squared then over here i have 12 r to the fourth power over 4r cubed

12 would cancel with 4 and give me a 3. r to the fourth power over r cubed this would cancel be r to the first power so this is plus three r then lastly i have eight r cubed over four r cubed this is going to cancel With this and give me a two r cubed will cancel with r cubed so this will be plus two so i get 2r squared plus 3r plus 2. and again very very easy to check all i'm going to do is take this answer here 2r squared plus 3r plus

2. this is the quotient and we'll multiply it by the divisor 4r cubed and i should get the dividend back So 4 cubed times 2r squared is 8r to the fifth power 4r cubed times 3r is plus 12 r to the fourth power and then four r cubed times two is plus eight r cubed so what i get is exactly the dividend that i started with eight r to the fifth power plus twelve r to the fourth power plus 8r cubed 8r to the fifth power plus 12r to the fourth power plus 8r cubed let's

take a look at another one we have negative 3x cubed Plus 24x squared minus 4x and then we're dividing by negative 8x squared so i'm going to take negative 3x cubed plus 24x squared minus 4x this is going to be my numerator right that's the dividend and i'm dividing this by the divisor it's going to be my denominator negative 8x squared and so once i have this i'm just going To split it up into three different parts here so negative 3x cubed over negative eight x squared and then plus 24 x squared over negative eight

x squared and then i have a minus here but i think it's a little better if we put plus negative 4x over negative 8x squared just so we don't make a silly sign mistake All right so let's take a look at this piece by piece so the first thing is i have negative 3 over negative 8 can't really cancel anything other than the negatives right negative over negative is positive and then i have x cubed over x squared so this is going to cancel this will be x to the first power so for kind of

for the first part we'll have 3 8 Times x or 3x over 8 you could write it that way too doesn't really matter then the next part we have a positive over a negative so i know that's going to be a minus 24 over 8 is 3. so this will cancel become a 3. and we know it's negative 3 but i've already accounted for that right there then x squared will cancel with x squared so basically this is just minus 3. then for the next part here i have Negative 4 over negative 8. negative over

negative is positive and then 4 over 8 this is going to become a 2 right here right because 8 divided by 4 is 2. now when i look at x and x squared this is going to cancel completely so the numerator is now a 1 and then down here i'm going to take one of these factors of x away so this would be x to the first power so essentially what you'd have is plus 1 over 2x and again if we want to check this we'll take this quotient here and multiply it by the divisor

and it should give me this dividend back so let's try that out real quick so we'll do 3 8 x minus 3 plus 1 over 2 x And we're multiplying this by the divisor which was negative 8 x squared so let's see what this gives us so negative 8x squared times 3 8 x so let me write that out so you can see what's going to cancel so obviously this eight will cancel with this eight i'll basically just have a negative one there negative one times three is negative three so that's negative three And then

x squared times x is x cubed then next i have negative three times negative eight x squared negative times negative is positive 3 times 8 is 24. so plus 24 x squared then lastly i have negative 8x squared times 1 over 2x so negative 8x squared times 1 over 2x this is going to cancel with this and give me a 4 and x squared over x this would cancel With one of these so basically i'd have negative 4x there so minus 4x and we could scroll back up and see that that is our dividend negative

3x cubed plus 24x squared minus 4x all right let's take a look at another one we have 24a to the fifth power b plus 2a to the fourth power b squared minus 18 a cubed and then this is divided by Negative 6 a b squared so let's kind of shortcut this and instead of writing this as a fraction and then kind of splitting it up let's just split it up right away so i'm going to take this term right here this 24 a to the fifth power b and write it over the divisor here negative

6 a b squared then plus i'm going to have 2 a to the fourth power b squared over that divisor Negative 6 a b squared and then i have that minus let me put plus negative 18 a cubed over again negative 6 a b squared anything i can do to kind of give you a shortcut because a lot of times you're going to be doing this on a timed exam and you don't want to take two steps when you can do something in one So let's simplify each part kind of separately 24 over negative 6

would be negative 4. so this would cancel with this and it'd be negative 4. then 8 to the fifth power over a this would cancel and this would be a to the fourth power then b over b squared this would cancel and this would be b to the first power so kind of the first part here would be negative four A to the fourth power over b then if i move on over here i see that i have 2 over negative 6. i know this is going to be negative so let me just go ahead

and put a negative out in front that's going to go right there 2 over 6 this is going to be a 1 and this is going to be a 3. right 6 divided by 2 is 3. then we have a to the fourth pi over a this is going to cancel with one of these so instead of 4 this will Be 3 and then b squared will cancel with b squared so what i'll have here i have the minus and then i'll have a cubed in the numerator over in the denominator i'll just have a

3. right i've already accounted for that negative all right then next i move on over here i have a negative over negative i know that's going to be positive and then 18 over 6 is 3. A cubed over a this is going to be a squared and then b squared i don't have anything to kind of match up with that so that's just going to stay all right so then it's going to be plus we have 3 a squared over in the denominator b squared so here's our answer negative 4 a to the fourth power

over b minus a cubed over 3 plus 3a squared over b squared so i'm not going to check This one just in the interest of time but go ahead and pause the video if you're interested take your quotient here the negative 4a to the fourth power over b minus a cubed over three plus three a squared over b squared multiply it by the divisor okay multiply it by the divisor which we know is negative six a b squared and you will get the dividend back 24 a to the fifth power b plus two a to

the fourth power b squared minus 18 a cubed so go ahead and Pause the video if you want take a look at that and then we're going to do one more problem all right for the last one i have 1 8 x to the fourth power y squared plus four fifths x cubed y minus one-half y cubed this is divided by 12 x y now before we kind of look at this i want you to recall that if you're dividing a fraction so something like three-fourths by a number or another fraction how do we do

it we take the first Fraction we multiply by the reciprocal of the second so if i had 3 4 divided by let's say 10 i would say this is 3 4 times the reciprocal of 10 which is 1 10. and you could write this as 10 over 1 if that makes it easier for you so instead of sitting here and saying okay i have 1 x to the fourth power y squared over Eight and this is divided by 12xy i know in the end i'm going to write this over 1 and this is going to

be equal to you know x to the fourth power y squared over eight times the reciprocal of this guy which is one over 12xy i can just start out i can make this much simpler and write this as kind of a multiplication problem I can see this and say well instead of dividing by 12xy i can kind of say this is times 1 over 12xy it's the same thing if i were to multiply 20 by one half it's the same thing as if i divided 20 by 2. right same thing so doing it this way

is just going to make it simpler for us we would have X to the fourth power y squared over 8 times 1 over 12xy again if i did it using division after i do my multiply the first guy by the reciprocal of the second i end up with this anyway so i might as well start out doing that again if you have something that normally takes two steps if you can cut it down to one it's just going to put you one step ahead All right then plus we have four x cubed y over five

times one over 12xy we have minus when i'm going to put plus negative we have y cubed we could put one y cubed doesn't matter over 2 and again times 1 over 12 x y all right so let's take a look at this now so what can we cancel well here i'm cross canceling i can Cancel one of the factors of x this will be x cubed with this x here i can cancel one of the factors of y this will be y to the first power with this y and then nothing else i can

really do so in the numerator i'm going to have x cubed y in my denominator i'm going to have 8 times 12 which is 96. then moving on to this part here i can easily see that 4 would cancel with 12 And leave me a 3 down here one of the factors from x cubed would cancel with this x here this would be x squared y would cancel with y so i would have x squared over 5 times 3 or 15. so this is plus x squared over 15. and then for the last part here

i have y cubed over y so one of the factors of y cubed would cancel with this so this Would be y squared and then nothing else i can really cancel so you'd have negative y squared i could put minus y squared over 2 times 12 is 24 and then times x so 24 x so this is what we end up with as our answer x cubed y over 96 plus x squared over 15 minus y squared over 24 x and again If you want to check this take your answer here multiply it by the

divisor remember when we started out this problem was dividing by 12xy so that's your divisor and again i modified it just for the sake that we were working with fractions and it made it easier but when you do that you should get that original dividend back hello and welcome to algebra 1 lesson 34. In this video we're going to learn about dividing polynomials so in the last lesson we kind of got started with dividing polynomials we learned kind of the simplest case where we divided a polynomial by a monomial so i want you to recall

that if you had something like 6x squared minus 2x and we divided this by 2x remember we could write this as a fraction so 6x squared minus 2x over 2x And then we could split this up right kind of using that same logic as when we work with fractions with whole numbers right this is a common denominator so we could say this is 6x squared over 2x and then instead of minus i'm going to put plus negative 2x over 2x then we can simplify each part separately so six over two is three x squared over

x is just x to the first power and then negative two x over two x this Would just be negative one so essentially what i'd have here is three x minus one as my answer so let me write this as 3x minus 1. now another important thing we know how to do that at this point remember if i want to check i take the quotient which is this guy i multiply it by the divisor which is this guy and i should get the dividend back which is this guy So 2x times 3x minus 1 should

produce 6x squared minus 2x so 2x times 3x is 6x squared and then 2x times negative 1 is minus 2x so we get that back exactly all right so that's not very difficult hopefully you have mastered the concept of dividing a polynomial by a monomial once you've done that it's really not much more difficult to divide a polynomial By a non-monomial polynomial so in other words a polynomial that consists of more than one term so for example if i saw something like 8n squared minus 42n plus 10 and i divided this by n minus 5.

what would i do well i can't use the same technique that's not going to work anymore we need to turn to long division right so the long division that we use with whole numbers we can use kind of a similar process for Polynomials so let me just do a really really quick rundown with long division of whole numbers that way it kind of refreshes your memory a little bit let's say i had something like 4620 and i wanted to divide that by five remember the steps for long division you're going to take your dividend which

is this number off to the left this is the dividend And you're going to start by putting that as we say under the house so here's your 4620 and then it's divided by five the five is known as the divisor and that goes off to the left so once you have this set up you recall we have that d which stands for divide m which stands for multiply s which stands for subtract b which stands for bring down and then r Which stands for repeat or remainder and the way you memorize this you say dad

mom sister brother rover right you think about the family right a mom and a dad and then a brother and a sister and a family dog so divide multiply subtract bring down repeat or remainder so when we get to polynomial division here in a second it's going to be the same thing there's a few slight differences but basically we're Following that format so you start out by saying how many times will five go into this leading term four well it's not going to right so what you do is you kind of expand that selection and

say how many times will 5 go into 46 well 5 will go into 46 9 times 9 times 5 is 45 subtract and we get 1. right so we did the division then we multiplied then we subtracted and now we just need to bring Down so we bring down the 2 and then we repeat right so 5 goes into 12 twice so we multiply 2 times 5 is 10 subtract we get 2 bring down and then repeat so now we divide again 5 goes into 24 times 4 times 5 is 20. subtract we get 0

nothing else to bring down so we're done so we get 924 as the result right and that result is called the quotient let me label that that's called the Quotient so we're going to follow a very similar process to divide polynomials so go ahead and start out by setting this up in the same way that you did that long division problem with whole numbers so again this is your dividend and this is your divisor so this goes under the house so 8 n squared minus 42n plus 10 And that goes under the house and then

off to the left we have n minus 5. what's going to be different here remember your steps you divide you multiply you subtract you bring down and then repeat or remainder in the first step here you're only thinking about the leading term of the divisor so you're going to ask leading term into Leading term so you only want to know leading term into leading term okay that's that's one of the differences you're not including the -5 part there okay so just forget about that now what is this going to be well it's just going to

be 8 n squared divided by n that's all it is you just Need the answers to that question so i always do that off to the side if i can't figure it out in my head and it's easy to simplify this we know that n squared over n would just be n to the first power so this is 8n so that's the answer now just like with long division with whole numbers i write my answer up here but here's the main mistake that a lot of people make and it's not that big of a deal

but i see a lot of students that Will put 8n right here it doesn't go there because you have to think about place value this is where n squared goes this is where n to the first power goes and this is where n kind of to the power of 0 or just a constant would go so we have n to the first power so i'm going to shift that over here i'm going to write my answer right here 8 n To the first power and you can liken that to what we did here i didn't

write the 9 all the way over here in front of the 4 because this isn't 9000 right this is 900 and so we want to maintain place value even though we're not working with whole numbers anymore so this is kind of where n to the first power goes so that's where i'm going to put the 8 in when i get that as my answer all right so now i Want to multiply so 8n times n is 8n squared and i also want to multiply by the minus 5. so 8n times negative 5 is minus 40n

so that's the main thing to remember everything else is pretty much the same because when i multiply i multiply by everything but when i do the division step it's only leading term into leading term so that's one of the things you just have to remember and it's just one of those Things you're just going to have to practice and after you practice 10 or 15 of these you're just going to remember and then once i've done these two steps the next step is to subtract so i'm subtracting this whole thing away now there's a few

different ways to do this what i always do and what i'm going to do in these videos for a while is i'm going to put parentheses around this guy and i'm going to put a minus sign there now You can do it this way you'll say 8n squared minus 8n squared is 0. negative 42n minus a negative 40n minus a negative is plus a positive so that would be negative 2n you could do it that way or you can put a minus sign out in front of these parentheses and then you could just say that's

a reminder to remove the parentheses and to change the sign of each term okay so this would be minus 8n squared this would be plus 40n and then I can add 8 n squared minus 8 n squared is 0 negative 42n plus 40n is negative 2n you're going to get the same answer either way so it doesn't matter how you want to do it then after the subtraction we bring down so plus 10 and now i can erase that zero i don't need that now i can just do again the repeat step so i go

back up to the Top so it's leading term into leading term okay i'm not involving that negative 5 there it's always just the leading term into the leading term so what is negative 2n over n well we know that the ends would cancel and i'm left with just negative 2. so negative 2 goes up here then i multiply negative 2 times n is negative 2n negative 2 times negative 5 is plus 10. now we subtract and we know the same thing over itself is going to be 0 but for the sake of completeness let's go

through and put parentheses around this guy and put a minus out in front and then that reminds me hey i need to change the sign of each term so this will be plus this will be minus and then i can erase this just a reminder for me so negative 2n plus 2n is 0 10 plus Negative 10 or 10 minus 10 is 0. so i get a 0 there nothing else to bring down so this 0 tells me that i don't have a remainder now let's talk about something else we know how to check division

when i have my answer when i have my quotient all i need to do is multiply it by the divisor so this 8n minus 2 multiplied by n minus 5 and that should Give me the dividend back so 8 n minus 2 multiplied by n minus 5. it only takes a second to check and you know if you're doing your homework and your homework's graded why not get the right answer right so 8n times n is 8n squared then 8n times negative 5 is minus 40n then negative 2 times n is negative 2n then negative

2 times negative 5 is plus 10. so this equals 8n squared Negative 40n minus 2n is minus 42n and then plus 10. so you can see we get exactly the dividend back 8n squared minus 42n plus 10. all right so that was pretty easy let's take a look at another one so we have x squared plus 16 plus 10x divided by x plus eight so one of the things i haven't mentioned yet is that each polynomial when you're doing the division has to be in standard form so this is out of order right the 10x

Should be over here the 16 should be over here so when i set this up in long division format i'm going to correct that so i'm going to say i have x squared plus 10x plus 16 and this is divided by x plus 8. so if you see a polynomial that's not in standard form you need to fix that so for example if this would have been 8 plus x i would have flipped that and Made it x plus eight right so that's kind of the first thing you want to do when you're setting this

up now we just go through the normal way right so let me write this down again we have divide multiply subtract bring down repeat or remainder right dad mom sister brother rover so again i start out by saying leading term into Leading term okay i'm not involving that plus 8 okay not doing that so what is x squared over x we know that one of these would cancel with this and i just have x to the first power again this is the spot for x squared this is the spot for x to the first power

this is the spot for a constant so we're going to put our answer here because It's x to the first power so that's going to go right there and then we're going to multiply remember when i multiply i multiply by everything so x times x is x squared x times 8 is plus 8x and then i subtract i subtract this guy away now you do it like that or again use this as a reminder to put a minus here and a minus here and then you can remove all the parentheses and you can add either

way You want to do it x squared minus x squared is 0. 10x minus 8x is 2x now we're going to bring down so plus 16 and then we're going to repeat so we go back up to the top so we're going to go leading term into leading term so what is 2x over x well the x's would cancel and i would just have a 2. so plus 2. Again multiply 2 times x is 2x 2 times 8 is plus 16. so we could see we wouldn't have a remainder here right anything minus itself is

0. but again for the sake of completeness we're going to subtract this away so i'm going to change the sign of each term this will be minus and that will be minus and i can erase this 2x minus 2x is 0. 16 minus 16 is 0. So again there's no remainder so our answer here is x plus two so x plus two and it's very very easy to check so again if i wanted to take the quotient x plus two and multiply that by the divisor x plus eight i should get my dividend back x

times x is x squared x times 8 is plus 8x 2 times x is plus 2x 2 times 8 is plus 16. So if i combine like terms i get x squared plus 10x plus 16 which is exactly this just looks a little different because this is not in standard form and this is right but if i flip the position here of the 10x and the 16 i would get exactly that all right let's take a look at one more where there's not a remainder and then i'm going to go through what happens when there

is a remainder we'll just look at one example of that So we have negative 6x to the fourth power plus 29x cubed minus 5x squared minus 43x plus 10 and then i'm dividing this by 3x squared minus x minus 5. all right so let's go ahead and set this guy up so this is my dividend so that's going to go under the house so negative 6x to the fourth power plus 29x cubed minus 5x squared minus 43x plus 10. again that goes under the house And then off to the left we have 3x squared minus

x minus 5. so again let me write the steps up here so we don't forget so we have divide we have multiply we have subtract we have bring down and then we have rapido remainder right dad mom sister brother rover so we start out with the first step leading term into leading term So 3x squared into negative 6x to the fourth power so negative 6x to the fourth power over 3x squared so this negative 6 will cancel with 3 and give me negative 2 x to the fourth will cancel with x squared and give me

x squared so negative 2x squared is the answer where am i going to write that am i going to write it here over the x fourth spot no am i going to write it here over The x cubed spot no i'm going to write it over the x squared spot because that's where it goes so negative 2 x squared all right now i'm going to multiply that's my next step let me erase this stuff and let me scroll down a little bit and i realize it's going to take this guy off the screen so let

me kind of drag this down right there And let me scroll down a little bit there we go okay so now again we're multiplying negative 2x squared times 3x squared is going to be negative 6x to the fourth power negative 2x squared times negative x is positive 2x cubed negative 2x squared times negative 5 is going to be positive 10 x squared Now i am subtracting this thing away remember usually what i do is i put parentheses around here put a minus out in front you don't need to do that if you just change the

sign of everything so this is minus it's going to become plus this is plus it's going to become minus this is plus it's going to become minus now i can just add negative 6x to the fourth power plus 6x to the fourth power is 0. 29x cubed minus 2x cubed is 27x cubed negative 5x squared minus 10x squared is minus 15x squared and then i want to bring down so bring down the negative 43x and then we're going to repeat so i'm going to go leading term into leading term so what is 27 x cubed

over 3x squared well i know 27 divided by 3 is 9 x cubed over x squared would be x to the First power so i'm going to put plus 9 x okay and then i'm going to multiply so let's erase this so 9x times 3x squared is 27 x cubed 9x times negative x is negative 9x squared 9x times negative 5 is minus 45 x again i am subtracting this away so just change the sign of everything so this will be minus Plus plus now i can just add 27x cubed minus 27x cubed to zero

negative 15 x squared plus nine x squared is negative six x squared negative 43 x plus 45 x is positive 2 x now we're going to bring down bring down this plus 10. then we want to repeat so i'm going to again go leading term leading term into leading term so what is negative 6x squared over 3x Squared well x squared over x squared is going to cancel negative 6 over 3 is negative 2. so this will just be minus 2 here and then one last time i'm going to multiply so negative 2 times 3x

squared is negative 6x squared negative 2 times negative x is plus 2x negative 2 times negative 5 is plus 10. So you can see it's the same thing minus itself that would be 0 but again to kind of make this complete if i'm subtracting this away i'm going to change every sign so this will be plus this will be minus this will be minus and then if we add it here negative 6x squared plus 6x squared is 0. 2x minus 2x is 0 10 minus 10 is 0. so that's 0. so there is no remainder

so let me erase everything so this is equal To that and so again if we want to check you're going to take this guy multiply by this guy so negative 2x squared plus 9x minus 2 would be multiplied by 3x squared minus x minus 5 and i should get this long thing up here in the top right the negative 6x to the fourth power plus 29x cubed minus 5x squared minus 43x plus 10. so let's see if we do in fact get that okay so let's begin we have negative 2x Squared times 3x squared that's

negative 6 x to the fourth power i have negative 2x squared times negative x that's positive 2 x cubed i have negative 2x squared times negative 5 that's positive 10x squared then i have 9x times 3x squared that's plus 27x cubed i have 9x times negative x that's minus 9x squared and then i have 9x times negative 5 that's minus 45 x then i have negative 2 times 3x that's minus 6x squared negative 2 times negative x that's plus 2x and then negative 2 times negative 5 is plus 10. all right so if i combine

like terms here i'm going to get negative 6x to the fourth power i have A cubed here and here so 2 plus 27 is 29. so plus 29x cubed then i have 10 x squared negative 9x squared and negative 6x squared so 10 minus 9 is 1. 1 minus 6 is negative 5. so this would be minus 5x squared then i'm looking at negative 45x and 2x negative 45 plus 2 is negative 43 so minus 43x And then lastly plus 10. this is exactly what we started out with as our dividend negative 6x to the

fourth power plus 29x cubed minus 5x squared minus 43x plus 10. all right so now let's talk a little bit about what to do when your division problem with polynomials has a remainder so i want to go back to a problem that we would have encountered in grammar school so something like 50 divided by 4. 50 is not divisible by 4 because when the division is concluded you have a remainder right so we could say that 50 divided by 4 we'd start out with saying 4 goes into 5 once 1 times 4 is 4 subtracting

you get 1 bring down the 0. 4 goes into 10 twice 2 times 4 is 8 subtracting you get 2. so this 2 is known as a remainder right so remember kind of in grammar school we Wrote this as 12 r 2 right that was the remainder that was the leftover amount so in other words to kind of break this down completely if i took 50 and i split it up into equal groups of 4 i could make 12 groups right 12 times 4 is 48 and then i would have 2 left over right so

48 plus 2 would get you back to 50. the remainder is kind of what i have left over i don't i don't have enough to make another group of 4. so That's how that comes about but we learned a couple of different things that we could do the first thing we kind of did was we put a decimal point here we bring that up here we put a zero and we continue our division 4 goes into 25 times 5 times 4 is 20. subtracting you get 0. so now i know the answer is 12.5 in

decimal form but that's not really going to help us with polynomial division the other way we could have written this Think about the 0.5 there we could have gone back to this let me erase all this and say that this remainder 2 we could say plus i have 2 that's divided by 4 or i have 12 plus 1 half right 2 over 4 is 1 half and we know that 12 plus 1 half or 12 plus in decimal form 0.5 is 12.5 that's exactly what we got when we continued the division So it's legal to

write it as 12 plus one-half or as a mixed number right 12 and one-half like that this is how we're going to write our answers when we're dividing with polynomials we're going to write it in that format you'll have your quotient plus you'll take the remainder and put it over the divisor now this is very very important when you check your result it's a little Bit more complex let me erase this real quick if i was to check the answer here what would i do well i could multiply this whole thing right here in this

format times 4 so i could say 4 times 12 plus 1 half 4 times 12 is 48 plus 4 times 1 half is 2 and that will get me back to 50. so i can do it that way another way that i could do this i could go back to kind of 12 r2 And i could say okay 12 the quotient times 4 is 48 add the remainder back in i get to 50. right so there's kind of two different ways that you can look at that all right so let's take a look at our

problem so we have 3k cubed minus 17k squared plus 27k minus 12 and i'm dividing by 3k minus 8. so again this guy goes under the house so k cubed Minus 17 k squared plus 27 k minus 12 and we're going to divide this by again 3 k minus 8. so i can keep writing this off to the side so division multiplication subtract bring down repeat or remainder so dad mom sister brother rover again first step leading term into leading term So what is 3k cubed over 3k so if i look at this i know

the 3s are going to cancel and i know k cubed over k is k squared so i'm going to put that over here this is going to be k squared then i'm going to multiply so k squared times 3k is 3k cubed k squared times negative 8 is minus 8 k squared i'm subtracting this line away so this Becomes minus this becomes plus three k cubed minus three k cubed is zero negative 17 k squared plus eight k squared is negative nine k squared then i bring down so bring down the 27k and i'm going

to repeat so now i'm going to do again leading term into leading term So what is negative 9k squared over 3k well this is going to cancel negative 9 over 3 is negative 3 and then k squared over k is k to the first power so we get minus 3 k now again i'm going to multiply and erase all this negative three k times three k is negative nine k squared negative three k times negative eight is plus 24 k Again i'm subtracting this away so what am i going to do this is going to

become plus and this is going to become minus so negative 9k squared plus 9k squared is 0. 27k minus 24k is 3k now i'm going to bring down so bring down the negative 12 and i'm going to repeat so let me scroll down just a little bit and so again i want to do leading term Into leading term so 3k goes into 3k we all know that's gonna be one right i don't need to write out 3k over 3k same thing over itself is 1. so plus 1 here and then i'm going to multiply so

1 times 3k is 3k 1 times negative 8 is negative 8. now i'm subtracting this away so this becomes minus this becomes plus and then look at what's going to happen Three k minus three k is zero negative 12 plus eight is negative four i don't have anything else to bring down here so i'm done negative 4 is my remainder this is the remainder okay so let's erase everything i so i'm going to put equals and then i have that negative 4 so i'm going to put plus negative 4 over 3k minus 8. Remember when

i divided 50 by 4 i had an answer that was 12 plus the remainder was 2 right it was 2 over the divisor of 4. i just simplify that 2 over 4 is 1 half so plus 1 half this is in the same format as this all i did was i said plus the remainder over the divisor so the remainder was negative four the divisor's three k minus eight same as what i did here the remainder was two And let me kind of write it as two fourths so it's not confusing the remainder was two

the divisor is four i just had one half because that's the simplified form but i'm going to leave it as two-fourths right now so that's how you would go about writing your answer if you have a remainder take the remainder put it over the divisor and then you're done now to check this again it's the same thing as when you check this You can take the quotient multiply it by the divisor so 12 times 4 is 48 and add that remainder of 2 that will get you back to 50. or the other thing i could

do is i can multiply 4 by this whole thing right so i'm going to show you both ways here just so that we're completely clear on how to check something so i can kind of just take this part right here and multiply it by this part right here And when i'm done just add negative 4 and that will give me this so in other words if i took 3k minus 8 and i multiplied it by k squared minus 3k plus 1. 3k times k squared is three k cubed three k times negative three k is

minus nine k squared three k times one is plus three k then negative eight times k squared is minus eight k squared negative 8 times negative 3k is plus 24k And then negative 8 times 1 is negative 8. so combining like terms here negative 9k squared minus 8k squared would be negative 17k squared then 3k plus 24k is 27k and then i have that minus 8 there so if i look at my answer here 3k cubed minus 17k squared plus 27k minus 8 it's the same thing as this the only difference is this minus 8

here so why do you think that is again the remainder was negative 4. so if i added a negative 4 if i said hey this minus another 4 this would end up being negative 12 just like it is right there and you get your original dividend back now let me show you the other way so you can also multiply 3k minus 8 times this whole thing k squared minus 3k plus 1 plus negative 4 over 3k minus 8. let me just kind of do this a different Way let me go we'll have k squared times

three k minus eight right i'm just distributing this to each term and then i'll have negative three k or i can put plus negative three k like that times three k minus eight and then we'll distribute this again so we'd have one times three k minus eight that's just three k minus eight and then let's distribute it one more Time what you'd have here is plus three k minus eight times negative four over three k minus eight so these two are going to cancel themselves out it's the same thing over itself and so i'm just

left with a negative 4. so if i put equals here let me kind of go through this k squared times 3k is 3k cubed k squared times negative 8 is minus 8k squared Then negative 3k times 3k is minus k squared negative three k times negative eight is plus 24k then i have three k minus eight so plus three k minus eight and then after we canceled this we ended up with just plus negative four so what you're gonna find here is that you get your dividend back exactly so three k cubed three k cubed

negative eight k squared minus nine k squared is minus 17 k squared then 24k plus three k that's plus 27k And then negative 8 plus negative 4 is negative 12. so i get 3k cubed minus 17k squared plus 27k minus 12 which is exactly the dividend that i started with hello and welcome to algebra 1 lesson in this video we're going to learn about dividing polynomials with missing terms so somewhere in your section on dividing polynomials you're going to get this kind of subsection that talks about dividing polynomials with missing terms it's not any more

difficult Basically what you're looking at is that in some cases we will get polynomials with missing terms and what i mean by that if you look at this kind of first example here we see 5x squared minus 3. so if my highest power on x is a 2 then i've got to go down kind of in order so in other words i need an x to the first power there and i don't have it that's what i mean by a missing term so we have a little trick for that we Rewrite this as 5x squared

plus 0 as the coefficient for the missing term the missing term was x to the first power x to the first power and then we have minus three remember zero times anything is zero so legally i can do that and these two things are the same i haven't changed the value i just made it look a little bit different remember that 0 is used in our number system As a placeholder and it's so different here just being used as a placeholder for example in a number like 1007 i have one thousand seven these zeros here

are just placeholders they're telling me that i have zero hundreds and i have zero tens it is the same thing i have zero x to the first power here right so the 0 is just nothing more than a placeholder but look at something like 4x cubed plus 1 again if i'm looking for what's missing i want you to think about this carefully the highest power on x is a 3. right we have x cubed so i'm missing x squared and i'm missing x to the first power so i can rewrite this as 4 x cubed

plus 0 x squared plus next i'd do 0 x to the first power and then finally plus 1. Right so i have 0 as a placeholder and 0 as a placeholder right i have 0x squared and 0x to the first power so i'm representing them but with 0 as the coefficient because again zero times anything is zero so the basic idea here is when performing long division with polynomials with missing terms we need to use zero as the coefficient for each missing term and if something's written in standard Form it's very very easy to see

what's missing right if i have four x cubed plus one i know i'm missing an x squared and an x to the first power if i had something like let's say seven x to the sixth power minus two well my highest power on x is the sixth power so i'm missing x to the fifth power fourth power third power squared And first power and a lot of times you're not gonna have that many missing powers you'll just have you know one or two here there like i had two here you know in this instance you'd

have the fifth power the fourth power the third power the second power and the first part would be five missing powers it'd be a lot for your teacher to throw at you you'll probably just get one or two so let's take a look at an example again this is Not any more difficult than dividing with polynomials when you have all the terms right it's the same kind of setup so i have 2x cubed minus 20x minus 6. that is my dividend right so remember that goes as we say underneath the house so 2x cubed i

have minus 20x next so i'm missing that x squared term so i'm going to put plus 0x squared as a placeholder and then minus 20x And then minus 6. again if you're missing that x squared term you just put 0x squared that's all you need to do all right so then we're going to set this up out here my divisor 2x plus 6. and once you've done that again it's the same thing that you've been doing throughout this section you're going to go leading term into leading term so 2x goes into 2x cubed how many

times so what is 2x cubed over 2x We know that these twos are going to cancel completely x cubed over x we know this is going to be x squared so i'm going to put x squared over this guy right here this is where i'm going and then i'm going to multiply so let me erase this real quick x squared times 2x is 2x cubed then x squared times 6 is plus 6x squared now Again i want to subtract away this polynomial and i know in the last kind of section where we just talked about

dividing polynomials i kind of got you to the point to where i didn't go through and do this extra step of putting parentheses around it putting a minus in front and then erasing all this and changing the signs again once you get comfortable just change your signs and move on right Now i'm just going to add 2x cubed plus negative 2x cubed is 0. 0x squared minus 6x squared is minus or negative 6x squared scroll down get a little room going all right now i'm going to bring down my next term and my next term

is a negative 20x and then again i'm going to go leading term into leading term so what is negative 6x squared Over 2x we know that this is going to cancel with this and give me a negative 3 and x squared will cancel with x that'll give me x to the first power so i'm going to have a minus 3x i'm going to have a minus 3x and i'm going to multiply negative 3x times 2x is negative 6x squared negative 3x times 6 is negative 18x And again i want to change the sign of each

term here because i'm going to subtract this bottom one away so this will be plus and this will be plus so negative 6x squared plus 6x squared that's going to cancel and be zero negative 20x plus 18x that's going to give me negative 2x and then i'm going to bring down this last term here which is negative 6 and Then my leading term 2x goes into negative 2x how many times so that's like saying negative 2x over 2x is what and we can all see that this is going to cancel with this and give me

a negative 1 and x over x is 1. so this is going to end up being -1 here so negative 1 times 2x is negative 2x negative 1 times 6 is minus 6. if i subtract the same thing away from itself i get 0. right this would become plus This would become plus negative 2x plus two x is zero negative six plus six is zero so there's no remainder so i get x squared minus three x minus one all right so remember i like to check things and if you don't like to check things just

skip ahead in the video it's no problem it usually takes about two three minutes to check something and i understand that you're going to get on timed examinations where you don't you just Can't check it but particularly when you're doing homework and your teacher is grading it you want to check it make sure you got the right answer you don't want to blow a problem right so let's go ahead and just multiply this guy right here times this guy right here your quotient by your divisor should give your dividend back so i'll have 2x plus

6 that's multiplied by x squared minus Three x minus one so two x times x squared is two x cubed two x times negative three x is minus six x squared two x times negative one is minus two x six times x squared is plus 6x squared 6 times negative 3x is minus 18x and then 6 times negative 1 is minus 6. so if i combine like terms here i got like terms here i've got like terms here i'd have 2x cubed Negative 6x squared plus 6x squared is 0x squared right so i don't have

to write that so i could put plus 0x squared kind of like i did when i set up the division or i'll have to put it right i can leave it just like it's going to be up here then i have negative 2x minus 18x that's negative 20x and then lastly minus 6. and that completely matches this right here 2x cubed minus 20x minus 6 so we know we Have the correct answer all right let's take a look at a note we have 7x cubed plus 28x plus 35 and this is divided by 7x plus

7. so again let's take this guy right here we have 7x cubed now the next one is 28x this is to the first power here so i'm missing my x squared term again so plus 0x squared again if you're missing something just write the coefficient of zero and then give yourself a representation of the Variable raised to that missing power then plus 28x then plus 35. okay then this guy over here 7x plus 7 that's going to go off to the side and then we divide like we normally do leading term into leading terms so

7x cubed over 7x this is going to cancel with this and give me a 1. x cubed over x is x squared so that's going to go right here And then i'm going to multiply x squared times 7x is 7 x cubed x squared times 7 is plus 7x squared so i'm going to change the sign of each term here now this is going to become minus this is going to become minus and i'm going to add 7x cubed plus negative 7x cubed is 0. 0x squared minus 7x squared is negative 7x squared let's bring

down the next term which is plus 28x And then let's go leading term into leading term so i have negative 7x squared over 7x and i'm going to cancel this with this that's going to give me a negative 1 and then x squared over x this is going to cancel with this and give me x to the first power so this is going to be negative x to the first power and then i want to multiply negative x times 7x is going to be Negative 7x squared negative x times 7 is minus 7x and again

change the sign of each term here so this has become plus and this has become plus so negative seven x squared plus seven x squared is zero 28x plus seven x is 35x so this is 35x and then this comes down this is plus 35 then 7x goes into 35x five times right we can all see that 35 would cancel with 7 and give me a 5 x over x is 1. so this is going to be plus 5. 5 times 7x is 35x 5 times 7 is 35. if i subtract 35x plus 35 from

that i'm going to get 0. right this would become negative and so this so 35x plus negative 35x is 0. 35 minus 35 is 0 so there's no remainder here so we get x squared minus x plus 5. all right let's do one final problem the Only difference here is i'm going to put the missing term in the divisor instead of the dividend so i have 2x to the fourth power minus x cubed minus 5x squared plus 3x minus 3 we're dividing by x squared minus 3. so let's set this up my dividend is 2x

to the fourth power minus x cubed minus 5x squared plus 3x minus 3 Then over here my divisor is x squared minus 3. so x squared i'm missing my x to the first power so i'm going to put plus 0x and then i have minus 3. so again leading term into leading term so what is 2x to the fourth power over x squared well we know that this is going to cancel with this and give me 2x squared right this would cancel out so 2x squared again i'm going to write that over the Term that

has x squared in it and then i'm going to multiply 2x squared times x squared is 2x to the fourth power 2x squared times 0x is 0x cubed then 2x squared times negative 3 is minus 6x squared so we're subtracting this guy away so that means i'm going to change this to a minus i'm going to change this to a minus And i'm going to change this to a plus so let's go ahead and add 2x to the fourth power minus 2x to the fourth power is 0. negative x cubed minus 0x cubed is negative

x cubed then negative 5x squared plus 6x squared is 1x squared or just x squared all right so now i'm just going to bring down this 3x and i'm going to go leading term into leading term So what is negative x cubed over x squared well we know that this is going to cancel out with two factors of x from that numerator so this would be negative x to the first power so minus x and then we're going to multiply negative x times x squared is negative x cubed negative x times 0x is minus 0x

squared Negative x times negative 3 is positive 3 x so let me change the signs here so this will be positive positive and then this guy's going to be negative so negative x cubed plus x cubed to zero x squared plus zero x squared is just x squared so then three x minus three x we could put plus zero x and then we can bring down this negative three And now once again i'm going to do leading term into leading term x squared over x squared is 1. so i'm going to put plus 1 here

and i'm going to multiply 1 times x squared is x squared 1 times 0x is 0x and then 1 times negative 3 is minus 3. so i have the same thing over itself if i'm going to subtract here if i subtract the same thing away from itself i get 0. X squared minus x squared is 0 0x minus 0x is 0 negative 3 plus 3 is 0. so this becomes 0. so 2x squared minus x plus 1 is our answer hello and welcome to algebra 1 lesson 36 in this video we're going to learn about

finding the gcf otherwise known as the greatest common factor so hopefully you learned how to find the greatest common factor in Pre-algebra if this is new to you i'm going to do enough of a review here to where you can catch on very quickly so i want you to recall that the gcf again that's the greatest common factor or some people like to say the greatest common divisor this is the largest number that each number of a group is divisible by so let me give you a quick example let's say you had two numbers let's

say 12 And 18. and i said what's the largest number that each number here 12 and then 18 would be divisible by well what you would do is you would think about the factors of each number and then you would find the largest one that's common to both so for 12 if you think about the factors of 12 you use a factor tree here you'd have 4 times 3 3 is prime 4 is 2 times 2. So i can think about 12 as 2 times 2 times 3. so 12 i'm going to think about as

2 times 2 times 3 and then for 18 if i use a factor 3 i could do 9 times 2 2 is prime and then 9 is 3 times 3 3 is prime so 18 is 2 times 3 times 3. so what's common to both well if i look here i have one factor of 2 here and then a second factor of 2 here over here i only have one factor of 2. So really if i was to build a list here kind of a list i would think about just putting one factor of 2

in because that's what's common to both i have 2 here only one here one is common to both so i'd put a two there then if i kind of move on to kind of the next number that's involved i have a three here and then i have one three two threes here so one three is common to both so times 3. So my list would contain 1 2 and 1 3. i'd multiply them together to get 6. and so the gcf is the number 6 right and you can see if you took 12 and you

divided by 6 you get 2 right remember if we talk about something being divisible by something then that means it divides evenly there's no remainder so 12 divided by 6 is 2 no remainder 18 divided by 6 is 3 no remainder so that is the largest number That each number of this group the 12 and the 18 would be divisible by all right so let's go over the official procedure for finding the gcf and once you kind of get good at this you don't need to do all the work that i just did or even

the work that you do in this kind of procedure in most cases if the numbers that you're working with are small enough you can pretty much eyeball it and see what the gcf is going to be so i want you to factor each number Completely so you can do that with a factor tree or you can do it using kind of any method that you've learned then you want to list every prime factor that is common to all numbers all numbers of the group okay so it's got to be common to everything then for prime

factors that are common to all we list the least the least amount of times it appears so once you've done this you're just going to multiply the numbers of the list Together to form the gcf all right so let's take a look at an example here so we want to find the gcf of 20 45 and 60. so kind of the best way i've found for students to kind of completely understand this i just build a little table so my table will consist of the prime factors for 20 and 45 and 60. so i'm going

to put equals And then i'll list my prime factors here so to do that i'm just going to use a factor tree in each case so for 20 i can start by saying 4 times 5. i know 5 is prime so i'm going to circle that and then 4 i know i can write as 2 times 2. 2 is prime so i'll go ahead and circle both of those so then the next one i have is 45. so 45 and i can start out by saying 9 times 5 And again 5 is prime so let's

circle that and then for 9 i would split that up into 3 times 3. 3 is prime let's go ahead and circle both of those and then the last one is 60. so for 60 i can do 6 times 10 6 is 2 times 3 and 2 and 3 are both prime so let's circle those and then for 10 we do 5 times 2 And 5 and 2 are both prime so we'll circle those all right so i'm going to write this in this table in such a way to where it's completely obvious what's common

to everything so for 20 the smallest prime factor is 2. so let's go ahead and start with that so we have 2 times 2 and then i have times 5 but i'm going to put that way out here i'll put that way out here For 45 i don't have any factors of 2. so i'm going to kind of mark those out to show that it's not here not here and then i have one two factors of three so let's put that there so times three times three and while i'm doing this i'm gonna go ahead

and mark this out to show i don't have that here right it's not gonna be common to everything and i also have a factor of 5 in 45 so Times 5. now for 60 i have 2 factors of 2. so 2 times 2. i have 1 factor of 3 okay but i don't have a second factor of 3. so i'm going to mark that out and then i have one factor of 5. one factor of 5. so when we kind of look at this table here it makes it completely obvious what's going to be common

to everything so starting out with kind of 20 i have Two factors of 2 but i don't have any factors of 2 and 45 even though i have two factors of 2 and 60 2 wouldn't go in the list right when i go to build my gcf because it's not common to everything okay moving on i don't have any factors of three up here even though i have two factors of three here and one factor of three here because there's no factors of three in The prime factorization of 20 it doesn't go on the list

either it's got to be common to everything the only number that's common to every prime factorization is 5. so i want to highlight that there's a 5 in each prime factorization and there's only one of those so my gcf here would just be the number five and go ahead and take each number here and divide it by five you'll see that You don't get a remainder and you won't be able to find any number that's larger than five that each number of the group is divisible by right 5 is the largest one and so that's

our greatest common factor or a lot of you will hear this called the greatest common divisor all right let's take a look at a null so we want to find the greatest common factor of 120 168 and 216. so again i'm just going to factor each One so 120 and i'm going to start out because it says a zero at the end i know it's divisible by 10 so i'm just going to go 12 times 10. i'm going to start out with that 12 i can do 4 times 3 and 3 is prime 4 is

going to be 2 times 2 and 2 is prime for 10 i'm going to write that as 5 times 2. 5 and 2 are both prime All right so that one's done so let's do 168 now so what are two factors of 168 to kind of get started it's an even number so i know it's divisible by 2 but also i know that the final two digits here 68 would be divisible by 4 so therefore the number is so what is 168 divided by 4 well that's going to be 42. so i know this would

be 42 times 4 to start and four is easy that's two times two both of those are prime we'll circle Them for 42 we know we can do seven times six right that's easy from watching football right your team scores you know six touchdowns with 6 extra points they have 42 points so then 7 is a prime number we'll circle that and then for 6 that is 3 times 2 and both those are prime so we'll circle those okay so now we want 216 Again it's an even number and also 16 is divisible by 4

so that means 216 would be so let's go ahead and start out with 54 times 4. 54. times 4. 4 again is 2 times 2 and 2 is prime so we'll circle those and for 54 i know that's 9 times 6 and nine three times three three is prime let's circle those and for six three times two again three and two are both prime so Let's circle those okay so now that we've got this set up let's go ahead and make a table let's go ahead and make a table just like we did in the

last example so we're going to make four lines here okay so we'll have 120 168 and 216. a little line there okay so the prime factorization for 120. I have 2 times 2 times 2 2 times 2 times 2 and let me kind of look at 168 now so i have 2 times 2 times 2. so i have 3 factors of 2 there as well and then in 216 i have 3 factors of 2 as well so 2 times 2 times 2 and then kind of going back to 120 i have one factor of 3.

so one factor of 3. and 168 i have one factor of 3. In 216 i have 1 2 3 factors of 3. so 1 2 3. so what i'm going to do to make this obvious i'm going to mark out what we're missing right only one factor of 3 is common to everything now moving on i have a factor of 5 and 120 and then i'm done so this one's closed out i'm just going to put a little x next to it So for 168 i don't have a factor of 5. so let me just

kind of notate that with this and then in 216 i also don't have a factor of five and then i could stop here if i wanted to because once i've closed one of them out that means anything that appears in the other prime factorizations won't matter because it's not in one of them and it's got to be common everything but just for the sake of completeness In 168 i have a factor of seven so let me write that let me kind of mark this out over here and in 216 i don't have anything else to

put in so let me just put that these are closed out as well let me kind of scroll down and get a little more room going all right so again it's completely obvious what the list is going to be right with the way that i've written this table and If you're struggling with this kind of concept of the gcf you might want to take the extra time to kind of spread everything out and look at what's common so i know that everything has one two three factors of two so two times two times two is

eight i could write it as two times two times two or i can just put eight ten times the next thing that's common to everything is going to be a 3 and i have 1 factor of 3 that's common to Everything now here's where students get confused they see 2 additional factors of 3 down here that's not common to everything so you can't put it in it's got to be common to everything so i'm going to put times 3 5 is not common everything and 7 is not common in everything this is what's common to

everything it's 8 times 3 and that is 24 and so that is your greatest common factor for the Numbers 120 168 and 216. so hopefully those last two examples gave you a general understanding of how to find the greatest common factor if you need more practice with this go back into the pre-algebra lessons and take a look at the video that we made on finding the greatest common factor again this is also called finding the greatest common divisor we're kind of moving forward and thinking about how we're going to use This for algebra in addition

to finding the gcf and again that's greatest common factor of numbers we also need to be able to find the gcf when variables are involved now the process for finding the number part is the same the variable part is a little bit different because you don't need to sit there and make a factor tree i know you don't need to do that every time for numbers but when they get kind of big You really kind of do need to do that so the variable part just includes any variable that is common to all the exponent

on the variable is the smallest that occurs in the group so for example let's look at the greatest common factor for 5x squared 25x cubed and 30x to the fifth power so what am i looking at here well i'm going to think about the number parts i have a 5 i have a 25 and i have a 30. i don't need to go through and factor That we all should be able to see at this point that the greatest common factor would be 5. 5 doesn't factor it's a prime number 25 is what it's two

factors of 5. it's 5 times 5. and then 30 is going to be 5 times 6 or 5 times 2 times 3. so if i look at these what's common to everything it's just a five right so i can eyeball that and see that the number part here is going to be five But what's the variable part well all i do is look at what's common to everything here i have x squared i have x cubed and i have x to the fifth power what is x squared it's x times x what is x cubed

it's x times x times x what is x to the fifth power it's x times x times x times x times x so what's common to everything well since i only have two factors of x Here that's as many factors that's going to be common to everything i have 2 here i have 3 here but only 2 would be in common with this i have 5 here but again only 2 would be in common with this so in other words all i need to do is look for the smallest exponent that appears on that variable

that is common to all the smallest exponent occurs right here it's a two so i just Use that when i build my gcf so this is basically just going to be 5 x squared that's going to be my greatest common factor all right for the greatest common factor of 6 n m cubed 24n squared and 27n again let's think about the number parts first so can we eyeball this and do it i can maybe you're not at that level yet but very quickly you will be i know right away that 27 is only going To

have 3 as a prime factor right it's 3 cubed this right here is 3 cubed for 24 it's divisible by 3. 24 divided by 3 is what it's 8. this is 8 times 3. and i could sit there and write it as 2 cubed times 3 but again i don't need to do that at this point for 6 i know that's 2 times 3. if i look at this what's going to be common to everything it's not going to be 2 because there's No 2 in the prime factorization of 27 so i can just not

think about that it's only going to be 3. this has a factor of 3. this has a factor of 3 and this has 3 factors of 3 but what's going to be common is only one of those so the number part is just going to be a 3 okay now when we get to the variable part again if i think about this you first ask the question is that variable common to everything so i have an n here i have an n squared Here and n here so the answer to that is yes then what's

the smallest exponent that occurs in that group well here i have n to the first power n squared and n to the first power so an exponent of one would be the smallest right so n to the first power or just n would go into the gcf now i have an m cubed here but i don't have an m here or here it's got to be Common to everything and so there's not going to be a variable m that's going to go into this so your gcf is just going to be 3 n all right

let's take a look at another one so we have the greatest common factor of 30 x y cubed z 24 x to the fourth y cube z 36 x y cubed z squared and 54 x y cubed z to the fifth so what is the gcf so let's think about the number part first So we have a 30. let's think about that so that's 3 times 10 10 is 5 times 2 3 times 5 times 2 24. let's think about that as 3 times all i have is a 2 here so 3 times 2 would

be 6 and then times another 4 and then 36 let's do again all i have is a 2 here so let's do 2 Times 18 18 would be 9. all i have is a 3 here so let's do 2 times 3 so that's 6 so 2 times 3 times 6. and i'm breaking this down to where i already know that all i'm going to have is a 3 times a 2 that's common to everything right so i can already see from these first few that the gcf is going to be a six in terms of

the number part right because this guy right here if you think about It is really what it's 6 times 9. so if i look at this nothing has a 5 in the prime factorization other than 30. so you can just mark that out so all i have here really is just 3 times 2 or 6. so i'm just looking for that and everything else and it has this has a 6 this has 3 times 2 or 6. this has 3 times 2 or 6 and then this has a 6. so the gcf the number part

at least is going to be 6. Now what about the variable part what about the variable part well is x common to everything i have an x here here here and here so let me put an x in there now what's the exponent going to be again you look for the smallest one so this is the number one four one and one so it's just going to be to the first power or just x Now the next variable we have y cubed so y cubed y cubed y cubed and y cubed so everything is y

cubed so the smallest exponent is a 3 that's going to go in there so this is going to be y cubed and then for the variable z i have z z z squared and z to the fifth power the smallest exponent is a one so i just throw a z in there and so my greatest common factor is going to be 6 x y cubed z all right let's take a look at another One so we have 44q to the eighth power p squared r cubed 22 q squared r cubed 33 q cubed r squared

and then 132 qp again let's look at the number part and i think it's obvious at least to myself that the gcf for the number part is 11 right this is 11 times 4 this is 11 times 2 this is 11 times 3 this is 11 times 12. and i can eyeball that and see it and again over time you're going to be able to do that as Well if you can already and you can easily see that okay this has a 4 that's 2 times 2. this has 1 factor 2 but no factors of

2 here so this isn't going to work that's not going to work and nothing else is in these other than 11. so you can just cross these out as well you're only left with the possibility of 11 for your number part so let's get rid of this let's think about what the variable part Would be so i have q q q and q the smallest exponent is on this one right this is to the first power you've got cubed squared into the eighth power so we're going to go with just q then when i look

at the next variable that appears it's p so i have p squared no p here so you can just stop right it doesn't matter if There's a p down here if it's not common to everything it does not go in now the next variable that appears is r so r cubed r cubed r squared no r here not common to everything doesn't go in so your gcf your greatest common factor here is just going to be 11 q all right so let's take a look at 8 z cubed y to the 10th 24z cubed 20

squared x 12z squared So for the number part for the number part everything here is divisible by 2. so i kind of think about that in my head so okay this is divisible by 2 this division by 2 so is this so is this everything is also divisible by 4. so i can think about this as 4 times 2 4 times 6 4 times 5 and 4 times 3. now because if i look at this right here this is a 5 this is a 3. this is a 2 and This is a 6. so everything

about this everything else have another factor of 2 meaning is it divisible by 8 the answer is no so i can mark this out and then basically everything else crumbles right i can just mark all these out because the greatest common factor the number part is just going to be a 4 right because it's got to be common to everything so little tricks like that are going to speed up your work Dramatically so the number part is 4 and then for the variable parts i have z z z and z smallest exponent is a two

right z cubed z cubed z squared z squared so z squared and then i have a y to the 10th no y i can stop i have an x no x anywhere else so the gcf here is 4z squared hello and welcome to algebra 1 lesson 37. in this video we're going to learn About factoring out the gcf again this is known as the greatest common factor so basically in this lesson we're going to learn how to write a polynomial as a product so this process is known to us as factoring and essentially all we're

doing is we're reversing the distributive property so let me kind of start off with a brief review of this and at this point in algebra 1 you Should be very familiar with the distributive property we use it all the time to simplify so if i have something like a times the quantity b plus c we know that we can take a and multiply it by b and i'd have a b and then plus a times c ac and essentially all we're going to be Doing is reversing this i'm going to see something that's a b

plus ac and i'm going to go to a outside of the parentheses times that quantity b plus c and we'll see how to do that in a minute another example a times the quantity b minus c so we have a times b this is a b minus a times c or ac we can even do this with numbers involved so we have 5 times the quantity three plus two This is five times three five times three plus five times two five times two when you get an example with numbers it's easy to show that it's

mathematically valid right if i do 3 plus 2 first that's 5 5 times 5 is 25. if i do it this way 5 times 3 is 15 and then 5 times 2 is 10. if i sum 15 and 10 i get 25 as well so last one here we have five times the quantity three Minus two so five times three five times three minus five times two five times two and again if you wanna see that this is mathematically valid three minus two is one five times one would be five if i look here five

times three is fifteen again five times two is ten fifteen minus ten is again five all right so factoring is essentially the opposite of the distributive property so we are pulling out what's common to All terms and placing it outside of parentheses this is something you're going to do for the rest of your life it's not something that you're just going to do for you know one little section in algebra 1 and then you can kind of forget about it you will become very very good at factoring and kind of as you move forward the

factoring gets tougher and tougher the stuff we're going to look at in this lesson is going to be very very Simple very very straightforward all right so let's take a look at our first example and again what we want to do here is factor out the gcf the greatest common factor so if we're going to factor out the greatest common factor the first thing we need to do is find the greatest common factor so what is that here so the gcf is equal to what well if i look at this first term here i Have

8x okay and if i look at the second term i have 20. now there's no x involved in both terms so it's not going to be part of the greatest common factor so now i'm just thinking about 8 and i'm thinking about 20. so 8 i can think of as what it's 2 times 2 times 2 and 20 is what it's 2 times 2 times 5. so it's pretty easy to see that the Greatest common factor here is 4 right 2 times 2. all right so once we know that we kind of move on to

our next step before i kind of get into that let me one more time show you the distributive property with something we've already seen so let's say i have again a times the quantity b plus c so i take a i multiply it by b i get a b i take a i multiply it by c i get plus Ac now what if i was to start out with a b plus a c and i want to go into this format in other words i want what's common or the greatest common factor which in this

case is a to be outside of a set of parentheses how do i get this space and this space what do i do well there's kind of two ways you can think about it The easiest way to do it would be just to take each term here and divide by what you're putting outside of the parentheses in other words if i took a b and i divided by a i get b if i took ac and i divided by a i would get c and the reason that works is because remember division is the opposite

of multiplication So if i take a b and i divide by a i get b if i was to multiply a by b i get a b right back right it's just the opposite operation so the other thing that you can do if you don't like doing it that way you could take this a b plus ac and you could just think about okay i'm taking this a that's common to both and i'm placing it outside of a set of Parentheses so that's going to go there then i could say well what's left after i

take that a out well there's just a b there and then there's just a c there right that works just as well so now taking this example and applying it to this what are we going to see well if i have 8x minus 20 And i know my gcf is 4 so let's write 4 out there well again i can start off by just saying okay each term divided by the gcf 8x over 4 is 2x and then minus we'd have 20 over 4 that's going to be 5. so i've just factored my first polynomial

i took 8x minus 20 and i said this is equal to 4 times the quantity 2x minus 5. and again if you want to check this it's very very Easy all you have to do is do the multiplication okay so 4 times 2x is 8x 4 times negative 5 is negative 20. so bam right there i know i got the right answer now one thing i want to draw your attention to a lot of students will understand how to factor but they'll make a mistake and not get the correct gcf so for example let's say

i put 8x Minus 20 and i only factored out a 2 so i put a 2 here so then 8x divided by 2 is 4x 20 divided by 2 is 10. so minus 10. if your teacher asks you to factor out the gcf this is not the right answer this is wrong because i only factor out a 2. i have 2 that's common to each term inside the parentheses that can come out So you might end up getting the wrong answer even though you factor correctly because you didn't pull everything out that you could so

pay close attention to that now the other thing i want to do i just want to show you really quickly the other method that i kind of used over here so i could have written this as 4 times 2x minus 4 times 5. And again this makes it easier to visualize i'm pulling out a 4 and again that's going outside of a set of parentheses so that's going to go there and then what's left here it would be 2x and then minus here it would be 5. so again i get the same thing either way

so however you want to look at that those are the two basic methods you can use to take that polynomial and write it In factored form all right let's take a look at another one so we have 2p plus 5p squared so again the first question is what is the greatest common factor and it's pretty easy here because 2 is a prime number and 5 is a prime number so the only thing these have in common number wise would just be the number one so kind of moving on we see that they each have the

variable p one is p to the first power one is p squared so you use the exponent on the Variable that's smallest in this case that's going to be one so the greatest common factor is just p all right so to factor this guy out again i'm going to put p outside of a set of parentheses so inside what am i going to use to fill these two spaces again i can do it two ways i can divide 2p by p 2p by p that's going to give me 2 and then i can divide 5p

squared by p That's going to give me this will cancel with this 5b so 5p and then the other way again if you're more comfortable doing it this way you could also say okay well i have 2 times p i'm going to write this in a different color and then plus i have 5 times p times p so then it makes it completely obvious that i'm going to pull this out in each Case pull this out place it outside the parentheses and then what's left so i'd have a 2 here plus in here i'd have

5p so either way you want to do it you would get p times the quantity 2 plus 5p and if it makes you feel more comfortable because i know none of this is in standard form you could flip the order of this to begin so you could say 5p squared plus 2p is equal to and then you'd have p times flip the order of this 5p plus 2. and again you can check it with multiplication p times 5p is 5p squared p times 2 is 2p so you get 5p squared plus 2p all right so

here's another thing that's trip students up a lot so sometimes it is convenient to pull out a negative one the result is not the true gcf but it's common practice so Your teacher might say on the instructions to pull out the greatest common factor but it might be more convenient for you to pull out a negative one and this might not result in this section but kind of moving forward in the next section when we start talking about factoring by grouping you're going to see right away that sometimes you have to pull out the negative

of the gcf to get everything to Work out so let's take a look at this real quick so i have negative 20 v to the fifth power minus 15. so what is the gcf here the greatest common factor well if i look at negative 20 v to the fifth power and i look at negative 15 i know there's no variable involved so i can really write negative 20 as negative one times five times let's just Say four let's shorten it a little bit and negative 15 i can write as negative one times five times three

so the gcf the greatest common factor between the two is five so let's start out by doing it with that and then we're going to try negative 5 right because we have a common factor of negative 1 between the two so let's start out with just 5. so this is equal to i'm going to put my greatest common factor out in front And i have my parentheses space minus space okay so if i took negative 20 v to the fifth power and i divided by five this would cancel with this and i'd have negative four

so i'd have negative four v to the fifth power then minus next i'd have 15 divided by five and we know that's three we know that's three so that's factoring out the greatest common factor but again sometimes you Might not want to do that sometimes you might want to factor out a negative 5 because you might not want minus signs in here right so if i factored out a negative 5 or what i have then well if i think about this negative 20 divided by negative 5 is positive 4 then i have v to the

fifth power negative 15 i'm going to include that sign there negative 15 divided by Negative 5 is plus 3. so either way you do it if you went back and use your distributive property you get this back but by definition 5 is the greatest common factor so this would be the true answer but again in some cases you're going to need to factor out the negative of the gcf and so you'd want to do it this way all right let's take a look at another one we have negative 2y to the 8th power x cubed

Plus 10y to the 10th power plus 8y to the fifth power x cubed okay so what is the greatest common factor here well i know that there's a y in each term but there's not a next in each term so x won't be involved the variable part is pretty easy the lowest exponent on any of the variables is a five so it'll be y to the fifth power And then as far as the numbers go each one is divisible by 2. 2 is the smallest number and it's a prime number so it's going to be

2y to the fifth power so i'm going to pull that guy out i'm going to have 2y to the fifth power and that's going to be times i'm going to have three spaces here because i have three terms so again i could just divide each so negative two Y to the eighth power x cubed divided by two y to the fifth power negative two over two is negative one so i'll just put a negative out in front then y to the eighth over y to the fifth is y cubed and then x cubed is just

going to come along for the right here all right let me erase this kind of move to the next term you'd have 10 y to the tenth over 2 y to the fifth 10 over 2 is 5 Y to the 10th over y to the fifth is y to the fifth all right for kind of the last term here we have eight y to the fifth power x cubed this is over two y to the fifth power eight over two is four y to the fifth over y to the fifth is one so you would

get four x cubed and so we have our answer here in factored form this is 2y to the fifth power times the quantity Negative y cubed x cubed plus 5 y to the fifth power plus 4x cubed now again you might come across a situation where you want to factor out the negative of the gcf so if i factored out a negative 2y to the fifth power all the signs in here would just change right so if i multiply or divide by a negative it just changes the sign so this would be plus this would

be minus this would be minus right but again if i change this back to Plus then this becomes minus plus plus so for the next one i have u to the sixth power of v squared plus two u to the eighth power v minus u to the tenth power so what is my gcf here there's obviously not going to be any numbers involved i have u to the sixth power u to the eighth and u to the tenth of v is not going to be involved because it's not common everything so my gcf here my

gcf Is u to the sixth power so let's pull that out so let's pull that out and what's going to be left well if i have u to the sixth power v squared over u to the sixth power this is going to cancel and i'll just have v squared if i have 2u to the eighth power v over u to the sixth power this is gonna cancel with this i'll have u to the Second power or u squared so this will be plus two u to the second power v then minus if i have u

to the 10th power u to the 10th power over u to the sixth power we're going to cancel this out and this will be a 4. so this would be u to the fourth power and so i'd have my answer u to the sixth Power times the quantity v squared plus two u squared v minus u to the fourth power all right let's take a look at another one we have negative 5 u squared v minus 25 uv plus 45 u cubed v cubed so my gcf here if you look at the number parts it's

going to be 5 right you have 5 here 25 is 5 times 5 45 is 5 times 9 so it's going to be 5. And then for the variable part you have a u in each case and you have a v in each case the smallest exponent on each is a 1 so it'll be u v okay so if we want to pull this guy out let me write this down here you'd have 5 u v outside of parentheses so i'm going to divide each term here by this so negative 5 u squared v over

five u v We know that negative five over five is negative one u squared over u this is going to cancel be u to the first power v over v is one so you'd have a negative u basically negative u and then you would have minus we have 25 25 uv over 5uv so the uv in each case is going to cancel 25 over 5 is 5. so this is minus 5 here and then plus next we have this 45 again 45 u cubed v cubed over 5 u v so then 45 over 5 is

9 u cubed over u is u squared v cubed over v is v squared so this is going to end up being 9 u squared v squared okay so that's my answer 5uv Times the quantity negative u minus 5 plus 9u squared v squared now again i want to come back to the fact that we can also do the negative of the gcf let me show you a little trick for this one other way you can do it so let's say i decided to factor out a negative 5 uv so let's write this as negative

1 times 5 times u Times u times v and then instead of minus i'm going to put plus i'm going to put negative 1 times 5 times 5 times u times v and then plus one more time instead of 45 what can i write i can write negative 1 times negative 1 right negative 1 times negative 1 is positive 1. so i'm not really doing anything illegal then times 5 times 9 and then i need a little bit more room so times u times u times u let me move this over a little bit and

then times v times v times v okay so if i look at the negative of the gcf that's negative five so that's negative five and then for the variable part it's uv so i'm going to circle that in each case negative 5 uv So negative 5 and then uv so if you think about it here what would be left right if i pulled out the negative of the gcf well if i pulled out a negative 5 uv what's left in this case i would have a u only and then here i would have a plus

5 then here i would have a negative a Negative 9 u squared v squared okay so using this little trick i can do it the other way and you can see that all i did was change the sign of everything instead of this being negative it's positive so this being negative it's positive so this being positive it's negative so you can always use little tricks like negative 1 times negative 1 is positive 1 to kind of write this in a longer format i know some of you like to write it this way versus dividing each

term i've seen a lot of students have a lot of success with this and just really hate doing the division so if this works for you i think that's great use it if it doesn't then by all means you know use the division either way you get the right answer all right let's take a look at the final problem here i have 16 x Cubed y cubed z to the fourth plus 20 x to the fourth y squared z minus 2x cubed y squared minus 20 x y squared z squared all right so what is

the gcf well for the number part i know it's going to be a 2 right because everything is divisible by 2 2 is the smallest number so it's going to be a 2. for the variable part everything has an x the smallest Exponent is a 1. then everything has a y the smallest exponent is a 2 but not everything has a z this doesn't have a z here so the gcf would be 2xy squared now pull this out and we're going to have four terms here so if i divide this by 2xy squared so 16

x cubed y cubed z to the fourth over 2 x y squared and eventually you'll be able to do this in your head but for Right now i want to just make it completely clear 16 over 2 is 8 x cubed over x is x squared y cubed over y squared is y so this is going to give me 8 x squared y z to the fourth and then we'll have plus for the next term I'm going to have 20 x to the fourth power y squared z over again 2 x y squared so 20

over 2 is 10 x to the fourth over x is x cubed y squared over y squared is 1. so i'm going to have 10 x cubed z and then move on to the next term i have minus 2 x cubed y squared and this is over again the gcf of 2xy squared So negative 2 over 2 is negative 1 x cubed over x is x squared y squared over y squared is 1. so this would be negative x squared and then lastly we have this term here which is negative 20 x y squared z

squared over 2 x y squared okay so let's divide this by this and i'll get negative 10 and then x over x is 1 y squared over y Squared is 1. so essentially i'll just have negative 10 z squared so minus 10 z squared so now i'm done so let's erase this here and let me kind of put equals and let me move this down over here and i'll move this up okay so we've written this as 2xy squared times the quantity 8x squared y z to the fourth plus 10x cubed z minus x squared

minus 10z squared hello and welcome to algebra 1 lesson 38 in This video we're going to learn about factoring by grouping so before we kind of jump in and start talking about factoring by grouping i just want to revisit what we did in the last lesson so our last lesson as you'll recall was all about factoring out the gcf or again the greatest common factor so if i had something like let's say 5 x to the fifth power minus 15 x cubed how would i go about factoring out the gcf Well essentially the first thing

i'd have to do is find the gcf so between these two terms i have a 5 here and i have a 15 here now 15 is divisible by 5 right 15 divided by 5 is 3. so the number part for the gcf would be a 5. that's simple enough now for the variable part it's even simpler in each case i have an x and i just go with the smallest exponent that Appears in either term so the smallest exponent is a 3 so this would be 5 x cubed for the gcf now that's the easy

part once we've gathered our gcf what we're going to do we're going to put equals and we're going to place that gcf outside of a set of parentheses now what's going to go inside the parentheses we'll have two spaces for those two Terms and then our minus sign don't forget about that now how do i figure out what goes in each space well i want you to think for a minute about what you're doing we talked about this in the lesson this is going to be multiplied by this and it's going to give me this

so we have multiplication here the opposite of multiplication is division So if i want to figure out what this unknown is here all i need to do is take the original term which is 5x to the fifth power and divide it by this 5x cubed is 5x cubed and i will get my missing term here so 5 over 5 is 1 right that would cancel and then x to the fifth power over x cubed is x squared so this would be x squared here and you can see that works itself out Because 5x cubed times

x squared would get me back to the original 5x to the fifth power we're going to do the same thing for this term right here this 15x cubed i'm going to again divide it by the gcf so 15x cubed over 5x cubed because again this times this okay will give me back this so again i can find that unknown by dividing this by this Okay so now what we're going to do i'm going to cancel the 15 with the 5 this is going to be a 3 and then x cubed over x cubed is 1.

so this just be a 3 here and again check that with multiplication 5x cubed times 3 gives me 15x cubed so if you went back okay if you went back and you used your distributive property 5x cubed times x squared would give you 5x to the fifth power then minus you have a minus there 5x cubed times 3 would give you 15x cubed so you get back what you started with and again one more time for those of you who don't like to do the division i know it's confusing for some of you you could

write this out if you wanted to you could say okay this is 5 times let's say we do it like this x cubed times x squared Minus we could do 5 times 3 times x cubed now if i'm going to pull out a 5x cubed so let me kind of use a different color here if i pull this out here and i've got to circle it here and here you just look at what would be left so if i write a 5x cubed here what would go in the space in each case is what's left

so you'd have an x squared there so an x squared and then you've got your minus And then what's left over here well just a 3. so just a 3 right there so that's an alternative method for those of you who don't like to do the division i know it confuses some of you especially you know in your algebra one once you get to algebra two you've done this enough so it'll kind of make more sense for you but this is an alternative method if you just can't get the division down okay so let's look

at something else Real quick we're going to kind of keep this we're going to go to a different screen here and i want to show you something that's very similar to that but just a little bit more complicated just kind of one step up the ladder so let's say i saw something like x squared times 2x minus 5 and then i had plus let's say i had 3x times 2x minus 5. so at this point in your kind of algebra class you've probably never seen or heard anybody talk about a common binomial factor but that's

what we have here the first thing i want to point your attention to is that this is multiplication here stick something outside of parentheses you're multiplying so i have x squared that is multiplying this whole quantity 2x minus 5. so these are factors Same thing goes over here so i have 3 is a factor times x which is a factor times again the quantity 2x minus 5. so these are all factors now i have a common binomial factor binomial is just a polynomial with two terms of 2x minus 5 in each case what i can

do is i can actually factor that out the same way that i factored the 5x cubed out in this example here So let me kind of show that to you let me erase this real quick and i know it's a little confusing at first so the first thing i'm going to do is put brackets around this just to make it obvious what i'm going to do so x squared times again 2x minus 5 plus 3x times 2x minus 5. now if i was to pull this out of the brackets let me kind of scroll down

And i'm going to visually show that to you so let's say i drag this down here and i pull it out so i'd have 2x minus 5 out in front of the brackets now so let me put the brackets there what's going to be left in each case well i've pulled this out from here so it's not there you can kind of think about it as we did in the last example with the division you could say if i had X squared times 2x minus 5 and this was over the quantity 2x minus 5 we'll

have a common factor here of 2x minus 5. so that would cancel and i'm just left with x squared again you could just picture it as i'm pulling this out so i'm just left with x squared however it makes sense for you long story short i'm just going to have x squared here and then plus This is pulled out from here so i just have a 3x so plus 3x and instead of brackets i could just use parentheses any grouping symbols will do and what i have is i have 2x minus 5 times i have

x squared plus 3x now we're going to learn later on that we can further factor this and we can pull another x out right because that's common to each term here but for right now let's not even worry about that okay Let's just keep it as it is so that is what factoring by grouping is going to entail we're going to pull out a common binomial factor you're going to see that kind of moving forward but before i go on i want to show something to you i want to prove to you that this is

still equal to this and a lot of you will doubt that but i want you to notice let me kind of erase this real fast i want you to notice that this Is nothing more than the way we used to multiply these guys together right before we learned foil we would take this guy right here and we'd multiply it by each term of the second polynomial so that quantity 2x minus 5 would get multiplied by x squared which it is here then plus we'd have 3x multiplied by that polynomial again so if i did foil

on this 2x times x Squared is 2x cubed the outer would be what plus 6x squared the inner would be minus 5x squared and then the last would be minus 15x so if i combine like terms i would end up with 2x cubed and then plus x squared and then minus 15x and i'm going to show you that you're going to get the same thing up here let me kind of erase that real quick x squared times 2x is 2x cubed x squared Times negative 5 is minus 5x squared and plus 3x times 2x is

6x squared and then 3x times negative 5 is minus 15x so if i combine like terms here again i'm going to get plus x squared in the middle so 2x cubed plus x squared minus 15x again 2x cubed plus x squared minus 15x all right so let's scroll down now we're going to talk about what factoring by grouping is and how you execute that process All right so when a polynomial has four terms we can sometimes factor using a method known as factoring by grouping and when we say grouping we're just putting it into two

different groups so i'm going to show you that right now so we're going to rearrange the terms into two groups again two groups and since there's four terms in the polynomial we have two groups of two Terms each we're going to do that so that each group has a common factor and in some cases the common factor might be one or negative one so the next thing we want to do is factor out the gcf or again in some cases it might be the negative of the gcf from each group then if the result yields

a common binomial okay just like we saw in that example we looked at a minute ago we want to factor this out if the result Does not yield a common binomial factor you want to try a different grouping so you might need to rearrange your terms and set it up into another two groups of two and just try again all right so let's take a look at the first one so i have 15x cubed minus 25x squared plus 9x minus 15. so the first thing i always do is i try to think about if i

set it up into two groups of two just as they are could i pull something common out from Each so if i look at these two terms and i made them into a group i see they have a 5 that's common and an x squared that's common so i could pull that out if i look at these two they both have a 3 income so let's go ahead and try that setting this up into again two groups of two with this being 15 x cubed minus 25 x squared let's make that into a group then

plus We'll have a group of 9x minus 15. and again i already told you the gcf for each for this one it's what it's 5x squared or this one it's 3. so let's pull that out from each one so if i pulled out a 5x squared from here what would i have left inside 15x cubed divided by 5x squared would be 3x then i have the minus 25x squared over 5x squared is 5 And then plus if i pulled a 3 out from here 9x divided by 3 would be 3x then minus 15 divided by

3 would be 5. so we can see that immediately we have a common binomial factor we have 3x minus 5 and we have 3x minus 5. so we're going to pull this out okay we're going to pull this out so again imagine i have parentheses around the whole thing And i pulled out the 3x minus 5 what would be left inside well i'd have a 5x squared plus a 3 right because this was pulled out i can just kind of draw an arrow down here that's pulled out put outside that's what's left inside the 5x

squared plus the 3. so we have successfully factored this by grouping so we can do is use foil here And we can prove that these two would multiply together to give us that original four-term polynomial so 3x times 5x squared is 15x cubed then 3x times 3 is plus 9x then negative 5 times 5x squared is minus 25x squared then negative 5 times 3 is negative 15. and if i reorder the terms i get 15x cubed Minus 25x squared plus 9x minus 15 which is exactly what we started with now let me erase this real

quick and i want to show you that this would also work with a different grouping so let's say you decided to reorder the terms and you said that you wanted to look at i don't know 15 x cubed plus 9x let's say that's a group because you can see that there's a 3x that's Common right there's a 3x that's common and then you have another group that's negative 25x squared and minus 15. and what's common here i'm going to pull out a negative and then a 5. right so negative 5 is common to both so

if we do that what would we get so let's put parentheses around each let's put parentheses around each and again i'm going to pull out a 3x from here so 3x times Inside i would have a 5x squared plus a 3. and then from here remember i'm going to pull out a negative 5. so i'd have plus and i can have a negative 5 out in front and then inside i'd have what i'd have 5 x squared plus 3. now i didn't have the same binomial factor but what's interesting is that if i look At

this here and i pulled this out what i'm going to end up with is the same thing in the end i have what i have 3x minus 5 which is what i ended up pulling out last time times 5x squared plus 3. so i end up with the same two binomials right but it's a little bit different how i got there now i want to show you one last thing if you had chosen if you had chosen To do let's say 15 x cubed minus 15 let's say that's one group and then negative 25 x

squared plus 9x let's say that's another group this would not yield a common binomial factor so what's common here would be what it would be 15. so if i pull out a 15 here i would have x cubed minus 1 inside the parenthesis right 15x cubed divided by 15 would just Be x cubed and then negative 15 divided by 15 would be negative 1. then kind of thinking about over here i could pull out an x or negative x however you want to do that wouldn't matter either way so let's pull out a negative x

so let's put minus x times inside of here we'd still have 25x and then here we have a minus 9. So you can see that you don't have a common binomial factor here you have x cubed minus 1 and you have 25x minus 9 not a common binomial factor so this is one of the scenarios where you might have set it up like that and said okay well there's something common to each one that i can pull out you pull it out you don't have a common binomial factor so you have to try a different

grouping all right let's take a look at another one and we'll kind of speed through These now so we have 15 n cubed minus 9 n squared plus 10 n minus 6. so i always like to look at the way they're set up i start out with these two and these two so i'll say okay what could i pull out of here well i could pull out a 3n squared what could i pull out of here well i could pull out a 2. so mentally you can kind of look at it And say well if

i pulled out a 3n squared here what would i have left in here well i'd have a 5n and i'd have a 3. if i pulled out a 2 here i'd have a 5 in and it'd have a 3. so i already know that's going to work out right that's what you're going to do kind of moving forward you're going to kind of do the groups here right you're not going to go through all the the tedious process of Writing everything out and doing you kind of do it mentally and then figure out what group

is going to work so i'm going to pull out again a 3n squared from that first group and that's going to give me a 5 n minus a 3 and then plus i'm going to pull out a 2 that's going to give me a 5 n minus a 3. so i have my common binomial factor here Which is 5n minus 3 okay 5n minus 3. and if i pull that guy out what am i going to have so i'd have 5n minus 3 times what would be left if i pulled out a 5n minus 3

here i'd have 3n squared if i pulled out a 5n minus 3 here i'd have a 2. so plus 2. and so this is my factored form i have the quantity 5n minus 3 times the quantity 3n squared plus 2. and in the interest of time i'm not going to check It but if you went back and used foil you would get this four-term polynomial back all right let's take a look at another one so we have 35 mn plus 5m and then we have plus 28n plus 4. so again let's check these first two

and the last two so if i pulled out a 5m what would i have here well this would be 7n and this would be 1. if i pulled out what's common in both Which is 4 this would be 7n and this would be 1. so that's exactly what we need so let's go ahead and do that so we'll put equals again in this first one i'm pulling out a 5m and what's left inside 35 mn divided by 5m is 7n plus 5m over 5m is 1 plus what i'm pulling out of this group i'm pulling

Out a 4. so 28n divided by 4 is 7n and then plus 4 over 4 is 1. so you can see we have that common binomial factor of 7n plus 1. okay that quantity there so if i pulled that out i would have seven n plus one outside and then i would have five m that's left here five m plus from here this is gone so i just Have a four so i'd have the quantity seven n plus one times the quantity five m plus four and again if you use foil to check this you're

gonna end up with this exact same four term polynomial back right now sometimes you're gonna have a common factor for all terms present before we begin the process it's best to pull this out before you start Because if you don't you're going to end up with additional factoring in the end so if i look at these terms here everything is divisible by 8. 320 divided by 8 is 40. 40 divided by 8 is 5. 128 divided by 8 is 16 and 16 divided by 8 is 2. so we can pull out this common factor of

8 before we even start so in other words this is what this Would look like i'd put an 8 out in front i'd put some parentheses and then i'd say okay well i have 40xy and then plus i have 5y and then minus i have 16x and then minus i have 2. now once we've done that we just factor the inside part just like the 8 doesn't exist don't even worry about that just leave it outside the parentheses we're going to worry about that at the Very very end i'll show you what to do with

it so just work on this so again we can just go with the groupings that are present at first see if they work out so these two and these two so 40xy and 5y common factor of 5y that would be 8x plus 1 and then 16x and 2 well they're both negative so i could pull out the negative part and i could pull out a 2. so if i pulled Out a negative 2 here i would have 8x left i pulled out a negative 2 here and i have plus 1. so it looks like that

would work out right we'd have that common binomial factor again this is something you're going to want to do mentally if you go through and write it all out it's going to take you a little bit longer so i'll rewrite the 8 and then i'm going to split this up so 40xy plus 5y the gcf there is 5y So 40xy divided by 5y would be 8 x and then plus 5 y over 5 y is 1. then we have this minus 16x minus 2. i'm going to pull out that negative part also right they're each

negative i'm going to pull that out because i notice that here i have a positive okay so i want that so i can have a common binomial factor so that's the situation we're pulling out the negative of the gcf so I'm going to pull out negative 2 and let me just put plus negative 2 because that always looks a little better to me so plus negative 2 times inside of here 16x over 2 would be 8x and again i've already taken care of the sign so i'm just thinking about the numbers as if they were

positive and then 2 over 2 is 1. so plus 1. Okay so let me kind of close the parentheses here and again you can see you have a common binomial factor 8x 1 8x plus 1. so that's what we're going to factor out so i have the 8 out in front that's going to stay there and then times this common binomial factor that's 8x plus 1 and then times what's going to be left inside here well what i'm going to have if 8x plus 1 was gone so if i just line that out i'd have

a 5y Then plus if 8x plus 1 was gone it'd have a negative 2. so you put plus negative 2 or you just put minus 2 doesn't really matter and close the parentheses there and we have our answer we have 8 times the quantity 8x plus 1 times the quantity 5y minus 2. now this is factored completely i'm going to show you what would happen if you didn't do this step when you begun you would have had something that had Additional factoring that needed to be done so if i look at these two groups here

what can i pull out well i can pull out a 40y so i pull out a 40y and i would have what i would have 8 x plus 1. then over here i could pull out a negative 16 and inside i'd have Again 8x plus 1. so if i factor the common binomial factor of 8x plus 1 out i'd have 8x plus 1 times what's left inside would be 40y this would go away and then you just have the negative 16 right because this would be gone so minus 16. and if you look at the

8x plus 1 that's fine there's nothing else i can take out of there but if i look at this right Here if i look at 40y minus 16 there's something i can pull out of there right 8 is common to both so if i report my answer like this on my test my teacher is going to go over here and she's going to put wrong because i didn't factor completely there was something else i could have pulled out okay you got to keep that in mind when you're doing a test or homework you might have

factored but you didn't factor completely so the answer is wrong If i pull out an 8 from here 40 divided by 8 is 5. so this should be 5y so this 8 is going to go out here if i pull out an a from here this would be 2. so this is how we got our original answer but again i pulled out the 8 to begin with so i didn't have to go through that additional step in the end so if you don't want to pull it out and you want to do additional factoring in

the end That's fine but you have to look for you have to always be looking for additional factoring that needs to be done because that's the quickest way to get either a wrong answer or at minimum points off on your test or homework by not factoring completely right so the answer here is 8 times the quantity 8x plus 1 times the quantity 5y minus 2. all right let's take a look at another one we have 60 a cubed z minus 15 a squared yc Plus 45 a cubed c minus 20 a squared y z so

again if i look here carefully i notice that 5 is common to everything before i even start and also a squared is common to everything before i start so before i do anything i can pull that out so i can pull out a 5 a squared and rewrite this whole thing here 60 divided by 5 is 12. a cubed over a squared is a so a z Minus 15 divided by 5 is 3 a squared over a squared is 1 so this should be yc plus 45 divided by 5 is 9 a cubed over a

squared is a and then times c and then minus 20 over 5 is 4 a squared over a squared is 1 and then y z okay so now i just go through and factor by grouping just as i normally would so again i'm going to look at what's already in order so these two as A group and these two as a group would that work out i have 12 and i have three so the greatest common factor there will be 3 i have a z and i have y c so nothing i can pull out

for the variables so if i did that i would end up with what i would end up with a 4az and if i pulled out a 3 there that would be minus yc and so here i would end up with what 9 and 4 nothing i can really do there Other than 1 ac and y z nothing i can really do there so 1 is the only thing i could pull out so you'd have 9ac minus 4yz so we would not yield a common binomial factor here so we would have to reorder our terms to

get a common binomial factor so let's go ahead and try a different grouping so i noticed that this has a z and this has a z this has a 4 this has a 12 12 is 4 times 3. so let's try writing let's try writing 5a squared times the quantity 12az then i'm going to put minus 4yz so this could be a group and then minus 3yc plus 9ac so now let's look at the way it's set up here so and this one's a little complex so i'm going to actually have to write it out

so i'm going to put 5 a squared times What's common here i have a 4 and a z right so i could pull out a negative 4z also but let's see if we need to do that let's just start out with positive 4z so what's going to be left inside the parentheses 12 divided by 4 is 3 z over z is 1 so just 3a then minus 4 over 4 is 1. z over z is 1 so just y now i have this negative out in front with these two it's a negative 3yc and it's

a positive 9ac so i can start by Just factoring out a negative 3 c so i'm going to put negative 3 c out in front and i like to put plus negative that's just me so i'm going to put plus negative 3 c and then inside if i pulled out a negative 3c i would just have a y and then if i pulled out a negative 3c all that's going to happen here is its sign is going to change so this would be negative and then you'd have a 3 right 9 over 3 Is 3

and then the c would cancel so you just have an a now i don't have a common binomial factor yet but we're close right we have a y and a y we have a 3a and a 3a it's just the signs that are the problem so if i was to factor out a negative here if i was to factor out a negative here Then this sign would change and this sign would change so positive y and positive y negative 3a and negative 3a so that would work out i just change the order remember you can

change the order it won't matter and let me just change the order here to y minus 3a i know this is a little complex when you first see it but this is some of the Problems you're going to get thrown at you right we have to you know rework things and change the signs and if you get something that's close like we just got it's usually just a sign that you're off by so try a different sign on each one all right so now we can go ahead and pull out our common binomial factor which

is y minus 3a so i'll have my 5a squared out in front Not going to touch that inside i'm pulling out the y minus 3a so y minus 3a and then what's left if i pulled this out from here if i canceled it basically i would have a negative 4z and then i would have a minus 3c this would be canceled if this was canceled so i have my answer 5a squared times the Quantity y minus 3a times the quantity negative 4z minus 3c and again if you want to use foil here for these two

and then multiply that result by 5a squared you're going to see that you get this original four term polynomial 60a cubed z minus 15 a squared yc plus 45 a cubed c minus 20 a squared y z and then one additional thing you might want to do and i'm not sure if your teacher requires you to do this or not but a lot of times they don't like any of the Leading terms in here to be negative so you could further factor this and factor out a negative one so you could pull this guy out

and this guy out and just change the sign so i would just put a negative one all the way out in front so i just put a negative sign out here and so this would turn into a positive and then this would turn into a positive and so you could also write the answer as negative 5 a squared Times the quantity y minus 3a times the quantity 4z plus 3c and again if you go through and you multiply negative 5a squared times the quantity y minus 3a and then you could do foil with that result

times this you're going to get your original four-term polynomial back the 68 cubed z minus 15 a squared y c plus 45 a cubed c minus 20 a squared y z Hello and welcome to algebra 1 lesson 39 in this video we're going to learn about factoring trinomials with a leading coefficient of one so before we kind of get started i want you to recall that a trinomial is a polynomial with three terms so in this lesson we will focus on how to factor a trinomial of the form we have ax squared plus bx plus

c so again you have one two Three terms now this is going to be the scenario that we're going to deal with today where a is always going to equal 1. so observe that a is the coefficient on the x squared term b is the coefficient on the x term right the x to the first power term and c is just a constant so that's something i want you to memorize i want you to write it down on a sheet of paper i want you to look it over because we're going to Refer to that

throughout this lesson and i don't want you to be lost i'm talking about b and i'm going to talk about c and so again b is the coefficient for the x to the first power term and c is just the constant at the end of the trinomial so we've already learned the basics of factoring and we're going to kind of take it one step further so far we kind of know you know if i see something like 3x cubed minus 7x that i can pull out an x that's common in both so i could pull

it out and write it outside of parentheses and i would have 3x squared minus 7. all right we know how to do that at this point we also know how to do that if we have a common binomial factor we learn that with factoring by grouping but this is going to be something that's a little bit different we're going to Take a trinomial and we're going to factor it into the product of two binomials right remember a binomial is a polynomial with two terms so in other words we're going to be reversing the foil process

that we learned so for example if i have something like x plus 3 times x plus 5 using foil we know that we can do the first terms x times x is x squared We can do the outer terms x times 5 is plus 5x we could do the inside terms 3 times x is plus 3x and then we can do the last terms 3 times 5 is plus 15. now when we combine like terms here in the middle we would end up with x squared plus 8x plus 15. now the question is If i

started out with this this x squared plus 8x plus 15 and i asked you to go to this and you didn't have all this information how would you go about doing it well the key here is to take this part right here so let's take the x squared plus 5x plus 3x plus 15. and let's write it in a different way to make something obvious we're going to factor out an x from These two middle terms so x squared plus i'm going to stick the x outside of a set of parentheses it's common to both

here so 5 plus 3 and then plus 15. so the reason i'm doing that is to show you that these two integers here that we have inside the parentheses here the 3 and the 5 These two guys are going to sum to give you your middle term in the trinomial so these sum to give you 8. that's what we have here so this remember this is the b so we have ax squared plus bx plus c so b in this case is going to be eight so again notice that those two integers sum to give

you eight now i want you to also notice that those same two integers Multiply to give you 15. right c is 15 in this case so two integers whose sum is b and whose product is c that's what you're looking to find here so let me erase this real quick and let's say i started out with x squared plus 8x plus 15 and i said go ahead and factor this into the product of two binomials well start with what you know you know that you have Two sets of parentheses like this and it's very very

easy to think about what goes here and here remember the first thing you do is you multiply the first terms together the f and foil so this times this would give you this so x times x gives you x squared and that's always going to be the case when your coefficient for the square term is A one right the first two positions are going to be that variable that you're working with raised to the first power so in other words if this was p squared plus 8p plus 15 well then the first two positions in

this case would be p and p all right it's the same thing just changing the variable up do the same thing though or if i had something like Z squared plus 8z plus 15 then this would be z and this would be z right very very straightforward okay so once we've figured that part out that part's a given so the question is how do i find out what goes here and here well again i want two integers so let me write this out two integers where the sum Is going to be b and the product

is going to be c so we already know at this point that those two integers are 5 and 3. right they sum to b remember this is b and the product is c remember this is c so 5 times 3 is 15 again 5 plus 3 is 8. so those are the two numbers we're looking for once we find them we can just write them in any order i could put plus five And plus three or i could switch that up and put plus three and plus five and you have successfully factored this right we

know through foil x times x is x squared x times five is plus five x three times x is plus three x combine like terms there you get plus eight x and then we have plus three times plus five that would be positive fifteen and so you get this trinomial x squared plus 8x plus 15 back so let's Just quickly run down the official procedure so for factoring a trinomial again with a leading coefficient of 1 so that a is equal to 1. you're going to look for any common factors first so just like we

saw with factoring by grouping you know if you had something like let's say 2x squared plus 6x plus 8 i have a common factor of 2 and i could start out by just pulling That outside the parentheses so 2 times the quantity x squared plus 3x plus 4. i would start out by doing that and then i would factor what's inside the parentheses into the product of two binomials so once i've done this once i've looked for any common factors and if they exist pulled them out i want to again find two integers whose sum

is b remember that's the coefficient for the x to the first power term and whose product is c Remember that's your constant at the end if no two integers exist we have a prime polynomial so at this point we're not going to be able to factor that all right let's take a look at the first one so we have p squared plus 18 p plus 80. so again when you first start out just go ahead and put two sets of parentheses i notice that there's nothing common to everything so there's nothing to pull out the

start And so i know that here and here is a given right if this is a 1 right there then it's just going to be p times p very very easy to get the first ones then after that i'm again looking for these two positions right here so give me two integers whose sum is this and whose product is this so what you need to do is start thinking About the factors of 80 well you could make a factor tree if you wanted to and sometimes that's necessary or you kind of just on your scratch

paper go okay well i got 1 and 80. that's obviously too far apart to sum to 18. i've got 2 and 40. again that's too far apart 80 is not divisible by 3. i've got 4 and 20 again too far apart i've got 5 and 16 That's not going to work for a sum then i've got not divisible by 6 not divisible by 7 but it's divisible by 8 8 and 10. so i've found two numbers whose product is 80. 8 times 10 is 80 and whose sum is 18. 8 plus 10 is 18. once

you've found that you can go ahead and stop so i have my combination here and again i can write this in any order i can put plus 8 here plus 10 here or i could reverse it And i could go you know plus 10 plus 8. doesn't matter now i like to check things those of you who watch my videos know that so let's spend a little bit of time here and just check it so if i have p times p that's p squared then p times 8 that's plus 8 p then 10 times p

that's plus 10 p and then 10 times 8 that's plus 80. and you can see that these two middle terms The sum is going to be this right here 18p that's what you're looking for this number plus this number gives you this number this number times this number gives you this number so again two integers whose sum is b whose product is c all right let's take a look at another one so we have p squared minus 11 p plus 28. so again i don't have anything to factor Out here that's a common factor to

start so i'm just going to go ahead and write my parentheses and i know what goes in the first position of each i have p squared so that means i have a p here and a p here p times p is p squared now for this part right here i need two integers whose sum is now negative 11 and whose product is positive 28. so because we have a negative involved It's a little bit more complicated because now not only do we have to think about the factors of 28 we also have to play around

with the signs to get the addition and the multiplication to work out so let's just think about the factors of 28 so we have 1 and 28 those are too far apart to ever produce negative 11 no matter what the signs are we'd have 2 and 14. if i did 14 minus 2 that would Give me 12. so that's not going to work and if i did 2 plus 8 is 10 so it's not divisible by 3 it is divisible by 4. i can do 4 and 7. looks like that might work if i had

a negative 4 and a negative 7 negative 4 plus negative 7 is negative 11. right sine is the same keep the sign the same and you just add absolute values so i Would have negative 11 from that negative times negative is positive negative 4 times negative 7 is positive 28. so that is what we're looking for we're looking for a negative 4 and a negative 7. again once you figure out what it is you just stop this is just you're listing a list of factors just to figure this out it's not like your teacher's going

to grade you on you know coming up with all The factors of 28. so let's go ahead and erase this and we can check our results so p times p is p squared the outer p times negative seven is negative seven p the inner negative four times p is minus four p and then lastly we have negative four times negative seven that's plus 28 and again if i combine like terms here i would have p squared minus 11p Plus 28 exactly what i started with right there so although it's a little bit more work a

little bit more complex if you have a negative sign involved it's still pretty easy overall all right let's take a look at 3r squared minus 24r minus 27. so one thing you'll notice here is that everything is divisible by three because you might look at this problem and say well you know if i'm looking at ax squared Plus bx plus c well the coefficient on the squared term here it's r doesn't matter if it's r z or q whatever it is the coefficient on the variable that's squared is not one here so if you're in

a section where the coefficient's supposed to be one that should clue you in that hey i need to do some additional work before i can start and you can see that everything is divisible by 3 so we're going to pull That out so i'm going to put a 3 out in front inside the parentheses i'm going to have 3r squared divided by 3 that's r squared negative 24r divided by 3 that's minus 8r negative 27 divided by 3 is negative 9. i'm just going to think about what's inside the parentheses here so when i set

up my kind of two binomials i'm going to get three times we'll have A set of parentheses and then another set of parentheses so nothing has changed r squared here i'm gonna put r and then r and now i need two integers whose sum is now negative eight and whose product is negative nine okay two integers that fit that profile so let's think about factors of just 9. forget about the fact that it's negative just factors of 9. you've got 1 and 9 and you've got 3 and 3. Well it can't be 3 and 3

because there's no combination that's going to get you a sum of negative 8. it doesn't matter how you play with the signs so that's out so i need to play with 1 and 9. now i know that the middle term is negative and the last term is negative i can only get that if i have different signs right because we know that a positive times a negative will yield us A negative so that's how we got that through multiplication so one is positive one is negative now we think about the fact that we have one

and nine which one would i make negative which one would i make positive well obviously 9 would become negative right i put negative 9 and positive 1 and that would get us exactly what we need right negative 9 plus 1 would give us negative 8 negative 9 times 1 would give us negative 9. so we're good to go there so My integers would be negative 9 so i'll put minus 9 and positive 1 so i put plus 1. and again i like to check things so let's erase all this and let's do r times r

that's r squared and then r times one that's plus r negative nine times r that's minus nine r and then negative nine times one that's minus nine If i combine like terms in the middle i will have 3 times r squared minus 8r minus 9 and then if i continue i just multiply 3 times r squared that's 3r squared 3 times negative 8r that's negative 24r and then 3 times negative 9 is minus 27 and again that's exactly what we started with right there all right let's take a look At another one we have 2x

squared plus 18x minus 20. again the first thing you should look at is okay all these numbers are even and the coefficient for the squared term is not a one so again that's a dead giveaway if you're in this section where the leading coefficient is supposed to be one that you got to do something first right so i know that everything's divisible by 2 i can pull that out And in my parentheses i'll have x squared plus 9x minus 10. now i'm going to factor what's inside the parentheses here so let's go ahead and set

this up as two times we'll put our parentheses for the two binomials and i know that in this position and this position i have x squared so it's going to be x times x right x times x is x squared now how do i get these two Middle terms again two integers whose sum is 9 whose product is negative 10. again we have a negative involved so it's a little bit more complex just think about the factors of positive 10. forget about the negative for right now the factors of 10 you have 1 times 10

you have 2 times 5 not divisible by 3 4 6 7 8 or 9. so we can stop there so obviously 2 and 5 are not going to work right there's no combination with the signs that would get you to a sum of 9. what about 1 and 10 well if i had a positive 10 and i had a negative 1 that would work positive 10 plus negative 1 gives me a positive 9. positive 10 times negative 1 gives me a negative 10. so i've found my two Integers so plus 10 and then minus 1.

very very easy and again i like to check things so let's go ahead and do 2 times x times x is x squared outer x times negative 1 is minus x inner 10 times x is plus 10x and then 10 times negative 1 is minus 10. if i combine like terms here i have plus 9x and let me scooch this down a little bit and if i multiply here i'd have 2 times X squared that's 2x squared 2 times plus 9x is plus 18x and then 2 times negative 10 is minus 20. so 2x squared

plus 18x minus 20 and that's exactly what we started with all right for the next one i have 5x squared plus 65x plus 180 and again if i am in a section where the leading coefficient is supposed to be 1 and it's not it's a dead giveaway that i can pull something else out everything here is divisible by 5. So if i pull the 5 out i'd have x squared plus 13x plus 36. all right so now i'm going to factor what's inside the parentheses here so let's go ahead and write this as equals 5

times we'll put the parentheses for the two binomials and i always know that okay if it's x Squared here this will be x and this will be x all i need to figure out is the two integers that go in those positions so give me two integers whose sum is 13 and whose product is 36 so let's think about that for 36 we have 1 times 36 and obviously that's not going to work for the next one we could do 2 times 18 that's not going to work We could do 3 times 12 and that's

not going to work we could do 4 times 9 and that will work 4 plus 9 is 13 4 times 9 is 36 so plus 4 and plus 9. and there you go you have 5 times the quantity x plus 4 times the quantity x plus 9. and again if you want to check this which i like to check everything x times X is x squared x times 9 is plus 9x 4 times x is plus 4x 9x plus 4x would be plus 13x and then 4 times 9 is plus 36. and then multiply that

by 5 and you will get this back 5 times x squared is 5x squared 5 times 13x is plus 65x 5 times 36 is plus 180. it's exactly what we started with right there all right let's take a look at m squared plus 9m minus 9. So this is one where we're going to have a problem so again you go ahead and start out is there anything common to everything no there's not let me go ahead and put my parentheses there m squared is the first term so i know i have m here and m

here now for this spot in this spot so i need two integers whose sum is nine and whose product is negative nine so let's think about the factors of just Nine you have what you have one and nine and you have three and three that's it so i can't use three and three that's not ever going to work 1 and 9 if i had negative 1 plus positive 9 that would give me positive 8 and it would multiply to be negative 9 but i don't want positive 8 i want positive not if i used positive

1 and Negative 9 well that's going to give me negative 8 and again i want positive 9. so there's not really anything else i can do i have to use two different signs because the product here is a negative and a negative times a positive yields a negative i can't have two positives or two negatives involved so i have to have one of these two work out they don't and so this polynomial is Prime all right let's finish up with one that's a little bit tricky a lot of students see this and they get really

really confused they're like why is this in this section you have two variables involved but what you're going to do is you're going to end up treating one of the variables as just a number right so let me kind of explain this to you you have x squared minus 4xy minus 21y squared the first variable that you see the x squared Treat that like you normally would okay so when you set this up just go x here and x here nothing else has changed when you go to find these two positions here now what i'm

looking for is i'm looking for a sum of not just negative 4 i'm looking for negative 4y i'm looking for a sum of negative 4y Again the original variable the x squared don't include that but the other variable that's involved you will include that so in this case that's a y so you do negative 4y then i want a product a product of this whole thing here negative 21y squared so just start out by thinking about the number for 21 i have 1 And 21 it's not divisible by 2 it is divisible by 3. i

have 3 and i have 7. now i already know that if i work the signs correctly i can get negative 4 out of 3 and all right i can do that so let's think about that for a second if i had a negative 7 and a positive 3 negative 7 plus 3 would give me negative 4 negative 7 times 3 would give me negative 21. so i could put negative 7 and plus 3. now what about the y part we just throw it in there just put a y here and a y here and think

about it now negative 7y times 3y would be negative 21y squared negative 7y plus 3y would give you your negative 4y so everything works out there when you see something like this with two variables involved it's no more Difficult just have to kind of think about that second variable that you see as if it was a constant right as if it was part of the number that you're working with so in other words again when you look for your sum you take negative 4 and then you just include the y so i'm looking for the

sum of negative 4y and then when you think about your product include the whole thing i'm looking for a product of negative 21y Squared so again i like to check things x times x is x squared outer x times 3y is plus 3 xy inner negative 7y times x is minus 7xy these would combine together to be negative 4xy and then negative 7y times positive 3y would be negative 21y squared so again if i combine like terms here i get x squared minus 4xy minus 21y squared which is Exactly what i started with right there

hello and welcome to algebra 1 lesson 40. in this video we're going to learn about factoring trinomials with a leading coefficient that is not equal to 1. so in our last lesson we learned how to factor a trinomial that looks like this ax squared plus bx plus c Or a the coefficient for the x squared is a one now this is the easiest scenario so if i had something like let's say x squared plus five x plus six i would set this up as the product of two binomials i know that since the coefficient of

this is one it's very easy to get the first two so x goes here and x goes here right because x times x gives me that first term of x squared Then for these two missing parts here i just need to figure out two integers whose sum is 5 right that middle term there the coefficient for the middle term and whose product is that constant term 6. that's why we say a sum of b and a product of c right so this is b and this is c all right so once we figure that out

we just place those two integers in those missing spots so two integers that would multiply together to be six but sum to five would be positive Two and positive three so x plus two times x plus three those two quantities multiplied together would give you x squared plus five x plus six but this is the easy scenario to deal with hopefully at this point you've mastered that we're now going to look at the harder part which is to say now we will focus on factoring trinomials of the form ax squared plus bx plus c Where

a is not equal to one right so this will no longer be one so i can't just jump ahead and say okay this is x and this is x because you won't have that in each position anymore so that's going to be another unknown that you have to deal with so how do we do this so there's two main methods that pretty much everybody's textbook will rely on to show you how to factor A trinomial when the leading coefficient is not 1. now the first method relies on factoring by grouping this is the method i

primarily use i still use it today i think this method is fast easy and efficient the other method is reverse foil i do not like this method personally i just i find it to be just a complete waste of my time it's trial and error and even Though i'm pretty good at it it still ends up taking me longer in most cases i'm going to teach you both because your teacher might require that you demonstrate the ability to do both so i want to cover everything completely but you get the same answer either way and

this method is so much easier for me and i always use it i know there's some tricks out there on the internet you know whatever you find that works for you do it again i like this method it's Never let me down so i'm going to start out with this so the factoring by grouping method the first thing you're going to do and you do this always you factor any common factor this will save any additional work in the end so if i have a trinomial and there's a factor that's common to everything that i

can pull out i'm going to go ahead and do that and then work with what's inside the parentheses all right so the next thing i want to do Consider the trinomial ax squared plus bx plus c you want to look for two integers whose product is ac before you were just looking for the product of c but really if a is 1 1 times c is just c so it's it's the same thing you're just adding this right here because now it's not equal to one so the product is ac and the sum is b

so when you find those two integers You're going to use the integers to rewrite the middle term so that we have a polynomial with four terms and i know that might be a little confusing but once you see me do an example that's going to really resonate with you and show you how easy this is so then at this point we can use factoring by grouping all right so let's start out with the first example we have 5k squared minus 18k plus not So again this is my a this is my b and this is

my c just for reference sake when i start i always look to see is there anything common to everything something i could pull out well there's not i have a 5 that's prime 18 is not divisible by 5 so i can just stop so nothing i could pull out so again i want two integers whose product Is a times c and whose sum is b so b is negative 18 and ac 5 times 9 is positive 45. so we know we have to have two negative values right because if i have a positive sign here

and a negative sign here for the sum i've got to have two negatives two negatives will multiply together give me a positive two negatives will sum together to give me a negative so what are some factors of 45 And let's just think about positive factors and we'll make them negative so we have 1 times 45 obviously that's not going to work not divisible by 2 it's not an even number 4 plus 5 is 9 so it is divisible by 3. 45 divided by 3 is 15. so let's stop there we can see that 3 plus

15 is 18. we just need to make the signs work and to do that we just make a negative 3 and a negative 15. now in the last kind of section we were done at This point we got our two integers we threw them in and we were done it's a little bit more work here you've got to rewrite your middle term into a polynomial that's four terms so i'm going to take this and write it as 5k squared minus 15k minus 3k plus 9. and i did two things there that i want To explain

the first thing is that i rewrote the middle term as negative 15k minus 3k i have not changed the value of this polynomial negative 15k minus 3k if i combine like terms i'm going to get negative 18k that's perfectly legal we've seen that with fractions and all kinds of other stuff where we change what it looks like but the value is the same now the second thing is i put 15 next to 5 because i know they have a common factor of 5. i put 3 next to 9 because i know they have a common

factor of 3. so i did that because remember if you're factoring by grouping you want to put your groups together in a way to where you know you have some common factor you can pull out so that's why i did that all right so now we're going to factor by grouping so in the first group i can pull out a 5k so i would have 5k outside Inside i would have a k minus a 3. and then in the second group i would pull out a negative 3 so i'm going to put plus negative 3

times what's left in there you would have a positive k minus 3. so you can see you have a common binomial factor of k minus three Common binomial factor so let's factor that guy out so this is equal to five k minus three times k minus three if i took out the k minus 3 and put that here what would be left is a 5k minus 3. so that's why i have 5k minus 3 times k minus 3. and i'm done i've completely factored this guy as long as i didn't have a common factor When

i started i won't have one in here that requires additional work i should be done the only thing i have to do now is check 5k times k is 5k squared 5k times negative 3 is minus 15k negative three times k is minus three k and then negative three times negative three is plus nine so you can see i got exactly back what i had here if i combine like terms i'd get negative eighteen k in the middle and then i would have exactly What i started with there 5k squared minus 18k plus not okay

let's take a look at another one so we have 7a squared plus 17a plus 6. so again i'm looking for two integers whose product okay whose product is a c so again this is a this is b and this is c and i know this is ultra confusing because i now have a variable that's a okay i have a variable that's a But you can kind of ignore that typically we have what we have ax squared plus bx plus c so the coefficient for the square term in this case it happens to be an a

it's a squared is denoted with a when we talk about a generic formula so it's just whatever the coefficient on the square term is times whatever that constant is so 7 times 6 is 42 and then sum is B which is 17. right so whatever that number is okay including the sign for your middle term right in this case that's 17. all right so moving on what two numbers would multiply together to give me 42 and sum to 17. well we think about the factors of 42 we've got 1 and 42 we've got 2 and

24 then 4 plus 2 is 6 6 is divisible by 3 so 42 is 42 is 3 times 14 and that looks like that would work right 3 plus 14 is 17 3 times 14 is 42. so again at this point we're not done right before we were done we found our two integers we throw them in we're not done we're going to use that to rewrite that middle term so we're going to break it up into 7a squared plus 14a plus 3a Plus 6. and again notice that i put 7 next to 14 3

next to 6 right because i know i can pull a 7 out of here and a 3 out of here so that's why i did that all right so the gcf for these two would be 7a the gcf for these two would be 3. so then equals if i pull a 7a out of there i'll have 7a times a plus 2. if i pull a 3 out of here i'll have 3 times a plus 2. so if i factor out that common binomial Factor of a plus two right if i pull that guy out i'll

have a plus two and what's left i'll have a seven a plus a three so a plus two times seven a plus three and again check this with foil a times seven a is seven a squared a times three is three a two times seven a is plus fourteen a so fourteen a plus three a would be seventeen a and then 2 times 3 is 6. so You get 7a squared plus 17a plus 6. all right let's try negative 12 v squared minus 24v minus 9. so everything is negative here so i know i can

pull that out to start and make my life a lot easier i hate working with negatives everybody does i also notice that everything is divisible by three so let's pull out a negative three and we'd have four v squared plus eight v plus 3. so now i just need to worry about what the inside of the Parentheses here would factor to so again i'm looking at the a or the coefficient for the squared term and i'm looking for the c which is the constant so multiply those two together 4 times 3 is 12. so product

of 12 and then i'm looking for a sum of 8. so what are Some factors of 12 we have 1 and 12 it's not going to work we have 2 and 6 that would work so let's rewrite that middle term so we're going to have equals negative 3 times four v squared and i'm gonna put plus two v here and then plus six v here then plus three and then equals we'll put negative three out in front we will break this up the greatest Common factor here is 2v so we'll put times we'll have 2v

times if i take that out i'll have a 2v plus a 1 so then plus the gcf here is three so if i pull that out i'll have a two v plus a one and then if i factor out the common binomial factor here that's two v plus one what am i gonna have i'll have a Negative 3 out in front and then i'll have 2v plus 1 multiplied by 2v plus 3. so there's my answer negative 3 times 2v plus 1 times 2v plus 3. and again you can check it in the interest of

time we still have a lot to cover i'm not going to check it but just use foil for this and then multiply by negative 3 and you should get negative 12v squared minus 24v minus 9 back as your answer all right let's take a look at one more and Then we'll look at the reverse foil so we have 20 m squared plus 6m minus 36. so everything is divisible by 2. so let's pull that out to start so i'd have a 2 out in front and then 10m squared plus 3m minus 18 and then i'm

looking at what's inside the parentheses here so remember i want a product of 10 times negative 18 that's Negative 180 and a sum of three so we think about that we know that we're gonna have to have a positive and a negative so what are some factors of a hundred eighty well you've got one times 180 that's not going to work you've got 2 times 90 that's not going to work you've got 3 times 60 that's not going To work you've got 4 times 45 not going to work you've got 5 times 36 not going

to work you've got 6 times 30 not going to work and you've got 10 times 18 that's not going to work what else do we have we have 12 times 15. 12 times 15 would work for some way to work those signs out to where i can get 3. in my head i just look at 12 and 15 i know 15 minus 12 is 3. so positive 15 and negative 12. So let's rewrite this so we have 2 times inside of parentheses 10 m squared i'm going to rewrite that middle term that 3m as plus

15m and then minus 12m right if i combine those together i get plus 3m and then minus 18. okay so if i factor this by grouping i'll have two times inside the parentheses Here i would pull out from this first two a five m so five m times 2 m plus 3. then from this one we're going to pull out a negative 6. so minus 6 times you'd have 2m plus 3. So my common binomial factor is 2m plus 3. it's really easy to see i can pull that guy out so i'd have 2 times

2m plus 3 times what's left i'd have a 5m minus a 6. so 2 times the quantity 2m plus 3 times 5m minus 6. so again if you want to check this use foil here then multiply the result by 2 and you should get 20 m squared plus 6 m minus 36 back all right so now Let's learn an alternative method which involves reversing foil using trial and error and again i just want to state that i personally don't like this method i don't use it i find it to just be very tedious and very

time consuming i like using the factoring by grouping method but for yourself if you enjoy using this if this is more productive for you you know by all means we're just trying to get the correct answer in the end all Right so i want to start out with just a demonstration of foil and then showing you how we would go about reversing this so we know foil is first term so 2x times x is 2x squared so this is the f then outer terms 2x times 4 is plus 8x this is the o then enter

3 times x is plus 3x this is the i And then last 3 times 4 is 12. so plus 12 this is last so if we combine the outer and the inner we're going to get a middle term that's 11x so this is equal to 2x squared plus 11x plus 12. now what if i was to give you the 2x squared plus 11x plus 12 just like this And i take this information away so let's scroll down past it and cover it up and i tell you i want you to factor this into the product

of two binomials and i don't want you to use factoring by grouping i want you to reverse the foil process so the first thing is let's think about foil so we have the f that's the first terms so something times something gives me this guy right Here the 2x squared so what are the possibilities for that well since 2 is a prime number really you can only do 2 times 1 and x times x is x squared so 2x times x is basically what you could do now you could consider negative 2x times negative x

but everything is positive here so you really don't need to think about that you don't need to get that involved you could say or you know negative 2x times Negative x but we really don't need that right all the signs are positive so if i just write out 2x times x like that i know that 2x times x would give me 2x squared so that's done now the next thing i would look at i would skip the outer and the inner and i would look at the last the last term here and the last term

here will produce 12 when they're multiplied together So i've got to go through all the factors of 12 and see what combination would give me a product of 12 and a sum of 11x so that's where we talk about trial and error for your factors of 12 you have 1 times 12 you have 2 times 6 and you're going to have what 3 times 4. now you have the negative versions of these negative 1 times negative 12 negative 2 Times negative 6 negative 3 times negative 4 but again all the signs are positive so we

don't really need to think about those at this point so what i would do is i would go through again trial and error you'd have let me put this down here 1 times 12 2 times 6 and 3 times 4. so let me make all the scenarios up so you have 2x and x and then you have 2x And x so first i'm going to vary 1 and 12. now i'm going to put plus 1 and plus 12. so what i need to check here is the outer and the inner let me kind of move

this down a little bit so i can show you that so in other words the outside here 2x times 12 is 24x the inside here 1 times x is 1x if i combine like terms with that would i get 11x the answer to that is no so that's not going to be My solution so i would start with something else then now i can reverse the order here and try 12 and then try one but there's one little tip that i want to give you if your original trinomial did not have a common factor then

you're not going to have a common factor in any of the binomials so in other words this has a common factor of 2. i could pull a 2 out and i'd have x plus six So this isn't a possibility so we can eliminate that and once we've eliminated that we can cross this one off the list that's not going to work so now let's try two and six now if i put two here and 6 here again there's a common factor of 2. so that's not going to work so if i switch that up and

put 6 here and 2 here common factor of 2 that's not going to work so that's eliminated so now i'm just down to 3 and 4. so i can put 3 here and 4 here and Try that so the outer 2x times 4 would be 8x the inner 3 times x would be 3x if i add those two together 8x plus 3x is 11x so here's my factored form i have 2x plus 3 times x plus 4 those two quantities and we know that because we did this problem earlier right we saw it from kind

of this form 2x plus 3 2x plus 3 x plus 4 and x plus 4. so you can see that's a lot more tedious a lot more involved and i gave you an Easy scenario right the coefficient on the x squared term is a prime number and everything is positive so we have a lot less work with this versus another scenario which would be more complex okay let's take a look at 5x squared plus 26x plus 24. so again i'm going to set this up as the product of two binomials there and for 5 that's

a prime number so i'm thinking about what 5x times x again everything is positive so I really don't need to think about negative 5x times negative x i don't need to involve that so i'm just going to start out with 5x times x that's for the first part of foil right f so that's what that comes from now i'm going to do the l so for 24 something times something it's got to give me 24. so what are the factors of 24 well i've got what i've got 1 times 24 i've got 2 times 12

i've got 3 times 8 i've got 4 times 6 and then that's going to be it so if i start with 1 times 24 let me kind of look at these scenarios here so 5x and x let me make out a few different scenarios So i've got positive 1 positive 1 positive 24 and positive 24 and then i'll check 2 and 12 also so plus 2 plus 12 plus 12 and plus 2. so you kind of set it up like this and check it very quickly so all you're looking at is the outer so forget

about the x per second 5 times 24 is 120 the inner would be 1 so 120 plus 1 is 121 that's not going to Get you to 26. the outer here 5 times 1 is 5 the inner 24 times 1 is 24 5 plus 24 is 29 so that's not going to work so this one is eliminated then for this one 5 times 12 is 60. the inside 2 times 1 is 2 so that's 62 that's not going to work 5 times 2 is 10 12 times 1 is 12 10 plus 12 is 22 so

that's not gonna work so you can get rid of all of those scenarios None of those work and now we can go on to three and eight and let's try 4 and 6 if we need to okay so 5 times 8 is 40 3 times 1 is 3 43 is not going to work 5 times 3 is 15 and then 8 times 1 is 8 15 plus 8 is 23 that's not going to work so this is eliminated so now i'm down to the last possibility i have 5 times 6 that's 30 And then 4

times 1 is 4 that'd be 34. and then in this possibility 5 times 4 is 20 and then 6 times 1 is 6 that is 26 so it's the last one so it's plus 6 and plus 4. let's erase all this and again that's that's the problem with trial and error sometimes you get it right away sometimes you have to go through all the possibilities you know you get it in the end so 5x times x would be 5x squared 5x times 4 would be 20x 6 times x would be 6x 20x plus 6x is

26x and then 6 times 4 is 24. so what if we had 12x squared plus 9x minus 30. the first thing you notice is that this has a common factor of 3. so i could pull that 3 out and i would have 4x squared plus 3x minus 10. so if i try to factor this as it is without pulling this out that trick where i talked about you can't have a common factor in the binomial is not going to work so that's Why i do this to start make my life a lot easier so again

if i set this up three times and have this scenario so what would go here and here for 4x squared i could do a few different things let's start out by just looking at 2x and 2x right 2x times 2x would give me 4x squared then the possibilities for negative 10 i've got 1 times negative 10 i've got negative 1 times positive 10. I've got negative 2 times 5 and i've got positive 2 times negative 5. so those are my possibilities so let's run through these really quickly and a lot of these we're going to

be able to skip over and i'll show you why okay so if i want to use 1 and negative 10 any way i do it i'm going to have a common factor of 2 so i can eliminate That again same thing here if i want to use negative 2 and 5 i'm going to have a common factor of 2. so that's eliminated and that's eliminated either negative 2 or positive 2. because i've got to put it on one of the sides and i'd have a common factor so this is eliminated right away so i've got

to go to a different one i've got to go to 4x times x let's try that So 4x 4x 4x 4x and then my x is over here so then again the possibilities for negative 10 i've got negative 1 times 10 i've got positive 1 times negative 10. i've got 2 times negative 5 and i've got negative 2 times positive 5. but again because of the common factor thing there's only so many arrangements i can use so i can't do a 10 right here there'll be a common factor of 2 but i Can do a

negative 1 and i can do a positive 10. so 4x times 10 would be 40x negative 1 times x would be negative x 40x minus x would be 39x that's not going to get me 3x i can get rid of that the next scenario we could do plus 1 and negative 10. again i can't put a negative 10 over here because it would have a common factor of 2 with that 4. so 4x times negative 10 Is negative 40x 1 times x is x that would be negative 39x that's eliminated so then the next one

we have two and negative five so it's got to be minus five and plus two again i can't put the two over here because it would have a common factor with the four and this right here doesn't have a common factor in it it does in the original one but we already pulled it out so this should not have any common factor when We consider what's inside the parentheses here and that's what we're factoring okay so again 4x times 2 would be 8x negative 5 times x would be negative 5x so 8x minus 5x would

be 3x so this is what we're looking for right here so let's erase everything and put 4x minus 5 times x plus 2. so again i know that a lot of you will use this method and You'll like it and that's fine again i dislike it i think it's a little too tedious i know there's tips and tricks to speed it up even more but you know i just don't care for it so the end result here is 3 times the quantity 4x minus 5 and then times x plus 2. and again if you want

to check it 4x times x is 4x squared the outer 4x times 2 is plus 8x the inner negative 5 times x is minus 5x so this would give me plus 3x And then negative 5 times 2 is negative 10. so if we took 4x squared plus 3x minus 10 multiply it by 3 you get 12x squared plus 9x minus 30 which is what you started with hello and welcome to algebra 1 lesson 41 in this video we're going to learn about special factoring rules so back when we learned how to multiply polynomials we did

a lesson on special products and that lesson was meant to Greatly increase your speed when you had to find certain polynomial products that occurred very often so for example one of the ones we looked at we looked at x plus y that quantity squared and you know it's the first term squared plus two times the first term times the second term so 2xy and then plus the final term squared so Y squared so without using foil i'm able to multiply that instantly and you can look at a example that's not generic you could do something

like let's say x plus 5 that quantity squared so that would be x squared plus 2 times x times 5 so that's 10x and then plus that last term which is 5 squared that's 25. so i'm able to use that formula to very Quickly get the answer now we saw this on a few different things we also looked at kind of this with a minus sign so x minus y that quantity squared and the only thing that changed here was the sign so it was x squared minus 2xy plus y squared so as another example

with just numbers let's say we made this a little bit more complex and we said this is 4x minus 5 that quantity squared Well the first term here the 4 and the x would be squared so 4 and the x that's all squared then minus the first term times the second term times two so two times four is eight eight times five is forty then times x so minus forty and then we have our final term which is 5 that would be squared so plus 25. so to simplify this 4 squared is 16 and X

squared is just x squared so this would be 16x squared minus 40x plus 25. so a much quicker way to do that versus going okay well this is 4x minus 5 times 4x minus 5. you know first times first is 16x squared and outer is minus 20x and inner is minus 20x and last is plus 25 you know combine like terms here and get this over here so let's kind of reverse this process we Already know how to do the multiplication part we've already gone through that lesson so now what we're going to do is

learn how to take something in this format and put it in this format we're going to factor it so we're going to start out by factoring the difference of two squares so remember when we went to our lesson on special products we had the quantity x plus y times the quantity x minus y So it's the same thing here and here and the same thing here and here so in other words the first position of each is x the second position of each is y what is different is going to be your signs right the

plus and the minus so this turns out to be the x what's in the first position of each squared minus the y what's in the second Position of each squared remember the middle two terms are going to drop out right because the difference of the signs makes them cancel so you get x squared minus y squared so in this lesson we're going to go from this format to this format so the first question you would ask is how can we identify a binomial as the difference of two squares well the first thing is that both

terms Both terms must be squares and what do i mean when i say both terms must be squares well when we square something we multiply it by itself so in other words if i have 2 squared i know that's two times two or four we know that at this point but if i looked at the number four i've got to be able to identify that that was created or can be created from Two times two and i know there's other ways to get four it could be one times four it could be you know negative

two times negative two all these different variations but you're looking for numbers or variables that can be created from some number or some variable multiplied by itself so for example if i have the number nine that's three times three and again it could be other things it could be nine times one it could be negative three times negative three but You know let's keep things positive and keep everything very simple if i had a variable like x squared well this is x times x or i could make this much more complicated if i have a

variable like let's say x to the 10th power well this could be x to the fifth power that's squared right because if we keep the base the same and we multiply exponents this would be x to the power Of 5 times 2 which is 10. so x to the 10th power is really x to the fifth power squared or if i wanted to do something like a number and a variable let's say i did something like 25 y to the 8th power well 25 i know is 5 times 5 and y to the 8th power

i could write as y to the fourth power so if i square this whole thing here this would be 5 times 5 or 25 and y to The fourth power squared remember keep y the same multiply the exponents 4 times 2 is 8. so 5 y to the fourth power squared would give me 25y to the eighth power so this is what you're looking for when you're looking for the difference of two squares so both terms must be squares now the other thing is that the second term is going to be negative so your second

term Is negative now i say that and let me put a little notation next to it this can be changed around because remember if we have something with addition and subtraction we can flip things around into a different order so your teacher might try to trip you up let's say you saw something like x squared minus y squared you know this is the difference of two squares we already seen that this can Factor into x plus y times x minus y so another way you might see this presented your teacher might try to trip you

up and she might say okay well you have negative y squared plus x squared well it's not that the second term is negative now but remember i could reorder these i could change this into this and it would still factor the same way so just pay close attention to that you want one of the terms to be positive And one of the terms to be negative so those are the two conditions all right so let's start out with an easy one so we have a squared minus b squared so we have a squared term here

and a square term here and then we have a positive and we have a negative so that's exactly what we're looking for now we just follow the format we take remember when we had x squared minus y squared this was x plus y times x minus y so you can kind of think About it as okay what's being squared is going to go in the first position of each so what's being squared here is the a so that's going to go on the first position of each and then notice your signs you have different signs

plus and minus so you'll have plus and minus and then look at what's squared here you have y so that's going to go in your second position of each same thing here this b here squared so That's going to go here and here so a squared minus b squared is going to factor into a plus b that quantity times a minus b and so we've already seen this so many times with our special products formulas we already know that this goes back to this right so we don't even need to check it so let's take

a look at one that's a little bit harder so we have n squared minus 16. so we're Looking for two squares here so we know this is a square and then what about 16 well yeah this is 4 times 4. so i'm going to rewrite this for my convenience as n squared minus 4 squared and that makes it easier for me to follow that formula again i'm going to write this a few times x squared minus y squared is equal to x plus y times x minus y So then this is just equal to what

it's equal to n plus 4 times n minus 4 right following that format whatever squared here in the first position goes in the first position here and here then whatever squared here in the second position goes in the second position here and here and just remember to alternate your signs it's just that easy and again we know from our special products formula that n Plus 4 that quantity times n minus 4 that quantity would give us n squared minus 16. all right what about x squared minus 25 we know this is a square and then

25 we know that's 5 times 5. so again you can rewrite this x squared minus 5 squared and again just follow your formula if you want i'll just rewrite it every time so x squared minus y squared is equal to what x plus y times x minus y So then this is equal to what it's x plus five times x minus five right whatever squared in the first position goes here and here in the first positions of each whatever squared in the second position goes here and here in the second position of each and then

again you know how to check this you have your special products formulas we know that the quantity x plus 5 times the quantity x minus 5 Would give us x squared minus 25. all right let's take it up a notch we'll look at one that's a little bit more complicated so we have 9b squared minus 16. so then this one right here we know 9 is 3 times 3. obviously b squared is a square and then 16 we know that's 4 times 4. so again to follow that kind of formula i can write this as

3 b that quantity squared minus 4 Squared and then i can write this remember whatever is in the first position that's squared is going to go on the first position of each and then whatever's in the second position that's squared that's going to go in the second position of each so this becomes the quantity 3b plus 4 times the quantity 3b minus 4. and again if you want to check this you know your formula already it'd just be first times First 3b times 3b is 9b squared and then you know that the middle terms would

drop out so the outer and the inner would go away you just need to think about the last you have plus 4 times negative 4 that's going to be negative 16 so you would get 9b squared minus 16 back all right let's take a look at another so we have 125 x minus 80 x cubed now a lot of you will see this cubed here and say well this isn't the Difference of two squares i want you to remember something whenever you factor you want to think about what's common to everything before you even start

so i would think about okay both terms here have an x right at least one this one's obviously got three x times x times x but this one has one and this one has at least one then both of these guys are divisible by five This is 16 times five and this is 25 times five now when you see that you can see that 25 is 5 times 5 and 16 is 4 times 4. so this looks like the perfect kind of trap question you get on a test what they want you to do is

they want you to factor out the gcf first so the gcf is 5x so if i pulled that out what i'd have left 125x divided by 5x is going to be 25 And then minus then 80x cubed divided by 5x would give me 16x squared now this isn't in the format that we're used to we're used to a variable here or something like that but we still have something that's squared minus something else that's squared right so that's exactly what we're looking for let me erase all this and again we're putting our 5x out in

front and then we'll set up two parentheses Here and so 25 is what it's five times five so we would have five here and five here and then we have a plus and a minus 16x squared we can think of as 4x times 4x so i'll put 4x here and 4x here and if you go through and you do foil 5 times 5 is 25 the outer which is negative 20x and the inner positive 20x would cancel and then the final 4x times negative 4x Is negative 16x squared so that checks out and if we

multiply this by 5x we would get back to 125x minus 80x cubed so now let's take a look at something else that's very common we're going to look at some perfect square trinomials so perfect square trinomials so in other words a trinomial again is a polynomial with three terms and it factors into a binomial that's squared so in other Words this guy right here we already know the formula for this this is x squared plus 2xy plus y squared so if i factored it into this really i could write this as the quantity x plus

y squared right these two would be the same so if we see something like this in this format and we notice that we have the first guy squared and the last guy Squared and then the middle is 2 times both of the terms in the squared binomial so 2 times x times y well we have a perfect square trinomial and we can factor it using this formula so another occurrence of this we have x minus y times x minus y and so again this is x minus y that quantity squared we already know this turns

into x squared we have a minus sign here then 2xy then We have a plus and then y squared so same thing just when you have a negative there that sign right there is going to be negative if i have the x plus y that quantity squared all the signs are positive but everything else all the terms are the same all right so let's take a look at the first example so we have x squared plus 6x plus 9. so again the first thing i'm going to look at is to see if this is a

square So yeah this is x squared and then is this a square well this is 3 squared now the next thing i'm going to look at is the middle term if i multiplied 3 which is what's squared here times x which is what's squared here so 3 times x is 3x times 2 do i get the middle product i do 2 times 3 is 6 6 times x is 6x So then i can factor this guy as this what's squared which is x plus this which squared which is 3 that whole quantity squared right this

turns into x plus 3 that quantity times x plus 3. let's take a look at 25p squared minus 30p plus 9. now pay attention here because we have that minus sign so we know that if this ends up being factorable there's going to be a minus Sign in there you'll have something minus something else and then it'll be squared so let's check to see if we have squares here so we have 25 p squared so that's 5p that amount squared and then we have 9. that's 3 squared and then what about our middle product is

it 2 times 5p times 3 2 times 3 is 6 6 times 5 is 30 30 times p is 30 p so yes that all works out so we would write This we take what's being squared in the first term so that's 5p that goes in the first position and then minus we take what's squared in the last position that 3 and we write that so you get the quantity 5p minus 3 that's squared let's take a look at another one so we have 49x squared minus 84x plus 36. so this guy right here you

think about 49 that's 7 times 7 and then x squared is x squared so we'll Put 7x that guy squared and then for 36 we think about that as 6 squared right 6 times 6. so is the middle term 2 times the product of 6 times 7x 2 times 6 is 12. 12 times 7 is in fact 84 and the x would come along for the ride so that would give you 84x now there's a minus sign here so when You set this up it's going to be this guy here 7x there's a minus and

then this guy right here 6. so this amount this quantity squared all right let's take a look at another one so we have 72x cubed plus 120x squared plus 50x so the first thing i notice right away is that everything has an x right that's common to everything now everything is also divisible by 2. so i can pull out a 2x before i even Begin and what that's going to give me is 36x squared plus 60x plus 25 and i notice right away that 36 is what it's 6 times 6. so this would be 6x

that amount squared 25 i know is 5 squared and then for your middle term 2 times 5 is 10 10 times 6 is 60. x comes along for the ride so i would get 60x now All i need to do again make sure you have that 2x out in front i would put 6x in the first position i have two plus signs here so this is going to be plus and then i'll put 5 in the second position and then this is squared so this is going to factor into 2x times the quantity 6x plus

5 squared all right so lastly we're going to look at factoring the difference Or also the sum of cubes now this is one that's going to be confusing for you at first because when we looked at our special products lesson we saw something that was similar but it's not the same and i'm going to show you that in a minute let me go through and just kind of use the distributive property on this and show you what you get x times x squared is x cubed x times x y is plus x squared y And

then x times y squared is plus x y squared negative y times x squared is minus x squared y negative y times x y is negative x y squared and then lastly negative y times y squared is minus y cubed now what's going to happen is x squared y minus x squared y those are going to cancel plus xy squared minus xy squared those are going to cancel and what i'm left With is x cubed minus y cubed so this is one that's going to dramatically speed up your time to do homework or take a

test so you have to memorize it it's it's something you just have to practice write down a flash card just kind of go through now here's what i'm going to tell you this is not the same as when we looked at the quantity x minus y Cubed remember i can't just distribute my exponents and say this is equal to x cubed minus y cubed no wrong remember that okay these are not the same i taught you this formula already and it's really easy to remember if i have a minus here i have x cubed i

have a minus here and then remember i Have a space a plus a space a minus and then a y cubed so what goes in these two blanks here this is a three and this is a three remember that from the exponent and then each term goes in there each one is squared only once so in other words i'll do x squared y here and i'll do x y squared here so this is what this is equal to and This is what this is equal to they are not the same so please understand that because

a lot of students will see something like this and they'll try to factor it into this it's not the same okay so please understand that this formula was something we talked about when we looked at special products so let me erase this real quick and i want to show you The sum of cubes so this is the difference of cubes this again is the difference of cubes and we're going to look at this one is the sum of cubes so x times x squared is x cubed x times minus x y is minus x squared

y x times y squared is plus x y squared y times x squared is plus x squared y Y times negative x y is minus x y squared y times y squared is plus y cubed so everything's going to cancel except for the x cubed and the y cubed so let's go ahead and do that so negative x squared y positive x squared y those are going to cancel x y squared minus x y squared those are going to cancel so what i'm left with is x cubed plus y cubed so this one Because there's

an addition sign in it is called the sum of cubes all right when i raise them to the third power it's called a cube right i cube it so that's why it's called the sum of cubes you have two items there or two terms that are both cubed all right so the main difference here is going to be the sign this is just something you have to memorize When i have a minus i have a minus in the first guy and then i have all pluses in the second guy okay so you kind of have

to remember that the first one is the same the second one is going to be different and then the last one is always going to be positive no matter what so same different that's how i remember it if you look at this one same different last one's always positive Okay same different that's how i remember again put that on a flash card if you want to or maybe there's some other trick you got out there to remember it it doesn't matter at the end of the day if it's math you know whatever you can use

to memorize it as long as it works it works all right so let's take a look at m cubed minus one so m cubed is obviously m times m times m what about one remember one times Anything is just itself so if i have one times one that's one if i have one times one times one that's still one so this is one cubed so this follows the format of something cubed minus something cubed and so we can write it using that formula that we just learned and let me rewrite that one so if i

have something that's x cubed minus y cubed This will factor into this is how we're going to memorize it we have x and we have y what's the first sign remember first sign is the same then we have x squared and we have y squared in the end this is always plus what's the sign here remember same different so this is minus this will be plus and then you have x y so following this formula here all i need to do is think about okay This is what's cubed and this is what's cube so in

the first thing that's cubed it's going to go where it's got a position here it's got a position here that's squared and then it's got a position here and then that's it we're going to subtract this guy then we're going to add 1 times this guy which is just itself and then we're going to add That guy squared which is just one so we get the quantity m minus one times the quantity m squared plus m plus one and i know these are a lot more challenging especially when you see them at first it's like

hey how am i going to remember that if you do enough practice with this i suggest doing a lot a lot of practice because by the time you get to algebra 2 you're going to need this formula to get through a lot of the tests that you're Going to take especially like the sat and the act all these standardized tests that you're going to take where time matters right so if you know the formulas you can go through and crank these things out very very quickly and be assured that you have the right answer all

right so let's look at x cubed plus 125. so what's our formula again x cubed plus y cubed so the first sign is the same so we have a plus there so there's going to be x Plus y very easy to remember then the next one you have x squared the sign that comes after that is different this is plus so this is going to be minus then it's x y and then it's plus this one's always plus and then it's y squared okay so very very easy to remember so kind of following this format

if i look at this i have x cubed And i have 125 so 125 is 5 cubed so all i need to do is put what it's going to be x plus 5. x is what's cubed here 5 is what's cubed here so we have x plus y here x is what's cubed y is what's cubed so we have x plus y here we have x plus 5. then following this format we have x squared so we have x squared minus x y in this case we have x times 5 So 5x and then plus

we have y squared here so whatever is cubed that's now going to be squared so 5 is what's cubed so that's going to be squared 5 squared is 25. so we get the quantity x plus 5 times the quantity x squared minus 5x plus 25. all right let's take a look at one final problem so what is x cubed again minus y cubed so again the first sign is the same so x minus y Same sign then the next one that we're going to think about after the x squared is different so this is a

minus so now it's going to be a plus and then we would have x y and then this sign right here is always a plus so we have plus and then we have y squared so following this format we have a cubed A is what's cubed and we have 64. 64 is 4 cubes so 4 is what's being cubed so follow the format so whatever's being cubed in the first position in this case that's a minus whatever's being cubed in the second position in this case that's 4. then times whatever's being cubed in the first

position is now squared so that's a Squared then plus we have whatever's cubed in the first position times whatever's cubed in the second position so a times 4 or 4a then plus whatever's cubed in the last position in this case that's a 4 we're going to square it just like it is here so 4 squared is 16 close the parentheses and we have the quantity a minus 4 times the quantity a squared plus 4a plus 16. So just something to keep in mind these formulas are going to greatly speed up your homework and your tests

so something you definitely want to memorize what i always suggest for my tutoring students is write out the formula that you know and then just make the comparisons like we're doing here okay if i know x cubed minus y cubed is the quantity x minus y times the quantity x squared plus xy plus y squared i can just represent Everything and then go through and match up my formula and do it like that for a while and you know 20 or 25 problems later you won't even need to write the formula down you'll have this

stuff memorized hello and welcome to algebra 1 lesson 42. in this video we're going to learn about solving quadratic equations by factoring so a quadratic equation contains a squared variable so i have here such as X squared or it could be y squared or z squared or whatever variable you've chosen to work with in that equation squared and then i have here and no other term with a higher degree so in other words the largest exponent that you're going to have on your variable is going to be a 2 right you can't have x cubed

or x to the 6th power that's not going to be involved in a quadratic equation so generally we think About quadratic equations in standard form so we have ax squared plus bx plus c and then because it's an equation we have the equal sign and we set it equal to zero okay so it's set equal to zero now remember that a is the coefficient for the x squared term so this is a coefficient b is the coefficient for the x to the first power term and then c is a constant so if we look Here

we have one restriction so a cannot be equal to zero and why do you think that is well again this is the coefficient for the x squared term and for us to have a quadratic equation we've got to have a variable that is squared in that equation right that's part of the definition so we have ax squared plus bx plus c equals where a is not equal to zero and then a b and c are real numbers and i know i Haven't given an official definition for that yet but it's just any kind of number

that you can think of at this point so here are some examples we have x squared plus seven x minus three and this equals zero we have three x squared minus two x plus one equals zero we have nine x squared plus eight equals zero now you might see these two up here and say well those those kind of fit the format of you know ax squared Plus bx plus c right where a in this case is five b here is seven and c is negative three and then in this case a is three b

is negative two and c is one but what about this one where we have nine x squared plus eight equals zero it looks like i'm missing this b x term right i'm missing that but remember we defined b as being able to be any real number so that means b could be zero and remember if you multiply something by zero it Becomes zero so it would just disappear so nine x squared plus we could put zero x plus eight equals 0 0 times x is 0. so if i put plus 0 i can just get

rid of it and simplify this and put plus 8 equals 0. now they might throw something at you on a test and say is something like i don't know let's say six x cubed plus five x squared plus seven x minus two equals zero is that a quadratic equation The answer is no remember the highest exponent on your variable is going to be a 2 with a quadratic equation this one has a 3 here so that's not going to be a quadratic equation so this is not a quadratic output equation and again that's specifically because

you have an x cubed there so pretty easy overall when we think about the definition of a quadratic Equation and just think about when we say we're writing it in standard form think back to the standard form of a polynomial right if i have you know ax squared plus bx plus c when we think about this trinomial here it's written in standard form because the exponents go in descending order i start with the highest which is a 2 then i go to a 1 and then i go kind of to where i don't have a

variable where i Could say this is x to the power of 0 if i wanted to so that's how standard form is for a quadratic equation as well just remember that it'll make it easy for you to write it out each time now let's kind of move into the main part of the lesson we're going to start solving these quadratic equations using factoring so i want you to first recall that a solution for an equation is any value or could be values that makes the equation true when it replaces the variable or it Could be

variables if you have more than one so for example if we think back to when we're working with linear equations we have 3x minus 5 equals 10. i have here that this has a solution of x equals 5. why because if i plug a 5 in for x there i get a true statement if i take 3 and i multiply it by five minus five equals ten three times five is fifteen i get fifteen minus five equals ten And then i get ten equals ten so the left and the right side are the same value

and so this equation is now true right but it's only true if x is equal to 5. it's not true for any other value of x and then let's look at one other one we have that 2x minus 1 equals 11 has a solution of x equals 6. again if i put a 6 in for x i would have 2 times 6 minus 1 equals 11. we can see that would work out 2 times 6 is 12 minus 1 equals 11. we get 11 equals 11. again same Value on the left and the right side

so we know that x equals 6 is the correct solution all right so let's get our feet wet with a typical quadratic equation so you'll notice that it's set up for us in what we call again standard form so we have x squared plus 4x plus 3. so it's an exponent of 2 it's an exponent of 1 and you kind of think about your constant term there so it goes in descending order of powers and then again notice how we've set it equal to 0. so if something's equal to zero you're gonna see that we

can use a special property for this so the first thing i want you to do is i want you to factor this guy so we know how to factor a trinomial and that's what the left side of this is left side of the trinomial right side is zero so i'm going to set this up into two binomials and so i'm going to have x here and x here and these are Both positive so plus something and plus something else and just give me two integers whose sum is what is four right the coefficient for the

x to the first power we generally refer to this as b right remember we say ax squared plus bx plus c the coefficient for the x to the first power we generally refer to that as b so the sum is b And the product the product is c right that constant term which in this case is three so b here is four c here is three so give me two numbers whose sum is four and whose product is three well one and three right if you think about the factors of three three is a prime

number so without going into any negatives you just have three times 1. so 3 and 1 and 3 plus 1 does give you 4 So that would work out and you can check it through foil if you want x times x is x squared the outer would be x the inner would be 3x x plus 3x is 4x that would give you the middle and then 3 times 1 is 3. that would give you your last term there so we factored this correctly now once we've factored this correctly what do we do we have a

binomial times another binomial equal to zero so in order to solve this type of problem i'm going to introduce something known as The zero product property so the zero product property and this tells us that if a times b is equal to zero and when i say a times b i mean any real number a times any real number b so two numbers being multiplied together if that result is zero then What could be the possible answers for a and for b well a could be equal to zero or b could be equal to zero

or both a and b are equal to zero those three scenarios because when i multiply a number times zero i get 0 as a result So a could be any number that's not 0 and b could be 0. when i multiply them together i get 0. b could be any number that's not 0 a could be 0 multiply them together you get 0. a and b could each be 0. if i multiply 0 times 0 i still get 0. so that's where these three scenarios kind of come from so we can use this property to

get a solution for this equation here if i First start by realizing that x plus 3 that quantity is being multiplied by the quantity x plus 1 well if i plugged in something for x there some value and it made this factor here this binomial x plus 3 equal to 0. then when 0 is multiplied by this guy right here it's going to give me 0. so i've found a solution for the equation i've replaced the variable with a value That makes the equation true so the way we utilize this is we just set each

of these equal to zero separately we'll get two solutions and then we will check each solution in the original equation remember a could be zero b could be zero or they both could be 0. so let's start this out by just saying okay x plus 3 x plus 3 is equal to 0. Very very easy to solve that i just subtract 3 away from each side of the equation and i would get that x is equal to negative three so this is one proposed solution and then the other one the other factor here is x

plus one so x plus one we set that equal to zero we'd subtract one away from each side and we get that x is equal to negative one so we put x equals negative three or x equals negative one could be any of Those at this point so let's erase this real quick and i can show you why this works so let me let me erase all this so without going back into the very original x squared plus 4x plus 3 equals 0 which we'll do in a minute i want to show you in this

format here how this plays out i think it'll give you a better understanding so we have The quantity x plus 3 that is being multiplied by the quantity x plus 1. this is supposed to be equal to 0. so let's start out by just plugging in a negative 1 for x here so if i plugged in a negative 1 there and a negative 1 there what would happen well negative 1 plus 1 is 0. we found that out by setting x plus 1 equal to 0 and solving for x so this would be 0 and

then if i plugged in a negative 1 Here negative 1 plus 3 is 2. well there you go 0 times 2 is 0. so that's how that works right anything times zero gives you zero so all i need is one of these factors to be zero and i'll get a true statement so now let me show you what happens with x equals negative three so again let me just write x plus three that quantity here and this multiplied by again let me write x plus 1 that quantity there and i'm going to plug in a

negative 3 Now so i'm going to plug in a negative 3 here and then also here and you're going to see the same thing happen negative 3 plus 3 is 0. so i have one factor that's zero and then negative three plus one is negative two so this is negative two and zero times anything is zero so zero times negative two is of course zero so again where we get a true statement so let's erase everything here and i'm Going to show you that these two solutions work in the original equation that's not factored so

if i plugged in a negative 3 here and here we're starting with this one so we would have negative 3 that quantity you're plugging in for x and x is squared so you've got to use parentheses there so that quantity is squared plus 4 times negative 3 plus 3 and this should be equal to 0. Negative three squared is nine four times negative three is negative twelve so minus twelve plus three equals zero nine plus three is twelve twelve minus twelve is zero so this is zero equals zero so that checks out as a solution

and then x equals negative one let's do the same thing with negative one erase this so we'd have negative 1 squared that's 1 and then 4 times negative 1 is negative 4 so minus 4 and then plus 3 and this equals 0. 1 minus 4 is negative 3 negative 3 plus 3 is 0. so you get 0 equals 0. so that one checks out as well and so my solution is x equals negative three or it could also be that x equals negative one right each of those would work and typically in an algebra one

or later on in algebra 2 You're going to write this using something known as a solution set so it's set braces which look like this and inside of the set braces you would list your solutions for that equation so in this case you'd have negative 3 comma negative 1. but in algebra 1 this is still pretty much acceptable so either way you want to write your solution either with the solution set Notation which will cover greatly in algebra 2 or with this notation here that we've been using forever as long as you're displaying to your

teacher what the solution is x could be negative 3 or x can be negative 1 and those solutions are correct you should get full credit all right so let's take a look at another one and i want to just cover the full procedure with you on this one and the Very first thing you always want to do if you're going to solve a quadratic equation by factoring you want to start out by putting the left side of the equation as ax squared plus bx plus c the right side of the equation should be 0. so

in other words with this x squared minus 6x equals negative 5 i would add 5 to both sides of the equation and what's that going to give me it's going to give me x squared Minus 6x plus 5 and this is equal to if i have negative 5 plus 5 that gives me 0. so i've got to write 0 where students make a mistake is they forget they have this equal sign here and they just don't write anything after they do that so they're stuck and they're like well what do i do now you have

equals 0 there so once we've done that we're going to factor the left side completely In most cases that's going to be as the product of two binomials now sometimes it'll be slightly different if you have more factoring involved but usually this is going to turn out to be two binomials being multiplied together and it's set equal to zero so in this case i know it would be x here and x here we just have to figure out these two missing parts here and here now once you're done with that you're going to set each

factor equal to Zero just like we did in the last example and then you're going to check each solution in the original equation so all i need to do here is find two integers whose sum is negative six and whose product is five so for the factors of 5 i know that it's a prime number so it's 1 times 5 and that's not going to work because of the sign so i got to think about Negatives i could do negative 1 times negative 5 and that would sum to negative 6. so this would be minus

1 and minus 5. so once i've figured out what the factoring is i just set each factor remember x minus 1. that whole thing let me kind of do this in different parentheses that whole thing is a factor and then this guy over here let me do that in Different parentheses that x minus five that quantity is also a factor so each one of those is going to be set equal to zero so x minus one is equal to 0 or x minus 5 is equal to 0. so just solve each one for x and

you can do them mentally right this one's going to be 1 this one's going to be 5. but to kind of do this for the sake of completeness i'd add 1 to each side here this would be x equals one i'd add five to each side Here this would be x equals five and so we get x is equal to one or x is equal to five so let me erase this real quick and at this point you can just erase everything and just check in the original equation so if i plug a 1 in

for x i'd have 1 squared minus 6 times 1 equals negative 5. so 1 squared is 1 and then minus 6 times 1 is 6. so 1 minus 6 is negative 5. so negative 5 Equals negative 5. so yeah this one works out then for the next one i'm plugging in a 5 for x so let's erase all this so i'm plugging in a 5 here and also here 5 squared is 25 minus 6 times 5 is 30 this equals negative 5. 25 minus 30 is negative 5 so negative 5 equals negative 5 so that

works out as well so we can say here That x equals 1 or x equals 5. and again if you wanted to use the solution set notation you could put 1 comma 5 and show your solution that way all right so what if we had something like 5x squared minus 45x equals negative 90. well the first thing i want to do again is write this left side in standard form so ax squared plus bx plus c so my constant term is on the right side i want to move this over here so just add 90

to both sides of the equation so I'm going to add 90 here and here so that's going to give me 5x squared minus 45x plus 90 and remember it's equal to zero negative 90 plus 90 is zero so then i notice that everything here is divisible by 5. so i want to pull that out before i even start so if i pull that out i'd have x squared minus 9x Plus 18. okay so this equals zero and i'm just going to factor what's inside right this trinomial in here so i'll put the five times i'll

have two sets of parentheses and this equals zero so no this is x and this is x so give me two integers whose sum is negative nine and whose product is 18. so if i think about the factors of 18 you think about negative 6 and negative 3 right negative 6 plus negative 3 is negative 9. negative 6 times negative 3 is 18. so minus 6 and then minus 3. so once we have that we can solve this take each factor and set it equal to 0. now 5 is a factor here but there's no

variable involved 5 doesn't equal 0. so it wouldn't make any sense for me to say okay this factor of 5 equals 0. so you can kind of modify that and say that you want to set each factor with a variable equal to zero so then i would take x Minus six and set that equal to zero and then or we'd have x minus three equals zero let me scroll down a little bit and obviously we can eyeball that and say x can be 6 or x can be 3 right but again for the sake of

completeness add 6 to both sides of the equation that gives you x equals 6. add 3 to both sides of the equation that gives you x equals 3. so we get x equals 6 or x equals 3. so then again x equals 6 or x equals 3. all right so let's check this in the original equation so i'd have 5 times plug in a 6 for x that's squared minus 45 times plug in a 6 for x this equals negative 90. so 6 squared is 36 so i'd have 5 times 36 which is 180 minus

45 times 6 which is 270 This should equal negative 90 and it does right you get negative 90 equals negative 90. so this one checks out and then for x equals three let's erase all this put a three here and here three squared is nine nine times five is 45 and then minus 45 times three which is 135 and this should equal negative 90 and again it does you get negative 90 equals negative 90. so this works as a solution as well so x equals 6 or x equals 3. again you could Write this like

this as 3 comma six in your solution set notation all right let's take a look at another let's say you had seven x squared plus forty nine x equals zero so it looks like i'm missing a constant over here i could put plus zero there but really that's just the same thing right so is there a way to factor this well the answer is yes each term here is divisible by 7x so i can pull that out so i could say i have 7x times Inside of parentheses if i divided this by 7x i just

have x plus if i divided this by 7x i just have 7. this equals 0. well i have 2 factors here that are equal to 0. remember my original is a b equals zero if that occurs then a equals zero b equals zero or both a and b equal zero so i can set each factor here because they each have a variable involved equal To zero and i can solve so i could say 7x equals 0 or x plus 7 equals 0. now 7x equals 0 is very straightforward you divide both by 7 and you

get x equals 0. so x can be zero or if we have x plus seven equals zero subtract seven away from each side of the equation you get x equals negative Seven so or x equals negative seven so let's check both of these obviously zero works if i plug a zero in here i get zero times seven that's zero plug a zero and here i get zero zero plus zero is 0. that works out for negative 7 you'd have 7 times the quantity negative 7 squared remember if i'm plugging in something for x and it's

being squared i want what's being plugged in squared so negative 7 is being plugged in so i've Got to use parentheses to say i want both the negative and the 7 to be squared so then plus 49 times negative 7. and then this equals 0. and so what are we going to get here well negative 7 squared is 49. so you'd have 7 times 49 plus 49 times negative 7 equals 0. you don't even need to do the multiplication because these two numbers would be opposites i have positive times positive and i have negative times

positive so This would be negative this would be positive it would be the same number it would just be opposites this would be positive 343 and this would be negative 343. so you'd have 343 minus 343 equals zero so zero equals zero so yes x equals negative seven is a solution so you get x equals zero or x equals negative seven again you could write that using solution set notation as negative seven comma zero inside the set braces so let's look at a harder one We have seven x squared minus three x plus seven this

is equal to negative 33 plus 58 x so we're going to add 33 to both sides of the equation and then we're going to subtract 58x from both sides of the equation and so let's cross this out and this out and put a zero and then over on the left hand side we'll have 7x squared negative 3x minus 58x is going to be Negative 61x and then 33 plus 7 is 40. so plus 40 and this equals 0. so let's go ahead and factor the left side now we have one where the a the coefficient

of the x squared term is not 1. so this requires a little bit more work to factor but that's okay we already learned how to do that in a previous lesson remember what i'm going to do is i'm going to take two integers whose product is this Times this so 7 times 40 which would be 280. so product of 280 and whose sum is negative 61 right the b the coefficient for the x to the first power term so sum of negative 61. so let's think about some factors of 280 we have 1 times 280

that's not going to work we have 2 times 140 that's obviously not going to work not divisible by 3 you Could do 4 times 70. that's not going to really work and then you could do 5 times 56 now we could play with the signs and make that work the product is positive but the sum is negative so all i need to do is have two negative signs negative 5 plus negative 56 is negative 61. negative 5 times negative 56 is 280. so that will work 5 and 56. all i'm going to do is use

that to rewrite that Middle term so i'm going to have 7x squared i'm going to put minus 56x then i'm going to put minus 5x plus 40 equals 0. and now i'm going to factor using grouping so let's scroll down a little bit so out of this first group the 7x squared minus 56x i can pull out a 7x so that would leave me with x minus 8. out of the second group i can pull out a Negative 5 and be left with an x minus 8 this equals zero obviously the common binomial factor is

x minus eight so i'll have seven x minus five that quantity times x minus eight that quantity and this is going to be equal to zero so now all i need to do is set each of these factors equal to 0 and i'll have my answer so 7x minus 5 equals 0 or X minus 8 equals 0. so for this one i just need to add 5 to both sides of the equation i'll have 7x is equal to 5 divide both sides of the equation by 7 x equals 5 7. for this one i'll add

8 to both sides of the equation and i'll have x is equal to 8. so x equals 5 7 or x equals 8. all right so let's check these two proposed solutions so the first one is a little bit more tedious so let's go ahead and get that Out of the way so we have 7 times 5 7 squared minus 3 times 5 7 plus 7 equals negative 33 plus 58 times 5 7. so let's work through this guy so 5 7 squared would be 25 49. so you'd have 7 times 25 over 49. this

would cancel with this and give me a 7 down here so 25 7 is what that would be and then minus you have 3 times 5 7. nothing i can really cancel there so i'm Going to write 15 7 and then plus 7. before i go on over here let me just get a common denominator going so i could write this as 49 over 7. and then i could just perform the operation with the numerators 25 minus 15 is 10 10 plus 49 is 59 so this is 59 over 7 on the left side so

59 7 and this equals over here if i do the multiplication of 58 times 5 7 well 58 times 5 is 290 so i'd have negative 33 plus 290 over 7. now to get a common denominator here i'd multiply this by 7 over 7 and i'd get negative 231 so this would be negative 231 over 7. if i do negative 231 plus 290 i get 59. so you'd end up with 59 sevenths is equal to 59 sevenths so this one checks out and then the second one is that x equals eight that one's a lot

Easier to do so let's knock that out real quick so seven times you'll have eight squared minus three times eight plus seven this is equal to negative 33 plus 58 times eight so eight squared is what that's 64. so 7 times minus 3 times 8 is 24 plus 7. so 7 times 64 is 448 so 448 minus 24 plus 7. over here let's just do some of this 58 Times 8 is 464. so you'd have negative 33 and then plus 464. all right so let's put equals negative 33 plus 464 is 431. then over here

448 minus 24 is 424 then plus 7 is 431 so you get 431 equals 431. so that works out so we could say that x equals 8 is also a solution here and again you can write this in solution set notation where you'd put 5 7 comma 8 right your two solutions inside of the set braces but again in algebra 1 this is a perfectly acceptable answer Hello and welcome to algebra 1 lesson 43. in this video we're going to have an introduction to rational expressions all right so back in pre-algebra we learned the definition

for a whole number and also for an integer we never really talked about an official definition for a rational number so the quotient of two integers where the denominator is not zero because remember we can never divide by Zero is known as a rational number okay so this is a rational number so some examples you have one half so this is a rational number we have an integer of one over an integer of two or something like negative two-fifths you have an integer of negative two over an integer of five or as a last example

something like three-eighths an integer of three over an integer of eight and Just so you know every integer and every whole number can be written as a rational number by just simply writing the number over one right if i had the number let's say three i could just write three over one and now it's a rational number or if i had something like negative seven i could write negative 7 over 1 and now it's a rational number where you would run into a problem and You're going to see this later on if you have something

like let's say the square root of 5 so the square root of 5. i cannot write this number as the quotient of two integers no matter how hard i try i can't just say okay square root of five over one that's not going to work because this part right here the numerator is not an integer so this is known as an irrational number and we're going to study those a little bit later On in algebra 1. all right so kind of moving forward here i want to just say that similarly the quotient of two polynomials

okay the quotient of two polynomials with again a non-zero denominator because you can't divide by zero is called a rational expression this is what we're going to learn about today a rational expression so here are some examples You have 3x minus 4 over x squared minus 1. so you have a polynomial in your numerator a polynomial in your denominator so then this one right here you have 5x cubed over 2x minus 4. so you have a polynomial in the numerator over a polynomial in the denominator and then here you have 8x to the fourth power

minus 5x minus 3 over 2x cubed minus 7. i get a polynomial in the numerator over a polynomial in the Denominator so our first topic for rational expressions is simply to learn where they are undefined and this has to do with division by zero so since we can never divide by zero we must exclude any value for any variable that would make the denominator 0. very very important so how can we determine where the rational expression is defined well it's actually kind of easy we already have all the skills to do it we just want

to set the denominator equal to 0 We want to solve the resulting equation and then the solutions are the values that will make the rational expression undefined because if we set it equal to 0 and we solve for those values those values would make the denominator 0 and that's not allowed so here's an example we have 4x minus 2 over x squared minus 16. so we know that this cannot be zero cannot Be zero so how do we figure out what values would make it zero we just set it equal to zero so we say

x squared minus sixteen is equal to zero in the last lesson we learned how to solve a quadratic equation by factoring now if you get something in your denominator that you can't factor if you get a quadratic equation you can't factor you're not going to be able to deal with that until we get to Completing the square and solving quadratic equations using the quadratic formula and that's all the way at the end of algebra one so right now i'm just going to give you stuff you can factor so you just get used to this so

we have x squared minus 16 equals zero we know this is the difference of two squares so i can factor this as x plus four times x minus four equals zero and we Know from our zero product property that x plus four is set equal to 0 and then we put or x minus 4 set equal to 0. if we solve these two equations subtract 4 away from here and here i get x equals negative 4 or add 4 to both sides of the equation i get x is equal to 4. and if you check

these negative 4 squared is 16 16 minus 16 is 0 and then 4 squared is 16 16 minus 16 is 0. so they both work out so what we would say is that replacing the variable x here with a 4 or a negative 4 would result in a value of 0 for the denominator and that's not allowed so your restricted values here for x are 4 and negative 4 right x cannot be equal to 4 or negative 4. So before we kind of move on i just want to show you what this looks like graphically

if you were to graph 4x minus 2 over x squared minus 16 what you would get is a graph that looks like this and depending on what you set your maximum range to be it might be a little bit different some computers draw it a little differently but the main thing to notice here is that look at when x is 4 And when x is negative 4. remember we said x couldn't be 4 or negative 4. look at the fact that on your y-axis the graph is never going to actually touch that point i'm trying

to make this as straight as i can it gets very very very close to that but it never touches it and the same thing going down here it gets very very very close to that but it's never going to touch it and then same thing over here gets very Very close to it but it's never going to touch it very very close never going to touch it these things are called asymptotes and we're going to talk about these a lot in algebra 2 when we start graphing really heavily but for right now i just want

to give you a little insight into what this looks like so when x is 4 there's no value for y that's corresponding to that and then also when x is negative 4 there's no Value for y that's corresponding to that on that graph right again that's because x is not allowed to be 4 or negative 4. okay let's take a look at another one so we have 3x squared minus 5 over x squared minus 5x plus 6. so again the denominator cannot be 0. the only way to make it 0 is to plug in values

for x that result in this becoming 0. so what i would do is i would say okay x squared minus 5x plus 6 Set it equal to 0 find out what those values are and then those are going to be the values that are restricted so this is a factorable trinomial so i can factor this into the product of two binomials and this is pretty easy give me two integers whose sum is negative five whose product is six so it would be negative two and negative 3. negative 2 times negative 3 is 6 negative 2

plus negative 3 is negative 5. So then i would set each of these equal to 0 and i would solve individually so if i add 2 to both sides over here if i add 3 to both sides over here i get x equals 2 or x equals 3. so i put over here x equals 2 or x equals 3. you can check those real quick plug in a 2 here and here you get 2 squared that's 4 and then so this is 4 and then minus 5 times 2 is 10 plus 6 4 plus 6

is 10 10 minus 10 is 0. so that One checks out and then i can replace these put 3 squared here that's 9 negative 5 times 3 that's negative 15. so 9 plus 6 is 15 15 minus 15 is 0. that checks out as well so our restricted values here are 2 and 3. so x cannot be 2 or 3. when we get to algebra 2 i'll give you kind of a fancier way to write that I just want you to have an understanding of sets before we kind of jump into that okay let's take

a look at one more of these it's pretty simple overall so we have 4n squared minus n plus 5 over 3n cubed minus 7n squared minus 10n so again all i want to do is take my denominator whatever it is and set it equal to 0 and find out what the restricted values are so 3n cubed minus 7n squared minus 10n Again set this equal to zero and i'm going to find out what values of n are going to make this denominator zero and that's again what we can't have those are the restricted values so

to factor this i notice that i have an n that's common everything so let's pull that out to start so we'd have 3n squared minus 7n minus 10 and this equals zero and then to factor this i'm going to use the grouping method And i want you to give me two integers whose product is negative 30 and whose sum is negative seven well that's pretty easy if you think about 10 and 3 10 times 3 is 30 and i can manipulate the signs to get to negative 7. so if i think about negative 10 and

positive 3 that gives me exactly what i want so 3n squared then i'm going to put plus 3n And then minus 10n and then minus 10 this equals zero again i'm going to be factoring by grouping so out of this first group i can pull out a three n so i'm gonna have n times pull out a three n you'll have n plus one and then out of the second group i'm gonna put out a negative 10 i'll have n plus one this equals zero Factor out the common binomial factor of n plus one and

you'd have three n minus ten times this n plus 1 is equal zero and in this case i have three factors that need to be set equal to zero the first one is just n n equals zero if n is zero then the denominator is zero and you can go up and verify that if i plugged in a zero here here and Here i'd have zero minus zero minus zero that would be zero so for your restricted values you'd have 0. so then the other one is 3n minus 10 equals 0. set that equal to

0. we add 10 to both sides of the equation and we get 3 n is equal to 10 divide both sides by 3 and we're going to get that n is equal to 10 thirds so another restricted value would be 10 thirds Provided that it checks out and it is a solution to this equation 3n cubed minus 7n squared minus 10n equals 0. we'll check that in a minute i already know that it works so we're going to just list it there and then the last one we're going to set n plus 1 this factor

here equal to 0. n plus 1 equals 0. of course we subtract 1 from each side we get n equals negative 1 so we're going to list that as well So let's check those two real quick and i'll show you that they do in fact work for 10 thirds it's kind of tedious but 3 times 10 thirds cubed would be 1000 over twenty seven then this would cancel with this and give me a nine here so basically i have one thousand over nine one thousand over nine and then we have minus seven times ten thirds

squared 10 squared is 100 so this would be 100 and then 3 squared is 9. so then if i multiply that by 7 i would get 700 over 9. so 700 over 9 and then minus we have 10 times 10 thirds so this would be a hundred thirds let's go ahead and write that and this should be equal to zero now if i do 1000 minus 700 that's 300 so i'd have 300 ninths 300 ninths Which i could simplify because each is divisible by three and write it as a hundred thirds so 100 thirds minus

100 thirds is 0. so that checks out all right so the next one is negative 1. so if i plug that in here and here and also here what am i going to get negative 1 cubed is negative 1. so you'd have negative 3 minus Negative 1 squared is 1 so 1 times 7 is just 7. then minus 10 times negative 1 is negative 10 you'd have minus a negative 10 so plus 10. negative 3 minus 7 is negative 10 negative 10 plus 10 is equal to zero so this one checks out as well and

so we can say that our restricted values here are zero ten thirds and negative one right if n equals any of those values you're going to get a denominator of zero and again that's not Allowed all right so now let's talk about simplifying rational expressions so i want you to recall how we go about simplifying a fraction so if i have something like 6 over 8 what i'm looking to do is cancel common factors so six i could write as two times three eight i could write as two times four but really if i wanted

to really break it down i could put two times two times two it's really only Necessary to write two times four because the greatest common factor between these two is just a two so i cancel that greatest common factor between the two and i end up with three over two times two or end up with three fourths so now this fraction is considered simplified because the greatest common factor is now a one right there's nothing else i can remove something like five tenths right five is a prime number And ten i can write as five

times two and so i can cancel this 5 with this 5 and i'll end up with 1 over 2 or one-half so with a rational expression we're going to factor the numerator and denominator completely just like we did with the fraction and we're going to cancel any common factors all right so let's take a look at the first problem here and we're going to start out with r minus 3 over 2r squared minus 6r So again the first thing you want to do just like when you worked with fractions and you were told to simplify

them you want to factor your numerator denominator completely so in the numerator r minus 3 is not going to factor so you're just going to leave that as it is in the denominator the 2r squared minus 6r i can pull a 2r out and set it out in front of some Parentheses because that's common to each and then inside the parentheses i would have an r minus a 3. so now what do i have that's common i have r minus 3 here and r minus 3 here so it's common to both numerator and denominator and

when that happens you can cancel it now when you first start doing this you might be like well how do i go about doing that Remember these are factors so this is multiplication here this is a factor and this is a factor you might say well that's not a factor up here well yeah i could write it as times 1. then again i could do the same thing i could say this is a factor and this is a factor so now it's completely clear that i can cancel this factor of r minus 3 with this

factor of r minus 3 and the Simplified version is just going to be a 1 in the numerator that's what's left over a 2r in the denominator that's what's left there so 1 over 2r if we want to just quickly refer back to an example we did with fractions we looked at five over ten five dozen factors a prime number so we could write five times one if we wanted to over ten we write as five times two so what am i doing i'm canceling a common Factor of five and i'm left with one over

two same thing i did here i cancelled a common factor of r minus 3. so that's all we're doing here it's not more difficult just need to realize one key thing and i'm going to explain that to you right now this is what trips up a lot of students if i have something like let's say 2r plus 1 over 2 can i cancel this 2 with this 2 No i cannot this is the most common mistake i see when we first start working with rational expressions for some reason students start thinking that they could just

cancel or they see something like okay i have 5 r squared plus 3 over 15. well they'll go through and say okay well some of them will cancel this 5 with the 15 and put a 3 and then they'll say okay i can also cancel this 3 with This 3. no you cannot do that these are both wrong these are wrong why are they wrong well because i'm canceling common factors this is just addition here this is addition it's not multiplication this is addition it's not multiplication i can't cancel anything where i can cancel if

i saw something like let's say 2r plus 2 over 2 and i factored my numerator i notice that there's a 2 that's common each term There so i write this as 2 times the quantity r plus 1 over 2. now i have multiplication 2 is a factor and the quantity r plus 1 is a factor so i can cancel this factor of 2 with this factor of 2. again i could write this 2 as 2 times 1. so now this simplifies to what it simplifies to r plus one over one or really just r plus

one so we're looking to cancel common factors okay please don't make this Mistake the way you wanna remember it if you want a generic example it's in a lot of textbooks you might see something like you know a plus b over a is not equal to b because a lot of students will see that and say okay well there's an a here and an a here so i can cancel this with this and i'm just left with b that's wrong again if i had something like a plus a b Over a well then i could

factor the numerator as a times the quantity 1 plus b you see what i did there i had an a here and an a here and i pulled it out and put it outside of the parentheses and so a times 1 would be a that would give me that back a times b would be a b so all i did was factor that a out and put this over a and now in this particular case again these are factors now a is a Factor the quantity 1 plus b is a factor and so i can

cancel common factors so very very important that you understand that all right now one other thing that i want to address before we kind of move on in a lot of cases you're going to get a joint question so you might be asked to state the restricted values and to simplify so let's say you go through and let me kind of erase all This kind of work that we did let's say you go through and you identify that this simplifies to this you say okay i'm ready to state my restricted values now so i look

at the denominator it's 2r so i say okay 2r equals 0. solve that for r divide both sides by 2 i get r equals 0 and i say okay r cannot be 0 because that makes the denominator 0. well yeah that's right but you're only partially correct The original denominator is what you have to look at so although the simplified version would be defined for r equals three the original is not so although these two are equal in the traditional sense where we say okay we can simplify this and they're equal you kind of have

to put a side note and say it's still not defined for r equals three because in this original one if i look at this factored denominator which Is this right here i have this factor r minus 3 equals 0. if i add 3 to both sides r equals 3 there so if i plug a 3 into the original denominator i would get a value of 0. since we started out with that rational expression we really can't say that these two are equal because this one is not defined when r is 3. this one is right

if we just look at one over two r I can plug in a three right so two times three is six i'd have one over six so that's where students get confused so you almost need like a side note there to kind of explain that and say well they are equal but not if r is equal to three and the way you say that is you look at your original rational expression and you look at that denominator to give you your proper restricted values okay So if you stick with the original denominator you're always going

to be good to go you don't want to look at the simplified one because you might not list all of them all right let's take a look at another so we have 15b over 9b squared plus 15. so what can i do here again i want to factor the numerator denominator and for the numerator i know i could break 15 into 5 times 3. so let's go ahead and just do that 5 times 3 and Then i have that b so times b in the denominator i have 9b squared plus 15. so what's common each

term there would be a 3. right so if i pulled a 3 out i would have what i'd have 3b squared plus 5. now i'm going to cancel a common factor of 3 only y this is multiplication so i have a factor let me just write factors here and point to all of them and then this is a factor And this is a factor so what can i cancel i can cancel common factors so i can cancel this three with this three okay so once i've done that the simplified numerator is 5b and the simplified

denominator is 3b squared plus 5. that's all i can do don't make the mistake again like most students do they go okay i can cancel this b with this and that's a 1 and i can cancel you know oh look there's a 5 Here i can cancel this with this you can't you're out of common factors to cancel again it's got to be a factor a common factor between the numerator and denominator this 3 is a factor this 3 is a factor it's multiplying this 3b squared plus 5. again very common mistake and i want

you just to understand that moving forward so your simplified answer here is 5b over 3b squared plus 5. all right let's Take a look at another one we have n squared minus 81 over n squared plus 3n minus 54. so if i factor the numerator i notice that is the difference of two squares this is n plus nine times n minus nine these are being multiplied together so each is a factor this is over n squared plus three n minus 54. so let's go ahead and set up our parentheses here put an n here and

an End here and a typical kind of trick to speed yourself up if you get stuff like this on a test you know they're going to give you stuff that cancels 99 of the time so it's probably going to be that one of these is going to be a 9. it could be plus 9 or minus 9 they make it to where one of them is going to cancel so if i look at this real quick i see that i need two integers that sum to 3 and Multiply to negative 54. so if i think

about that i can do positive 9 so n plus 9 and then negative 6. the sum is going to be positive 3. the product is going to be negative 54. and then i see that this would cancel with this right again these are factors those factors are canceling and so what i'm left with is n minus 9 over n minus 6. let's take a look at another one so we Have x squared minus 11x plus 18 over x squared minus four so we notice right away the bottom is what the difference of two squares this

is x plus two times x minus two on the top go ahead and set this up so again kind of a trick is to realize that it's going to probably be a plus 2 or a minus 2 in one of those scenarios because they make these things To where something's going to cancel 99 of the time you get some problems where it doesn't but most of the time it does so i need two integers whose sum is negative 11 and whose product is 18. so i know that's going to be negative 2 negative 2 and

negative 9. negative 9 plus negative 2 is negative 11 negative 9 times negative 2 is positive 18. so i can cancel common factors here cancel this factor of x minus 2 with this factor of x minus 2 and what i'm Left with is x minus 9 over x plus 2. now let me just show you why it's useful to simplify these things let's say i said that would i rather evaluate when it looks like this or when it looks like this let's say i said x is equal to 5 give me the value for the

rational expression well here i can say 5 minus 9 is negative 4 over 5 plus 2 that's 7. so i get negative 4 7. Nice and easy very quick if i did it up here i would have 5 squared that's 25 minus 11 times 5 that's 55 plus 18. you think about how much more work this is and then over 5 squared is 25 25 minus 4 is 21. so let's do this 25 minus 55 is negative 30. negative 30 plus 18 is negative 12. so this would be negative 12 Over 21. then i've got

to simplify this both are divisible by 3 so this would be negative 4 and this would be 7. so you get the same answer either way but look how much longer it took so that's one reason why we go through the process to simplify these things it makes it easier if you need to plug in a value for your variable and see what that rational expression is equal to all right let's take a look at one final problem So we have 2a cubed plus 16a squared over 21a cubed plus 141a squared minus 216a so in

the numerator i can factor out a 2a squared and that's going to give me an a plus 8. in the denominator i can see right away that i can pull out an a but everything is also divisible by 3. so i can pull out a 3a so i'd have 3a Times you'd have 7a squared plus 47a then minus 72. so for the denominator let's factor this down here real quick i'm going to use the grouping method give me two integers whose product is 7 times negative 72 which would be negative 504 and whose sum is

47. So for this one because i don't know the factors of 504 for the top of my head i'd have to make a factor tree so 504 i know i could start out with divisibility by 4 so i could do 4 times 126 4 is 2 times 2. for 126 i could do 2 times 63 so 2 is prime 63 is 9 times 7 7 is prime and 9 is 3 times 3 and those are prime so i'm looking for a sum of 47 So let me think about this i can do 2 times 2

times 2 which is 8 times 7 which is 56 times 9. so 56 times 9 would give me 504 and then if i did positive 56 and negative 9 that would give me 47 as a sum so let's go ahead and do that let's erase all this and so i'll put this is equal to we'll have 2 a squared and times a plus 8 Over we'll have 3a times let me kind of factor this down here so 7 a squared plus i'm going to put positive 56a and then negative 9a and minus 72 if i

factored this guy i'd pull out a 7a from the first group and i'd have a plus 8. from the second guy pull out a negative nine and i'd have a plus eight So this is going to be seven a minus nine times a plus eight so then i can easily see what i can cancel here cancel this with this right common factor of a plus 8 and i can also cancel this a with one factor of a up here so what i'm going to be left with is a 2a in my numerator over three times

We have seven a minus nine hello and welcome to algebra one lesson forty four in this video we're going to learn about multiplying and dividing rational expressions so in our last lesson we started talking about rational expressions we learned how to simplify rational expressions and we learned how to find the restricted values for a rational expression Now we're going to move on and start talking about some operations with rational expressions and kind of the easiest ones to start with would be multiplication and division now if you know how to multiply and divide fractions the process

is very very similar so let's start out by looking at 1 3 times 2 7. so what we want to do here is we want to multiply the numerator times the numerator so 1 times 2 is 2 We want to place this over the denominator times the denominator so 3 times 7 is 21 so you get 2 over 21 there now remember when you're working with fractions you want the fraction to be simplified or as they say reduced to its lowest terms now you can look at 2 and you can look at 21 and you

can tell that there's nothing that's going to be common other than 1. and you can kind of start from over here remember we're allowed to cross cancel If i look at 1 and i look at 3 there's nothing in common there other than 1. if i look at 2 and i look at 7 nothing in common other than 1. then i can look crossways 1 and 7 2 and 3. nothing in common there other than 1. so we would say that 2 over 21 is simplified or again reduced to its lowest terms now let's take

a look at one where you can do some simplification we have 5 8 times 4 15. so what i can do is i can multiply Everything together first 5 times 4 is 20 over 8 times 15 which is 120 but then i have to go through and simplify so that doesn't make much sense so for most people what they'll do is they will cross cancel or cancel before they even start the multiplication process so you look between numerator and denominator of each fraction first look between 5 and 8 there's nothing you can Cancel 4 and

15 nothing to cancel then i would look crossways so i would look between 5 and 15. there's a common factor of 5. so 5 divided by 5 is 1 15 divided by 5 is 3 and i look cross ways again between 8 and 4 there's a common factor of 4. so divide this by 4 i get 1 divide this by 4 i get 2. so now if i multiply 1 times 1 is 1 over 2 times 3 that's 6 so i get 1 6 as the simplified form or again reduced to Its lowest terms and

again if you wanted to do 5 times 4 and get you know 20 over 8 times 15 which is 120 so it's the same value it just looks different this is just unsimplified 20 divided by 20 would give me one 120 divided by 20 would give me 6. so at the end of the day you get 1 6 again as the simplified form or as they say reduced to its lowest terms all right so what about Division so recall that to divide one fraction by another you take the first fraction with a fraction all the

way on the left and you leave that guy unchanged then you change your division to multiplication and you're multiplying by the reciprocal of the second fraction so if i take the reciprocal of two-fifths remember i take the denominator and i bring it and put it Into the numerator and i take the numerator and i put it down into the denominator so i just interchange or flip the positions denominator becomes numerator numerator becomes denominator so that's how you take the reciprocal now once i've done that all i do is i just multiply so i can see

that i can cancel a common factor of two between here and here so put this as a one and this is a one and now i just multiply one times five That's five over three times one that's 3 so this is going to be 5 3. all right so what about 7 9 divided by 14 15. so i take 7 9 that's my fraction all the way to the left i leave that unchanged and i multiply this guy by the reciprocal of this 15 goes into the numerator 14 comes down into the denominator so that's

how you take the reciprocal again this goes up here this goes down there you interchange So now what can i cancel between 7 and 14 i have a common factor of 7. so this will be 2 and this will be 1. between 9 and 15 have a common factor of 3. so this would be 3 and this would be 5. so now if i multiply here 1 times 5 is 5 over 3 times 2 that's 6 so we get 5 6 as my answer all right so hopefully you're familiar with multiplying and dividing fractions if

you're not at this point you might want to go back into a Pre-algebra lesson on multiplying and dividing fractions get up to speed on that and then come back to this point in the video so now let's start looking at some rational expressions if i had something like 10 9 times 6 over 8x so the only difference here is i have that one variable which is x nothing else is really going to change here as far as the process i'm still Going to look to cross cancel before i start my multiplication so what are the

common factors between 10 and 9 can't really cancel anything there between 6 and 8 i can cancel a common factor of 2. so i can make this a 4x and i can make this a 3. now i can do more cancelling because i can cross cancel between this 9 and this 3. so if i divide this by 3 i get 3. if i divide this by 3 i get 1. then i can do some more cross cancelling because 10 is Divisible by 2 and so is 4. 10 divided by 2 is 5. 4 divided by

2 is 2. so now we just multiply 5 times 1 is 5 over 3 times 2x 3 times 2 is 6 and then x comes along for the right so we get 5 over 6x all right so that was pretty easy let's look at one that's a little bit more complicated so we have 6r cubed minus 12r squared this is over 7 and then we're multiplying by 7 over 4r squared minus 8r so the trick to this is the first thing you want to do is you want to factor out the greatest common factor from

every numerator and every denominator now before i even do that i notice that i can cancel this 7 with this 7 right i can cross cancel so this is a 1 and this is a 1. now if i kind of factor out the gcf here in the numerator over here it would be a 6r Squared so i can write 6 r squared times the quantity i would just have an r minus a two now this is over one so i can write it over one or i can leave it as it is it doesn't really

matter because i have a numerator and denominator here let me just write it over one kind of to make things consistent so times here i have a 1 over the gcf here is 4r So i'm going to write this as r minus 2 as well now what else can i cancel and there's two things i can cancel the first thing is that i have a common factor of r minus two remember these two here are factors they are being multiplied together to give me this six r squared times the quantity r minus 2. so they

are Factors okay remember that over here i have factors 4r is being multiplied by r minus 2. these are factors and that would combine to give me 4r squared minus 8r so i can cancel common factors so i can cancel this with this and that will give me a 1 and a 1. now in addition to that i can do some more Cancelling what i can do six and four each divisible by two if i divide this by two i get a three if i divide this by two i get a two then i can

also cancel one common factor of r so this would be r to the first power and this would just be a one so what i'd end up with let me put equals here i would have three r everything else is a one it's times one And it's over one so then times over here i'd have 1 over i'd have 2 everything else is a 1. so essentially when i write my answer here this is going to be equal to 3r over 2. so just 3r over 2. all right let's take a look at another one

so we have 1 over m minus 2 and this is divided by m minus 5 over m squared minus 10 m plus 25. okay so the first thing i want to do is set this up remember if i'm dividing just Like if i'm dividing with fractions the first guy over here is going to stay the same so 1 over m minus 2 then i'm going to multiply by the reciprocal of this so this is going to go in the numerator m squared minus 10 m plus 25 and then this is going to go on the

denominator m minus 5. now let's take a look at factoring the top So we need two integers whose sum is negative 10 and whose product is 25 well negative 5 and negative 5 would be those two integers so let's write equals 1 over m minus 2 times i'm going to factor this into m minus 5 times m minus 5 and obviously i can cancel that with m minus 5 down here and to make this obvious when you first start doing you can just write this as times 1 Right anything times 1. so you can say

this is a factor and so is this so it's 1 times that all right so we cancel common factors cancel this with this and what i'm left with is 1 times m minus 5 in the numerator let me scroll down here that would be m minus 5. and this is over m minus 2 times 1 which is just m minus 2. so my simplified answer here is m minus 5 over m minus 2. okay let's take a look at another one that's a little bit more tedious so we have 4n over 2n squared minus 4n

minus 6 times 10n squared plus 2n minus 8 and this is over 25n minus 20. so all i'm going to do is just factor everything completely or as much as i need to to figure out what's going to cancel so for example in this numerator here 4n i'm not going to write 2 times 2 times n I'm just going to write 4n you can do that if you want but i think it's a bit unnecessary so then over here you have 2n squared minus 4n minus 6. so i'm going to start by just pulling a

2 out of there and that would give me n squared minus 2n minus 3. now i can further factor this so this would be n and then n so give me two integers whose sum is Negative two and whose product is negative three so that would be negative three and positive one so let me erase this part and i'll drag this up here okay so then this is times up here i have 10 n squared plus 2n minus 8. so let me factor that at the bottom because i made a little room for that i'm

just going to start by just Putting okay 2 times if i pull a 2 out of there because it's common everything 5 n squared plus n minus 4 then over here 25 n minus 20 i can pull a 5 out and that's going to give me 5 n minus 4. now let me just factor this guy down here so i have a little room 5n squared plus n minus 4. so 5n squared Plus n minus 4. so i need two integers whose product is negative 20 for the product and then i need a sum of

1. so for 20 forget about the negative for a minute i know 5 times 4 so i can play with that and get a sum of 1 if i had positive 5 and negative four that would work out so let's use those so we would have this is equal to five n squared plus five n Minus four n minus four if we factor this by grouping i could pull a 5n out of the first group that would leave me with an n plus 1 here out of the second group i can pull out a negative

4 that would leave me with n plus 1 here so the common binomial factor is n plus one so i would have five n minus four that quantity times n plus one that quantity so that's your Factored form all right so let's erase this real quick we'll put that factored form in here nice and convenient okay so now what we want to do is we want to look to see what we can cancel we have everything that's factored as much as we need to and so i see that i have a common factor of 5n

minus 4. again this is multiplication here everything is being multiplied this is multiplication so These are all factors so this factor of 5n minus 4 will cancel with this factor of 5n minus 4. this factor of n plus 1 will cancel with this factor of n plus 1. we can cancel this 2 with this 2. and is there anything left that i can cancel there's nothing up here other than a one remember when your numerator or your denominator cancels completely you write a one Down here i just have a five so i need to be

able to cancel a five with something over here i can't so that's simplified over here i have a 4n and i have an n minus 3. nothing else i can do so when we multiply 4n times 1 is 4n this is over we have the quantity n minus 3 times 5 so we just write that as 5 times the quantity n minus 3. or you can leave it in this form or you can write 4n over Use your distributive property 5n minus 5 times 3 is 15. okay let's take a look at a null so

we have 2x squared minus 7x plus 5 over 2x minus 5. this is divided by 2x squared plus 11x plus 12 over 2x squared plus 13x plus 15. so for the first one in the numerator 2x squared 2x squared minus 7x plus 5. let's see if we can factor that So i would need two integers whose product is 10 and whose sum is negative 7. so let me start by just writing 2x squared so if i think about the factors of 10 it's really 2 and 5. if i did negative 5 and negative 2 that

would work so i would do minus 2x then minus 5x then plus 5. if i want to factor this by grouping i'd pull out a 2x from the first group And i would have x minus 1. from the second group i would pull out a negative five so minus five and that would give me an x minus one and so if i pulled out the common binomial factor of x minus one i would have two x minus five times x minus 1. all right so let's erase all this and we'll kind of drag this up

to the top So put equals and we'll drag this up okay so this is over 2x minus 5. so we can cancel that right away if we want or we can just kind of wait till we're done and we cancel everything at once so that's what i'm going to do now we're dividing by this guy right here so let's just start by finding the reciprocal we're going to put multiplied by i'm going to take the denominator here And put in the numerator 2x squared plus 13x plus 15 and then over the numerator will go into

the denominator 2x squared plus 11x plus 12. okay so let's factor each one separately here so if i had 2x squared plus 13x plus 15. so two integers whose product is 30 and whose sum is 13. so for 30 i've got 1 times 30 i've got 2 times 15. can i play with the signs there no Because everything is positive right so i can't really use that i've got 3 times 10 and that would work if i did positive 3 and positive 10. so let's go with that so then this equals 2x squared plus 10x

plus 3x plus 15. and again we're going to factor by grouping so from the first group i'm going to pull out a 2x that would leave me with an x plus 5. From the second group i'll pull out a 3 that'll leave me with x plus 5. and so pull out the common binomial factor of x plus 5 and you would get 2x plus 3 that quantity times x plus 5 that quantity so let's copy this all right let's also look at 2x squared plus 11x plus 12. so 2x squared plus 11x plus 12. So

for this right here let's think about two integers whose product is 24 and whose sum is 11. so for 24 you do 1 times 24 i can do 2 times 12 i can do 3 times 8 so 3 and 8 would work so let's break this up we'll say this equals 2x squared plus 8x plus 3x plus 12. so in the first group i can pull out a 2x so let's pull out 2x And inside i would have what i would have x plus 4. in the second group i can pull out a 3. so

what i have left is again x plus 4. so the common binomial factor here is x plus 4. so let's go ahead and pull that guy out and i have 2x plus 3 times x plus 4. so this is 2x plus 3 times x plus 4. so now that we've factored everything i Know that's the kind of tedious boring part but once you've gotten through that the rest of it's pretty simple i'm just going to look for common factors and cancel remember everything is being multiplied so these are common factors so i can put this

as times 1 here and i can cancel 2x minus 5 that factor with 2x minus 5 up here and then i don't have any x minus 1s anywhere Else i can't do anything with that but i can cancel this factor of 2x plus 3 with this factor of 2x 3. and what i'm left with is x minus 1 so x minus 1 then times x plus 5 over 1 times x plus 4 so basically x plus 4. now you can leave it in this format if you want or your teacher might want you To use

foil and present this as a trinomial in the top that's fine either way it's the same thing so x times x is x squared the outer would be 5x the inner would be minus x so this is plus 4x and then the final or the last would be negative 1 times 5 which is minus 5 and again this is over x plus 4. so you get x squared plus 4x minus 5 over x plus 4. all right let's take a look at one final problem so i Have 2n squared minus 4n minus 6. this is

over 35n squared plus 8n minus 16. this is divided by 2n squared plus 18n plus 16 over 5n plus 4. so i want to just factor everything here and see what's going to cancel so if i start out in the numerator i have 2n squared minus 4n minus 6. and in this particular case we've already factored this in a previous problem so i'd pull a two out to start And what i'd have is i'd have n squared minus two n minus three so give me two integers whose product is negative three and whose sum is

negative two well that would be negative three and positive one so i would have n minus three n minus three and then n plus one and a lot of you will remember that again from our previous problem So this is going to be over i need to factor that down here because i'm going to need a little room to do that so let's let's kind of scroll down a little bit so 35 n squared plus 8n minus 16. so if i multiply 35 times negative 16 i would get negative 560. and that's what i'm looking

for as far as the product For my sum i want 8. now most of us do not work with 560 a lot so this is something to where you want to make a factor tree and kind of think about what the prime factorization is so i would just start with something like you know 10 times 56 okay nice and simple so that's obviously not close for me so then 10 is 5 times 2. 56 if i did 28 times 2. i want you to notice something here if i have 28 times 20 28 and 20

i can make that work so i could stop here for the sake of the factor tree let's just continue this is 4 times 7 and then 4 is 2 times 2. so let me kind of erase this now we have our two integers so we know it's 20 and 28 what do the signs need to be Well since the sum is positive i want the larger absolute value to be positive so i want to do positive 28 and negative 20. so let's erase this real quick let's say okay we have 35 n squared plus 28

n minus 20 n and then minus 16. okay so in the first group i could pull out a 7n so this would be 7n And what would be left is 5n plus a 4. in the second group i can pull out a negative 4 and what would be left would be 5n plus 4. so if i factor out the common binomial factor of 5n plus 4 i would have 7n minus 4 times 5n plus 4. so let's go ahead and copy this Okay we're going to paste this right here okay now the rest of this

is pretty simple now we're dividing by so i'm going to put a multiplication sign here and then i'm going to take the reciprocal of this so the 5n plus 4 is going to go up top in the denominator here i can factor out a 2 before i start so it's going to make things simpler and i'd have n squared plus 9 n plus 8. now give me two integers whose sum is 9 And whose product is eight well that would just be eight and one so we'll do n plus eight and then n plus one

now i've got this as factored as i can get it all i want to do now is look to see what i can cancel so i can cancel a common factor of n plus one i can cancel a common factor of five n plus four remember i could write this as times one so this would cancel with this And then anything else i can cancel i can cancel this 2 with this 2 and what i'm looking at in my numerator i'm going to have an n minus 3. in my denominator i'll have this quantity seven

n minus four times this quantity n plus eight now again in a lot of cases you can just write this and stop your teacher might not want you to you know kind of come down here in the denominator and do foil But she might want you to do it it doesn't matter it's the same thing either way so if she asks for it do it if not you can just stop so this is n minus 3 over 7n times n is 7n squared the outer would be positive 56n the inner would be negative 4n so

this would be plus 52n and then the last negative 4 times 8 is negative 32. so you get n minus 3 over 7n squared plus 52n minus 32. hello and Welcome to algebra 1 lesson 45. in this video we're going to learn about finding the lcd that's the least common denominator for a group of rational expressions so now that we've mastered how to multiply and divide rational expressions kind of the next thing we're going to learn is how to add and subtract rational expressions now before we can do that we need to learn how to

find the least common denominator for a group of rational Expressions so i want you to think back to when we worked with fractions in pre-algebra and we should all remember that adding or subtracting fractions requires a common denominator and you want to find the least common denominator because that's going to save you time and effort with regard to your simplification in the end so the lcd or least common denominator is the smallest number that is divisible by all denominators The lcd is the lcm of all denominators now let me kind of explain this a lot

of students get confused when you start talking about lcd and lcm and gcf and gcd they always get those confused for the first part of kind of their math career so the lcm is the least common multiple so you have two three four numbers whatever it is and you want to find the least common multiple when we talk about the least common denominator that's when We're working with fractions and we just take the denominators and we find the least common multiple of the denominators and that is the lcd so you can see how the lcm

and the lcd are related now the greatest common factor or the greatest common divisor those are the same thing okay if something is a factor it's also a divisor so when we talk about that we're looking for a number that is the largest factor or the Largest divisor of a group of numbers now the way they're different is that the gcf or the gcd is going to generally be a smaller number and the lcm or the lcd is going to be a larger number so let's revisit real quick with this example with adding fractions so

i have something like 3 10 plus 7 15. now we know that we can't just add across like we do with multiplication 3 plus 7 is 10 10 plus 15 is 25. you know i see students do that that's wrong that is Wrong we need to have a common denominator before we start and the most efficient common denominator is the least common denominator so the lcd or the least common denominator and this is equal to the least common multiple of the denominators so of 10 and 15. now hopefully you remember how to do this from

pre-algebra but if you don't You basically just factor each number so for 10 it's 5 times 2 5 is prime and so is 2. for 15 it's what it's 5 times 3 5 is prime and so is 3. now each prime factor from every prime factorization is going to go in your list when you build it the only thing is if you have a duplicate prime factor it's only going to go in the largest number of times it occurs in either of the factorizations So in other words i have five and i have two i

have five and i have three so five occurs in each prime factorization the largest number of repeats are the largest number of occurrences and any of the prime factorizations is one just have one here and i have one here so i'm just going to put one in one factor of five in when i build that lcm or lcd or however you want to think about it now i'm going to multiply by i have a 2 Here no 2 over here so i just throw it in there i have a 3 here no 3 over here

so i just throw that in there so the lcd again is the lcm the least common multiple of 10 and 15 and that's 5 times 2 which is 10 times 3 which is 30. so once i have found that i go through and i rewrite each fraction as an equivalent fraction where 30 is the denominator so in other words with 3 tenths I would multiply the numerator by three and also the denominator remember i'm just multiplying by a complicated form of one so i'm not changing the value i'm just changing what it looks like then

plus we have seven fifteenths and this is multiplied by two over two and so this is equal to three times three is nine over ten times three that's thirty then plus seven times two is fourteen over fifteen times two that's thirty And then 9 plus 14 is going to be 23. so we end up with 23 over 30 as my answer and that can't be simplified any further okay so we know how to find the lcd with regard to regular fractions it's going to get a little bit more complicated when we start thinking about rational

expressions because we have some messy variables involved right so it's a little bit more complex but the general idea is the same now before i move on i Just want to show you so that i settle all confusion that this is not the same as the gcf now notice how you have the number 10 and you have the number 15. now the lcd or the lcm is a larger number 30 is the smallest positive number that is divisible by both 15 and 10 meaning i divide 30 by 15 i get 2 no remainder i divide

30 by 10 i get 3 no remainder okay if we talk about the greatest common Factor or the greatest common divisor this is equal to what it's going to be a smaller number because we're looking for the largest number that each of these numbers is divisible by so it's kind of the same process you factor each number to start but all you're going to do is look for what's common to both okay so i can only put something in the Greatest common factor listing if it's common to everything in this case i don't have a

2 that's common to everything i don't have a 3 that's common everything the only thing that's common in both is 5 and so my greatest common factor is a 5. 10 divided by 5 is 2 no remainder 15 divided by 5 is 3 no remainder so you see that the greatest common factor or again the greatest common divisor is a smaller number and the lcd or lcm is a larger number so That's one of the tricks i used to remember is taught to me by a teacher back when i was in eighth grade but it's

something that you should think about because a lot of students really struggle with this definition between again the gcf and the gcd same thing and then the lcd the least common denominator which is the lcm of the denominators so let's look at one more example with fractions real quick just to kind of get our feet going we have 11 24 minus 9 20. so again if i want the lcd this is the least common multiple of these two denominators so of 24 and 20. so what do i do i factor 24 and i can do

4 times 6 4 is 2 times 2 6 is 2 times 3 for 20 i'll do 5 times 4 4 is 2 times 2. so now if i build my list every prime factor from each prime factorization is Going to go in there i'm just paying attention for duplicate prime factors so with 2 i have 1 2 3 of those here and have one two of those here so when i build my list how many am i going to put in here well i'm going to put three i go with the largest number of repeats

from either prime factorization that occurs here i have three of them so i'm going to put 2 times 2 times 2 or 8. Again where students make a mistake is they throw all of them in there and that's going to give you a common multiple but it's not going to be the least common multiple and that's what you're looking for to reduce the amount of work when you go to simplify all right so now the next thing is i have a 3 not common to both so i'm going to throw that in there and then

i have a 5 not common to both so i'm just going to throw that in there So now we just multiply 2 times 2 times 2 is 8. 8 times 3 is 24 24 times 5 is 120. so that's your lcd or lcm now if we want to define the gcf the greatest common factor or the greatest common divisor what would that be when we build that list it's only going to be numbers in the prime factorizations that's common to everything the only thing we Have here that's common to everything is 2. now i have

two factors of 2 here and again i have one two three factors of two here with the gcf it's got to be common everything so it's gonna be two factors of two that's gonna go into the gcf and so two factors of two is four so the greatest common factor is four again this is a smaller number than these two and then the lcd or lcm is a larger number than these two that's how you kind of get an understanding of it 24 divided by 4 6 no remainder 20 divided by 4 is 5 no

remainder with this 120 divided by 24 is 5 no remainder and 120 divided by 20 is 6 no remainder so you can see the difference between the two let's go ahead and execute this problem just for the sake of completeness we know what the lcd is so let's transform each fraction so we'd have 11 over 24 we'd multiply this by 5 over 5 And then minus 9 over 20 and we'd multiply this by 6 over 6 so i'd end up with 55 over 120 minus 54 over 120 and this would give me one over 120

as my answer and i can't simplify that any further so again similarly when we work with rational expressions we need to have a common denominator to add or subtract it's going to be the same process again it's going to be messier because we have a Lot of variables involved so to find the lcd for a group of rational expressions we're going to factor all denominators completely just like if i'm looking for the lcd for a group of numbers i'm going to factor the denominators completely i'm getting it down to the prime factorization of each number

then i'm going to list each different denominator factor when a factor occurs in more than one factorization list the largest number of repeats okay there goes again Among all factorizations so again this is the same process that we use when we're thinking about numbers if i have a prime factor of two in one prime factorization and a prime factor of two and another one i'm looking for the largest number of repeats between those two factorizations to put that in when i build my list all right then the last step is just to multiply the factors

to obtain the lcd so again same processes when we work with fractions But it's just going to be a little bit more messy a little bit more time consuming a little bit more tedious for the simple fact that we have variables involved all right so let's start out with these two rational expressions we have 13 over 21x cubed and we have 7 over 12x to the fifth power so i just want to find the lcd so i'm working with the denominators so the lcd is what it's the lcm Of 21x cubed and of 12x to

the fifth power so pretty easy for the number parts we already know how to do that right we would just say okay well 21 is what it's 3 times 7 and 12 is 4 times 3 4 is 2 times 2. all right so to build the number part what would i do i list every prime factor i just watch out for duplicates if i have a duplicate prime factor Meaning it occurs in more than one prime factorization looking for the largest number of repeats so i have a three here and here it occurs only once

in each so i'm just going to put one in here not two i have a seven here no seven over here i have two twos in here so i'm going to throw both of those in there and we know that's 4. so the number part is 3 times 7 which is 21 times 4 which is 84. so let's just go Ahead and write 84 there and then what about the variable part well do i really need to go through and say well i have x times x times x and then i have x times x

times x times x times x again i want the largest number of repeats so if it's the same variable now i'm just going to go with the largest exponent remember when we talked about the greatest common factor we went with the smallest exponent okay now we're Going with the largest exponent so the largest exponent is the 5 so it's going to be 84 x to the fifth power that constitutes the largest number of repeats for either of the prime factorizations so your lcd here is 84x to the fifth power all right let's take a look

at another so we have 3x minus 2 over 25x squared and then we also are looking at negative 7 over 10x to the 4th power so again my lcd My lcd is the lcm of these two denominators so 25x squared and then 10x to the fourth power so again we know how to do the number part we can almost do that in our head we know 25 is what it's 5 times 5 and 10 is five times two so i know that i have a duplicate prime factor of five it occurs twice here so one

and then two two factors of five here only one here so i go with the Largest number of repeats and that's in the prime factorization of 25 so i'm going to throw two of those in here 5 times 5 is 25 and then i'm going to throw a 2 in there so times 2 25 times 2 is 50. so the number part is a 50. now for the variable part i have x squared and i have x to the fourth power again i want to go with the largest exponent on x and the reason for

that is that's going to be the largest number of Repeats between the two prime factorizations if i had x squared it's x times x so two of those if i have x to the fourth power i have x times x times x times x this is the largest number this is the largest number again as given by the largest exponent so i just use that so this is 50x to the fourth power all right so let's take a look at a number we have 7x cubed minus 11 over 7x squared plus 21x and then we

have negative 4x to the fifth power minus 13 over 5x squared plus 15x now for this one you might start thinking about as is really complicated what do we do again you've got to factor everything completely so i want the lcd of these two guys the first thing i'm going to do is i'm going to factor them so from this first guy here i can pull Out a 7x so i can pull out 7x and then what's left inside is x plus 3. from this guy right here i can pull out a 5x so i

pull out a 5x and what's left is x plus 3. so now i'm looking at what's common so i have x plus 3 and i have x plus 3. so when i build my lcd my lcd i'm going to have a factor of x plus 3 in there it occurs only once in each so that's all i'm going to do Now each one has an x i have an x here and i have an x in each case it's only one so i'm going to put one x in there and then i have a 7

and i have a 5. each of those is a prime number 7 times 5 is 35. so what this is going to be it's going to be 35x 35x times x plus 3 that quantity and i would just leave it just like this in factored form you can go through and multiply it if you want but usually when you're Finding an lcd it's just better to leave it in factored form in case you need to do some canceling later on all right let's take a look at another one so we have 2x cubed minus 5

over 5x minus 10. then we have 11x minus 4 over 6x minus 12. so again i want to look at these two denominators and i want to factor them completely so for this one i could pull out a five and i would have x minus two For this one i could pull out a six and i would have x minus two so you can see you have x minus two and x minus two so for the lcd you're gonna have what you have a 5 and you have a 6 5 times 6 is going to

be what that's going to be 30 and then you have that x minus 2. it only occurs once in each factorization so that's all i need to do when i put it in here i've got one factor of x minus two then i've got one Factor of five and one factor of six because six is two times three and five is prime so five times two times three would give me thirty that's how i got that and then i've got my one factor of x minus two so you get 30 times the quantity x minus

2 and again i would leave this in factored form all right let's take a look at one more problem i think you get the general idea so we have n minus 5 i can't really factor that i'll just write n minus 5. I have 3n cubed plus 15n squared so i could pull out a 3n squared and i would get n plus 5 and then over here i have n squared minus 25 that's the difference of two squares so this is n minus five times n plus five so for my lcd what do i have

i have n minus five here and also here And then i have n plus five here and also here so each only occurs once this occurs once this occurs once so i would just put one in there and then this occurs once and this occurs once so i'll just put one in there and then we have this 3n square that's got to go in so let me kind of scooch this down so we like it to be in the front that's just more traditional And so our lcd is 3n squared times the quantity n minus

5 times the quantity n plus 5. hello and welcome to algebra 1 lesson 46. in this video we're going to learn about adding and subtracting rational expressions so when we're adding or subtracting rational expressions we need to first have a common denominator it's just like when we added or subtracted fractions back in pre-algebra so when there's a common denominator all We need to do is add or subtract the numerators and place a result over the common denominator all right for the first one we're going to look at we have u minus 6 v over 16

u squared then plus u plus 2v again over 16 u squared so you can see that we have a common denominator of 16 u squared so just like with fractions all we need to do is add the numerators and put the result over that common Denominator so just to show you real quick it's the same thing as if i had something like let's say one fifth plus two-fifths i have a common denominator of five so that just stays the same and then i just add the numerators one plus two is three and so i get

three fifths so here i'm gonna do the exact same Process it's just going to be a little bit more tedious because i have some variables involved right it's just going to be more work so i'm going to keep this part the same so 16 u squared that will remain unchanged then for the numerators i have u minus 6v so u minus 6v then plus u plus 2v kind of expand this out a little bit let Me write this in the center 16 u squared and i just want to combine like terms in the numerator so

u plus u is 2u and then negative 6v plus 2v is going to be negative 4v and again this is over the common denominator of 16u squared now in a lot of cases you'll stop and you'll say this is my answer right i'll box this up and say i'm done but there's a problem here remember when we report our answer with fractions we Want to see if we can reduce the fraction to its lowest terms right or simplify you're always expected to do that if i had something like 2 15 plus 3 15 have that

common denominator of 15 2 plus 3 is 5. now i wouldn't leave my answer this way i would say okay the greatest common factor between these two is 5 so i would divide the numerator by 5 and i would get 1. i would also divide the Denominator by 5 and i get 3. and so my simplified answer here is 1 3. so i'm going to do the same thing here and all i really need to do is just factor it to see what the gcf would be so in the numerator i can pull out a

2. so this would be 2 times the quantity u minus 2v in the denominator i have a 16. now i don't need to go through and write 2 times 2 times 2 times 2 i know 16 is 8 times 2. so i'll just write 2 times 8 And then times u squared so i can see that i can cancel this factor of 2 with this factor of 2 and so i have reduced this rational expression to its lowest terms so i have u minus 2v as my simplified numerator over 8 u squared as my simplified

denominator and there's nothing else that i can do here remember this is a subtraction sign you can't go through and start going okay this is Going to cancel with this i see that all the time or students will go okay this is going to cancel with this and that's going to give me a 4. that doesn't work that way it's got to be factors right so we cancel common factors here this is multiplication i am multiplying 2 by this u minus 2v by that quantity i'm multiplying 2 by 8 u squared so this is multiplication

these are factors so i Can cancel common factors when i think about u minus 2v these aren't factors okay so i can't go through and just start canceling things so it's very important that you understand that we cancel common factors alright so again the answer here is u minus 2v over 8u squared all right let's take a look at another we have 2m minus 5n over m minus 3n and then we're adding to this m minus 4n over m minus 3n so again we Have one with a common denominator of m minus 3n we're just

going to leave that as it is and we're going to add the numerators so the numerators we have 2m minus 5n plus m minus 4n over again that common denominator of m minus 3n so just combine like terms here so 2m plus m is 3m this is 3m And then negative 5n minus 4n is negative 9n and this is over m minus 3 n so again i want to think about what i can cancel in some cases you're not going to be able to cancel anything in other cases you're going to be able to cancel

a lot so if i look at m minus 3n i can't factor that at all if i look at 3m minus 9n i can pull a 3 Out so let's just do that just pull out what you can and see what you get so i'm going to pull out a 3 and i'll get m minus 3n inside and we can see that now we have a common factor between numerator denominator of m minus 3n again if you want to you can write this as times 1 so that's completely clear and then cancel this with this

and what i'm left with is 3 over 1 or Just 3 as my simplified answer okay so now let's take a look at one with subtraction so we have v plus 8 over 24 v minus 12 and i'm subtracting away 7v minus 6 over 24v minus 12. very important that you realize that this whole thing is being subtracted away so that means i want to put my numerator inside of parentheses when i go to do the subtracting and we'll do that in a Second now i have a common denominator that common denominator is 24v minus

12. so i'm just going to put 24v minus 12 down here and this is over i have v plus 8 minus i'm subtracting this whole thing away please don't make the mistake of just putting minus 7v minus six that is wrong i am subtracting away the seven v and the negative six So there's kind of two things you can do you can change this to plus okay this the plus and then you can change the sign of each term so this can be plus negative 7v and then since this is negative 6 it would be

positive 6. you can do that or you could do it kind of the slower way and just put a minus sign out in front of parentheses so 7v minus 6. but again if you see a negative outside of some parentheses what do you do in order To drop the parentheses you've got to change the sign of each term inside remember multiplying by that phantom negative 1. so we can do plus negative 1. again negative one times seven v is negative seven v and then negative one times negative six is plus six so any kind of

way that you look at it it's going to be minus seven v minus seven v and plus six okay so if we combine like terms v minus 7v is Negative 6v and then 8 plus 6 is 14. so plus 14. this is over 24v minus 12. now again we want to look to see if we can simplify anything before we report our answer i know in the numerator i can pull out a 2 i could also pull out a negative 2 if i wanted to which is just kind of up to me let's just pull

out a 2 and i'd have Negative 3v plus 7. down here let's also pull out a 2 and i'll have 12v minus 6. now i can't really factor this any further so i really don't need to do anything else down here i would just cancel this common factor of 2 between numerator and denominator and then i will report my answer as negative 3v plus 7 over 12v minus 6. all right so the harder Scenario occurs when we add or subtract rational expressions without a common denominator it's the same thing as if we added or subtracted

fractions without a common denominator we've got all these extra steps so with fractions if i have something like let's say 1 4 plus two thirds i find my lcd in this particular case the lcd is 4 times 3 or 12. i Change each fraction into an equivalent fraction where 12 is the denominator so i multiply this by 3 over 3. remember that's a complicated form of one so it just changes the way it looks not the value right so it's still worth one fourth in the end then plus i'll have two thirds i'm going to

multiply this by four over four again just a complicated form of 1. so 1 times 3 is 3 over 4 times 3 is 12. Then plus 2 times 4 is 8 over 3 times 4 is 12. then we have a common denominator so we just put the common denominator there we add the numerators 3 plus 8 is 11. so i get 11 12 there so it's the same process for adding and subtracting rational expressions without a common denominator it's just the fact that you have variables involved you have to factor stuff it just gets very

very messy and very very tedious Depending on you know what level problem you're looking at all right so the first thing we want to do is find the lcd again the least common denominator we learned how to do that for a group of rational expressions in our last lesson if you don't know how to do that at this point you know if you start watching more of this video and you get very very confused please stop go back to that video and watch that Get fully up to speed on how to find the lcd and

then come back right it's very very difficult to do this if you've missed a step the next thing you're going to do is you're going to rewrite each rational expression as an equivalent rational expression where the lcd is its denominator just like we did with the fractions when we were adding them a minute ago all right then we want to add or subtract numerators and then simplify So very very simple overall procedure-wise just again if you're working with something that's you know very long and tedious it just gets kind of irritating sometimes so let's start

out with this first one we have 7 over 3k cubed plus 12 k squared plus 3 over k plus 4. so the first thing i want to do is i want to find my lcd so the way i do that is i factor each denominator So if i factor this denominator i could pull out a 3k squared and that would leave me with what i pull that out from here i'd get k and then plus if i pull that out from here i would have a 4. now this over here is k plus 4. so

really if i look at the lcd it's going to be what it's going to be 3k squared times k plus 4. so 3k squared times k plus 4. so this is okay i don't need to make any Changes to that rational expression so i can leave it as it is 7 over 3k squared times the quantity k plus 4 then plus for this guy over here we have 3 over k plus 4. now what would i need to multiply k plus 4 by to make the denominator the same as this well this is what i'm

missing this factor of 3k squared so i would multiply the denominator by 3k squared and to make that legal i need to also Multiply the numerator by 3k squared because i want to multiply by 1. right 3k squared over 3k squared is 1. just a complicated form of 1 but it's still 1. now if i multiply here 3 times 3 k squared is 9 k squared let's scroll down a little bit so this equals i'll rewrite this first 7 over again 3k squared times the quantity k plus 4. As i just said 3 times 3k

squared is 9k squared and this is over 3k squared times the quantity k plus 4. so 3k squared times the quantity k plus 4. so now i have a common denominator right this denominator is the same as this denominator so i can just add the numerators and so basically what i'd have is just 9k squared plus 7 over that common denominator of 3k squared Times the quantity k plus 4. and in this particular case there's nothing i can do to simplify this numerator here is prime there's nothing i can really pull out so i'm not

going to be able to really cancel anything here again don't make the mistake that i see so many students do they'll go okay well this 9 can cancel with this 3 and this is a 3. no that doesn't work like that you cancel common factors this this is Not a factor this is 9k squared plus 7. so you can't cancel there if i had this being multiplied by let's say 9 k squared something like that well yes since this is multiplication here these are factors these are factors and then these would be factors if that's

the problem that i had so then i could cancel this 9 with this 3 and this would give me a 3. if that's the problem we have we don't have that though so there's nothing i can do to simplify me Erase all this there's nothing i can do to simplify it so 9k squared plus 7 over 3k squared times the quantity k plus 4 is our simplified answer okay let's take a look at another one so we have 7x over x plus 1 and we're subtracting away 3 over x minus 3. so the lcd here

is very very simple each of these is prime right so i can't factor x plus one and i can't factor x minus three so i just multiply the two together to Get the lcd it's just like if i had fractions where i had one-fifth plus let's say two-sevenths right each of these is prime so to get the lcd i just multiply five times seven and i get 35. well same thing here each is prime multiply the two together and get the lcd all right so in order to change 7x Over x plus 1 into a

rational expression with this as its denominator i need to multiply by what i'm missing in x minus 3. so an x minus 3 is going down here and so i've got to also do that to the numerator so that it's legal then minus up here i have a 3 over x minus 3. and what i'm missing here is the x plus 1. so i'm going to multiply by i have x Plus 1 down here and again to make it legal i multiply by x plus 1 up here so now if we crank this out what

i'm going to get is 7x times x is 7x squared 7x times negative 3 is minus 21x and then i'm subtracting away this right here so 3 times x is 3x 3 times 1 is 3 so plus 3. now again i'm subtracting all of this away so i'm going to put minus and i'm going To put parentheses around this and what that's going to tell me is that to remove the parentheses again i've got to change the sign of each term so don't make the mistake again i'm just putting minus 3x plus 3. both of

those need to have their signs changed so this is going to be minus 3x changes from positive to negative and then minus 3 again changes from positive to negative so this is over the common denominator Of x plus one that quantity times x minus three and now we just want to combine like terms so seven x squared nothing i can do with that negative 21x minus 3x is minus 24x and then minus 3 and this is over the quantity again x plus 1 times x minus 3. now here comes part of the problem sometimes you're

going to get something like this and it's going to be extremely tedious Right so i've gone through all that work and now i've got to factor this to see if i'll have a common factor with the denominator to simplify so i've got to think about two integers whose product so i want a product of what 7 times negative 3 so of negative 21 and then whose sum is going to be negative 24. now for 21 you basically only have what you have 1 Times 21 and you have 3 times 7 that's it i can't play

with 1 and 21 i can't play with the signs and make that sum to negative 24. i also can't play with the signs with 3 and 7 to make that sum of the negative 24. so this is a prime polynomial up here in the numerator i can't factor it and so what i'm going to say here is that this is simplified we have 7x squared minus 24x minus 3 over the Quantity x plus 1 times the quantity x minus 3. all right let's take a look at another one so we have 5x plus 3 over

3x minus 3. then we're adding to this 2x minus 7 over x squared minus 1. so again to find the lcd i want to factor the denominators for this one i could pull out a 3. so i would have 3 times i would have x minus 1 x minus 1. for here it's the difference of two squares so some of these you should Start to remember like x squared minus 1 x squared minus 4 you know things like that x squared minus 16. start to memorize these because as they come up you want to be

able to see that okay if i see a binomial here and i got a minus sign first guy is squared the second guy is a perfect square i know i have the difference of two squares so x squared minus one will factor into x plus one Times x minus one now i've already got an x minus 1 here so i don't need to put it into the list of the lcd what i do need to do is throw in this x plus 1 because it's not currently present so my lcd will be 3 times the

quantity x minus 1 times the quantity x plus 1. so let's go ahead and change each rational expression to an equivalent rational expression Where the lcd here is its denominator so i'll have 5x plus 3 over i'm going to write this as 3 times the quantity x minus 1. and what i'm missing here is this x plus 1. so i'm going to multiply by x plus 1 over x plus 1. then plus over here i have two x minus seven over i have x plus one times x minus one those two quantities i'm missing a

three here so times three Over three now very important since i'm multiplying by three remember to use parentheses don't just multiply 3 times 2x and be done so i'm going to put parentheses around it because i've got to use my distributive property okay so let's put equals here let's scroll down and get some room going and so i'm going to use foil here 5x times x is 5x squared the outer 5x times 1 is 5x the inner 3 Times x is 3x so 5x plus 3x is 8x so plus 8x and then the last 3

times 1 is 3. so plus 3. then plus again use your distributive property 3 times 2x is 6x then minus 3 times 7 that's 21. okay so this is all over that common denominator of 3 times the quantity x plus 1 times the quantity x minus 1. so now let's go ahead and simplify the Numerator we'll have 5x squared nothing i can do with that i have 8x plus 6x that's 14x so plus 14x and then i have 3 and negative 21 that's minus 18. so 5x squared plus 14x minus 18. and then again this

is over that common denominator of 3 times the quantity x plus 1 times the quantity x minus 1. now can i factor this trinomial in the numerator So 5 times negative 18 is negative 90. so give me two integers whose product product is negative 90 and whose sum is 14. well let me think about that i've got what i've got 1 times 90 that won't work i've got 2 times 45 not going to work i've got 3 times 30. i've got not divisible by 4 i've got 5 times 18 that won't work i've got 6

times 15. That won't work not divisible by 7 not divisible by 8 is divisible by 9 it's going to be 9 times 10 and that's all we can really do so i can't take any of these and manipulate the signs to get a sum of 14 and a product of negative 90. so this trinomial up here is going to be prime and i can't factor it so what's going to happen is i'm just going to leave my answer as it is this is simplified 5x Squared plus 14x minus 18 over 3 times the quantity x

plus 1 times the quantity x minus 1. all right let's take a look at another one let's say i have negative 2x over x plus y then i'm adding to this 5x plus y over 2x plus 2y so what i want to do here again is start out by finding the lcd so it's the lcd this x plus y i can't factor that over here i can factor out a 2. so i Could write this as 2 times the quantity x plus y now i've already got an x plus y here i just need the

2. so the lcd is 2 times the quantity x plus y so because this denominator is already the lcd i don't need to change this rational expression i just need to change the one on the left so what i'm going to do i'm going to multiply negative 2x Over x plus y times what i'm missing here is a 2. so 2 over 2 then i'm going to add to this the 5x plus y over two times the quantity x plus y okay so negative two x times two is negative four x and then plus five

x plus y again over this common denominator of two times the quantity x plus y if i combine like terms here negative 4x Plus 5x is x so what this gives me is x plus y in the numerator over 2 times the quantity x plus y in the denominator and so what i can see is i have a common factor of x plus y again to make this obvious go ahead and put this inside parentheses and put times 1 right 1 times anything is just itself so now i can cancel this factor with this factor

right those are factors 2 is multiplying this so those two are factors one is multiplying this so again those two are factors so we cancel common factors what i'm going to be left with is 1 over 2 or one half as my answer all right so let's look at one where the denominators are opposites so we have three over four x minus five plus nine over five minus four x so remember we talked about this before if i see the same terms but they have Different signs so in other words i could rearrange this and

i could write this as negative 4x plus 5 if i want to so i have the same terms i have 4x and i have 4x but the signs are different i have a negative and i have a positive i have 5 and i have 5 but the signs are different right i have a positive and i have a negative right but same same values different signs so those two are opposites And to show that to you just take one of them and factor out a negative one so in other words you put a negative one

outside of parentheses and you change the sign of each term inside of the parentheses so negative 4x would be 4x and then positive 5 would be negative 5. now you can see that these two are the same with the difference of this being multiplied by negative 1 right so i have 4x minus 5 i have 4x minus 5. again this Is being multiplied by negative 1. what can i do here well i want you to recall that if i was working with the fractions let's say 5 over 7 and i'm adding to this let's say

1 over negative 7. what could i do remember if the denominator is negative and the numerator is positive the fraction is Negative so all i need to do is erase this put that here it's still the same value and i have that common denominator of 7. now 5 minus 1 is 4 over the common denominator of 7. it's the same thing here it's the same principle so really all i'm going to do i have the same denominator all i need to do is take this negative 1 move it into the numerator okay that's all i'm

going to do very very simple so I have 3 over 4x minus 5 and then plus i'm going to now make this negative 9. over 4x minus 5. okay all i did was move the negative from the denominator to the numerator again you can do that i just showed you that with regular fractions it's the same if negative 1 is multiplying this i can just move it up here and have the Negative 1 multiply the 9 that's how i got negative 9. okay so now i have a common denominator and all i need to do

is add 3 plus negative 9 is negative 6 so what i get here is negative 6 over 4x minus 5 very minimal work and you can see there's nothing that i could do to cancel anything so that's my simplified answer again negative 6 over 4x minus 5. All right so we saved the best for last here's one that you might get something extremely tedious i have z plus b over 18 z squared plus 12 zb minus 3 zb minus 2b squared plus 3z minus b over 36 z squared minus b squared now to find the

lcd i've got to factor these denominators first and this might take a little while now the first one i can factor by grouping it's a polynomial with four terms so let Me just kind of do this here and we'll figure out an lcd in a minute so i can pull out a 6z from the first group that would leave me with 3z plus 2b in the second group i can pull out a negative b negative b that would leave me with 3z plus 2b So i have a common binomial factor of 3z plus 2b and

so this can be factored as 6z minus b times the quantity 3z 3z plus 2b okay so let's erase this real quick so we've got this as the factored form of that now when i look at 36z squared minus b Squared what do you notice you've got two terms here and you've got a subtraction sign if you see that you've got some squared things you might want to think about the difference of two squares again they're going to throw that at you all the time 36 is a perfect square it's 6 times 6. z squared

is z times z b squared is b times b so really what this is this is six z squared minus b squared So this factors into six z plus b times six z minus b now what you can see is that you have this and this that are common so when i build my lcd what i'm going to have is one factor of 6z minus b i don't need two then one factor of 6z plus b then one factor of 3z plus 2b so that's my lcd Now what do i need over here over here

i would be multiplying by just 6z plus b so i would have z plus b and this would be multiplied by again 6z plus b now for this one over here let me put plus i would have 3z minus b times what am i missing well i'm missing the 3z plus 2b 3z plus 2b okay so let's go down here i know my lcd is this so let me write this over that so 6z minus b times 6z plus b times 3z plus 2b again just one that's pretty tedious overall so i'm going to use

foil z times 6z is 6z squared z times b is zb zb B times 6z is plus 6 zb if i combine these two i get plus 7zb so plus 7zb and then for the last b times b is b squared so plus b squared now over here 3z times 3z is 9z squared 3z times 2b is plus 6zb negative b times 3z is minus 3zb if i combine like terms here i'd have plus 3zb plus 3zb And then lastly i have negative b times 2b so minus 2b squared and again all over that common

denominator of 6z minus b times 6c plus b times 3z plus 2b all right so for like terms i have 6z squared and 9z squared that is 15z squared i have 7zb and 3zb that's 10zb i have b squared and negative 2b squared So that's minus b squared then this is all over again we have that 6z minus b times 6z plus b and then times 3z plus 2b okay so one last step just to check to see if the numerator can be factored i need to find two terms now because we have two variables

involved so i'm going to treat this last one like if it was a constant right remember this is usually my c term So i'm going to say a the coefficient here times this guy right here so 15 times negative b squared is negative 15 b squared so i need two terms that would multiply together to give me that and that would sum to give me 10 b okay z is my squared variable here so i'm going to treat this b part right here as if it's part of that coefficient of z right so 10 b

10 b this is what i'm looking for for the sum negative 15 b squared for the Product okay i know that's a little advanced we'll do a lot of that in algebra two but for right now just kind of follow me on this forget about the b part can you think of two integers that would multiply together and give you negative 15 and sum to give you 10 well you've got 1 times 15 that's not going to work you can't change the signs around and give you that and you've got 3 times 5. so nothing

you can really do With that so you can't get the number part to work out this part would work itself out automatically if you got the numbers to work out so this is going to be a prime polynomial and we're not going to be able to factor it any further so this is the simplified answer 15z squared plus 10zb minus b squared and this is over 6z minus b times 6z plus b and then times 3z plus 2b hello and welcome to algebra 1 lesson 47 in this video we're going to learn about complex fractions

so i know not all of you took my course on pre-algebra but hopefully before you got into the algebra 1 course you're currently in you've at least seen complex fractions at some point now what is a complex fraction let's just start out with that well it's a fraction that has a fraction in its numerator Its denominator or both the numerator and the denominator so for example here we have 5 6 plus 1 8 over 7 halves so there is a fraction there's actually two fractions being added together in this case in the numerator of the

complex fraction this is the numerator and there's also a fraction in the denominator you have 7 halves there and then you can kind of think about this as your main division or your main fraction bar so your main Division and why do i say main division remember if you have fractions you are dividing the numerator by the denominator something like 9 over 3 is the same as 9 divided by 3 which is 3 right it's the numerator divided by the denominator it is no different here with a complex fraction we have this numerator here although

it's a complex numerator 5 6 plus 1 8 is divided by this denominator here 7 Halves so how do we go about simplifying one of these well i gave you two methods in my pre-algebra course to do so the first one and one that we probably use most often is to simplify the numerator and denominator separately then you would just perform the main division so for example if i wanted to start out if i had 5 6 plus 1 8 let's just deal with this part first I would get a common denominator going and for

this one six factors into three times two eight factors into two times two times two so the lcd would be two times two times two times three right which is what it's 8 times 3 or 24. so i would take 5 6 and i would multiply by 4 over 4 and then i would take 1 8 and i would multiply by 3 over 3. Let me scroll down i'm going to take this off the screen for a minute but we'll come back to it so 5 times 4 is 20 plus 1 times 3 is 3

and this is over that common denominator of 24. so 20 plus 3 is 23 and then over 24. so if i simplify the numerator of the complex fraction i would get 23 over 24. so let's just erase all this and go back up to the top so i'm going To put equals out here i know i wrote a bunch of stuff but essentially what i'm doing is i'm saying the numerator of the complex fraction is 23 over 24. then the denominator of the complex fraction is seven halves so we think about it as 23 over

24 divided by seven halves and if it makes you feel more comfortable you can write it like this you can say okay well this is 23 Over 24 divided by seven halves we all know how to divide with fractions we take the first fraction or the left most fraction in this case you think about that as the top fraction and we multiply by the reciprocal of the fraction on the right or the second fraction so this would be 23 over 24 times flip this guy 2 over 7. in this case you take again 23 over

24 Times take the reciprocal of the fraction on the bottom 2 7 either kind of way you want to do it this is what we've been working with so far this is kind of the new way that we're going to see kind of moving forward in math so you know i want you to get comfortable doing it in this way i know you're already comfortable doing it in this way you know if the first few times you do this you've got to convert it from this To this so that it makes sense for you in

your mind then go ahead and do it but after you know 10 or 15 problems or so you want to try to merge into doing it this way because again as you move forward this is going to be more common okay so let me just erase this we know how that sets up and what i want to do i want to cancel any common factors before i multiply so now i can cross cancel a 2 between here and here And this would be 12. so now what i'm going to be left with is a 23

times 1 for my simplified numerator over 12 times 7 which is 84. so i get 23 over 84 as the simplified complex fraction now i taught you another method to do this in the lesson let me erase all this and let me just kind of erase this real fast the other method is to multiply the numerator and the denominator Of the complex fraction by the lcd of all denominators involved so in order to do this i'm looking for the lcd i want to look at all the denominators so i have a 6 here an 8

here and a 2 here that's all the denominators are involved in the complex fraction i'm not looking at just the denominator of the complex fraction i'm looking at individually each denominator that's involved so kind of in the numerator if i was to block this Out let me get a black color and just kind of block this out i have two denominators i have a 6 and an 8. so i'm thinking about 6 and i'm thinking about 8 and then kind of if i look at my denominator of the complex fraction let me block this out

for a minute i have a denominator of 2. so that's where i'm getting those numbers from i don't want you to be confused Now once i have those numbers i just find the least common denominator that i can form or again that's the lcm the least common multiple of these numbers so what is the lcm of 6 8 and two well i know that two doesn't factor i know eight is nothing more than two times two times two and i know six is what it's two times three so it looks like we're gonna have an

lcd of two times two times two or 8 Times 3 so that's going to be 24. now if i take 24 and again i multiply by the numerator and denominator of the complex fraction so i'm going to multiply this part right here by 24 and then this part right here by 24. what's going to happen i'm going to distribute this to each inside of the parentheses what am i going to get 24 times 5 6. Let's do that over here this is going to cancel with this and give me a 4 4 times 5 is

20. so let me write 20 over here and then plus i have 1 8 1 8 times 24 and this is going to cancel with this and give me a 3. 3 times 1 is 3 so i'd have plus 3. so then plus 3 and then over down here what i have is 24 times 7 halves so 7 Halves times 24 and we could see this would cancel with this and give me a 12. 12 times 7 is 84. so this equals 20 plus 3 is 23 over 84. so i get the same answer either

way it's just a different method and as far as which one you want to use just depends on you right if you have trouble finding the lcd or understanding what i'm doing here you might want to do it the first way To kind of break it down for yourself and then kind of move into this method as you get more comfortable if you're completely comfortable with what i just did then the lcd method is usually faster okay so why are we talking about complex fractions again well the reason that we're talking about complex fractions again

is because it comes up when you get to rational expressions somewhere in your chapter in rational expressions usually right after You learn how to add and subtract you're going to talk about complex fractions with regard to rational expressions so it's the same thing you have the same type of problem it's just more complicated because there's variables involved but you use the same methods to simplify so for example if i have 9x squared minus 4 over 2x to the fifth power over 3x cubed minus 2x squared over 2x to the fifth power Let's solve this guy

just using kind of the first method and what i want to do here is just simplify everything completely before i kind of set up the division so in the numerator here i have nine x squared minus four the trick here is just try to factor everything before you get started to see what can factor and the trick with rational expressions is always to factor as you're working right you want to end up Seeing what you can cancel nine x squared minus four i know that's the difference of two squares that is three x plus two

times three x minus two then this is over 2x to the fifth power now i'm dividing by or my main division is here this is the numerator of the complex fraction this is the denominator so i would be multiplying by The reciprocal of this so then the denominator here in this denominator of the complex fraction i know that gets confusing but this is your denominator for the denominator of the complex fraction or this guy that is going to come up here 2x to the fifth power then the numerator is going to come down here but

i see that i can factor out an x squared so let me just go ahead and do that real Quick so x squared times if i pulled that out 3x cubed divided by x squared would be 3x and then minus 2x squared divided by x squared would be 2. now before i do this multiplication what can i cancel i can cancel this factor of 3x minus 2 with this factor of 3x minus 2. remember this is multiplication here Multiplication so these are factors now i can also cancel this 2x to the fifth power with this

2x to the fifth power and what am i going to be left with well i'm just going to have a 3x plus 2 in the numerator over x squared in the denominator so not too bad overall and this is kind of an easier example and not that they get harder they just get much much more tedious right that's the trick it just gets much much more tedious and so you Need to take your time and go step by step so you don't make any silly mistakes and just one more thing before i kind of get

to the next problem again when you see your main division if you're uncomfortable setting it up like this go ahead and take this guy and say okay i have three x plus two times three 3x minus 2 over you know 2x to the fifth power Take your main division and put a division symbol out there keep this as it is so i would factor out the x squared and then have 3x minus 2. and then over 2x to the fifth power when i set it up like this it reminds me to multiply this guy right

here by the reciprocal of this guy i set it up in one step here but again it might take you two because you might be so used to Kind of doing that from pre-algebra and that's fine again 10 or 15 problems go by you want to kind of just convert to the new method because again as we move forward in math we're going to see it more and more as this versus what we saw back in pre-algebra all right let's take a look at another one so i have 3x minus 3 over 9 and then

this is over x over 3 minus 5 over x minus 1. so again what i'm going to do is use That kind of first method where i simplify the numerator denominator completely then i kind of perform the main division so the first thing i would do is i would just find the main division bar so this is the numerator of the complex fraction this is the denominator that's your main division okay so for the numerator is there anything i can do to simplify that well 3x minus 3 i could factor out a 3 and i'd

have x Minus 1 inside the parentheses this is over 9. 3 and 9 share a common factor of 3. so this can be cancelled with this this would be a 3 and this would be a 1. now this is over for this one i have to do a little bit more work and x over 3 minus 5 over x minus 1. now let me kind of scroll down a little bit and let me do this down here so x over 3 Minus 5 over x minus 1. what would the lcd be well it would just

be 3 times the quantity x minus 1. so i would multiply this by 3 over 3 and i would multiply this by x minus 1 over x minus 1. and what would that give me x times x is x squared minus x times 1 is x and then minus 3 times 5 is 15. and this is over that lcd of 3 times the quantity x minus 1. and i'm not going to Multiply it i'm just going to leave it in factored form for right now just in case it'll cancel with something later on you don't

want to go through a multiply and then have to go back and factor you just you know you want to save yourself a little bit of work can i factor x squared minus x minus 15 into the product of two binomials well this would be x and this would be x can you give me two integers whose Product is negative 15 and whose sum is negative one well fifteen only factors into one times fifteen or 5 times 3 so no that's not going to be possible so this is going to be prime and so i'm

going to drag this back up to the top and now i'm going to go ahead and set up my multiplication so i have the numerator here or the top part as x minus 1 that quantity over 3. And then i have my denominator down here which i'm going to flip it right i'm going to take the top part multiply it by the reciprocal of the bottom part so we would have 3 times the quantity x minus 1 on the top over x squared minus x minus 15 on the bottom so is there anything i can

cancel before i do the multiplication this 3 with this 3 and then that's it So what i'm left with is x minus 1 that quantity times x minus 1 again we could say that's x minus 1 squared if we want or your teacher might want you to go through and actually do the foil you know just to get some practice that's fine and then this is over x squared minus x minus 15. and just for the sake of completeness let's go ahead and do that if i have x minus 1 that quantity squared we can

use our special products Formula we know that would be x squared minus 2 times x times 1. so basically two x and then plus this guy squared which would be one so x squared minus two x plus one over x squared minus x minus fifteen again a lot of people will rather leave it in factored form so that you can show your teacher hey there wasn't anything that i could cancel but again that kind of goes back and forth as far as what he Or she may want you to do all right let's take a

look at another one so i have b squared over 4 minus a squared over 20 and this is over 1 over b plus 1 over 5. so again i have my main division here right here so this is your numerator this is your denominator right for the complex fraction so if i look at this i have b squared over 4 Minus a squared over 20 that's my numerator so i'm going to start out by just simplifying that guy now the lcd here would be 20 right 20 is going to factor into what 2 times 2

times 5 4 factors into 2 times 2. so essentially you need to multiply this by 5 over 5 and then you'd have a common denominator so this would be 5b squared minus a squared i don't need to do anything over here over that common Denominator of 20. and then this is over if i look at this guy right here 1 over b plus 1 over 5 the lcd is going to be 5b 5 i'm not going to be able to factor and i can't do anything with b either so i just have to multiply the

2 together so 5 times b so i'd multiply this guy 1 over b times 5 over five and i'd multiply one over five times b Over b so one times five is five plus one times b is b so you get five plus b or you could put b plus five doesn't matter over five b let me kind of reverse that and put b plus 5. all right so let's erase this real quick and again we're going to set up our multiplication so this top guy 5b squared Minus a squared over 20 is multiplied by

the reciprocal of this guy i'm taking this and i'm dividing by this so the first one or the top one now stays unchanged we multiply by the reciprocal of the bottom one so this is 5b over b plus 5. now can i do anything to factor this guy right here if i didn't have that 5 there i could do B plus a times b minus a be the difference of two squares but i do have the 5 there so that stops me from being able to do that so there's not really anything i can pull

out there now over here i know i can do some canceling because 5 and 20 have a common factor of 5. so this can cancel with this and i'll put a 4 here and a 1 here and then i'm basically done and all i want to do now is just multiply I can you know leave this in factored form and say okay this is b times the quantity 5 b squared minus a squared or you can go through and use your distributive property again that's going to be your personal preference or your teacher's preference again

this is over 4 times the quantity b plus 5. and again if we want to go through we can use our distributive property just to show both answers b times 5b squared is 5b cubed Then minus b times a squared is a squared b this is over 4 times b is 4b and then plus 4 times 5 is 20. so we get 5b cubed minus a squared b over 4b plus 20. all right let's take a look at one final problem kind of using this method and then we'll kind of look at the lcd method

to wrap up the lesson so if i look at 6 over x squared plus 2x over y squared then this is over 4 over X minus x over 2. so again this is the numerator for the complex fraction this is the denominator for the complex fraction this part right here is your main division okay your main division so for the numerator if i have something like 6 over x squared plus 2x over y squared the lcd would just be x squared times y squared nothing else we can do there so I'd multiply this by y

squared over y squared i'd multiply this by x squared over x squared and the numerator would end up being 6y squared plus 2x cubed over x squared y squared okay now for the denominator down here what do we have 4 over x minus x over 2. so the lcd here would be 2x so 4 over x times 2 over 2 minus x over 2 times x over x so this would be equal to 8 over 2x minus x squared over 2x so i'll write this over 8 minus x squared again over 2x So let me

erase all this real quick so we're going to set this up as a multiplication so this top part right here 6y squared plus 2x cubed over x squared y squared will be multiplied by the reciprocal of this guy which is 2x over 8 minus x squared now what can i do to simplify i notice that i could pull out a 2 from Here but it's not going to do me any good but i'm just going to go ahead and factor it anyway let me kind of scroll down a little bit so if i pull out

a 2 from here i'd have 3y squared plus x cubed this is over again x squared y squared times 2x over 8 minus x squared nothing i can cancel between numerator and denominator here nothing i can cancel between the numerator denominator Here the only thing i can cancel is one factor of x here with here right this would be x to the first power that's it other than that i'm pretty much done now i multiply 2 times 2 i would get 4 so i can put this as 4 times that quantity 3y squared plus x

cubed and then over then i would have xy squared Times the quantity eight minus x squared now your teacher may let you leave it in this format in most cases that's what i would do but she might also want you to get a little practice and use your distributive property so 4 times 3y squared is 12y squared then 4 times x cubed is plus 4x cubed and then over you have xy squared times 8 that's X y squared and then you have x y squared times negative x squared that is minus x cubed y squared

and for some of you you'll want to write this where 4 x cubed is in the front here so you can do that you can switch the order and say okay this is 4x cubed plus 12 y squared over switch the order here negative x cubed y squared Plus 8 x y squared all right let's take a look at a few that deal with the lcd method now i think for the majority of you you will prefer to use this method because it's a lot quicker to use in most scenarios now with that being said

it's a little bit challenging to kind of understand what you need to actually do when you first start utilizing this method but just like anything else if you practice this enough you'll get really good at it and again you'll Prefer this method because generally it's a lot quicker all right so let's take a look at a plus three over four and then over four over a plus three and then minus a plus three over thirty six so recall that the complex fraction has a numerator and let me kind of let me kind of label this

real quick this is the numerator and then it also has a denominator so This is your denominator and then recall that this is your main fraction bar right here so let me just highlight that this is the main fraction bar or main division again as i've said before when you have a fraction it's the numerator divided by the denominator so we have our numerator divided by our denominator so when the method that we've been using up to this point i would take this numerator And once it's simplified in this case it already is i would

multiply by the reciprocal of this simplified denominator right i'm just basically performing a division in the case of the method we're going to use right now we're going to kind of speed up the process a little bit and we're basically going to clear all these denominators the way we're going to do that is we're going to multiply the numerator Of the complex fraction and the denominator of the complex fraction by the lcd for all the fractions that are involved in the complex fractions so if i were to look at each denominator individually for all the

fractions i have involved i have this denominator that's 4 a denominator that is a plus 3 and a denominator that's 36. so to get the lcd i just think about the lcm the least common multiple of these three values again i'm thinking about four I'm thinking about 36 and i'm thinking about the quantity a plus three well this one's pretty easy because if i think about 4 it's what it's going to factor into 2 times 2. 36 has 4 as a factor right 36 i could factor into 4 times 9 right 4 again is 2

times 2 and 9 is 3 times 3. so for the number part of this guy i would basically have 2 times 2 times 3 times 3 or just 36 Right if 4 is a factor of 36 then 36 is going to be the number part then times i have that a plus 3 that quantity so this over here this a plus 3. so once we've found the lcd which is usually the most challenging part of this for students because they're not fully up to speed on that yet we want to multiply the numerator and denominator

of the complex fraction by that lcd and again your denominators for each Fraction are going to disappear so let me erase this real quick we know what the numerator denominator is at this point and i'm going to put a multiplication symbol right here and i'm going to be multiplying the numerator of the complex fraction by this 36 times the quantity a plus 3 and i'm going to do the same thing to the denominator And why is this legal before we move forward again this quantity over this quantity is equal to one right any non-zero value

over itself is one so 36 times the quantity a plus three over 36 times the quantity a plus 3 is 1. it's a complicated form of 1 but we are still multiplying by 1 and so we're not changing the value all right so to kick this off 36 times the quantity a plus 3 Multiplied by a plus 3 over 4 we could see that 36 would cancel with 4 and give us a 9. so essentially i have 9 times the quantity a plus 3 times the quantity a plus 3 again which i can just simply

write as 9 times the quantity a plus three and this guy is squared right or if you want to write it out you could do a plus three that quantity times a plus three does not matter okay then over Now this is going to distribute to each term so it's going to start out right there and what's going to happen is in that particular case this a plus 3 is going to cancel with this a plus 3 and then 36 times 4 is going to give me 144. then i have a minus so i have

this minus sign now i'm going to use my distributive property for this term And let me erase this part right here in this case the 36s are going to cancel and i'll have a plus 3 times a plus 3. so those two quantities again i can write a plus 3 of that quantity squared or i can write it out a plus 3 times a plus 3. okay so now what we want to do is just simplify so we've gotten to that stage now and again i want to talk about a common Mistake i've talked about

this in almost every video since i've kind of introduced this but i want you to understand this is multiplication here in the numerator i'm multiplying 9 times the quantity a plus three times the quantity a plus three so i'm multiplying those three factors together i don't have straight multiplication in the denominator right i have a subtraction sign here i have 144 that is subtracting away The product of a plus 3 times a plus 3. so these are factors but this 144 is not a factor of anything right it's subtracting away the result of this multiplication

again if i don't have straight multiplication in both the numerator and the denominator i can't go through and just cancel stuff right i see students all the time it'll say okay well i can cancel this with this and this with this and then okay well you know 144 divided By 9 is 16 so this with this and put a 16. you cannot do that if you had and let me just change this scenario up a little bit give you a different example something like 144 times the quantity a plus 3 times the quantity plus 3.

now these are all factors it's just straight multiplication here so in this case i can go through and cancel this with this and this with this And this with this and get 16 i'm going to end up with 1 16 there but that's not the situation we are presented with and so again we've got to act according to what we receive so this is a minus sign here 144 is subtracting away the result of the multiplication here so please understand that you can't go through and just cancel there okay so now that we've kind of

settled that what could we do to simplify here Well in the numerator i'm just going to keep that in factored form in the denominator i want to go through and use my special products formula to get the result for this i'm going to subtract 144 minus that result and then i'm going to try to factor that guy and see if i can then cancel with something in the numerator and i know this gets really really tedious but it's just something we have to do all right So i'm just going to copy the numerator 9 times

the quantity a plus 3 times the quantity a plus 3 and then over so the formula for this a times a is a squared then the middle term i know that i'd have 2 times a times 3 so that's plus 6a and then the final would be 3 squared or 9. now here's the kicker i am subtracting Away the result of this okay so that means i'm subtracting each and every term away or i'm subtracting the whole thing away so one good way is to put this inside of parentheses and then when you drop the

parentheses you're going to change the sign of each term so i'd have negative a squared minus 6a minus 9 and don't forget that you have the 144 out in front and then what i'm going to do here is i'm going to subtract 144 minus 9 that's Going to give me positive 135. so i'm going to write this as 9 times the quantity a plus 3 times the quantity a plus 3 again and then over we have negative a squared then minus 6a and again 144 minus 9 is plus 135. now the question would be can

i factor this and then cancel one of those factors with something up here well i'm going to Tell you that you can factor it but nothing's going to cancel we're going to go through and factor it anyway so the first thing i'm going to do to make this easy for you i'm going to factor out a negative 1. so the way that i do that is i just put a negative 1 out in front of parentheses and i change the sign of each term inside so this was negative now it's positive this was negative now

it's going to be positive and this was Positive now it's going to be negative so now it's very easy for me to factor this because i have a leading coefficient on a squared that's positive one and that's what we're used to and so give me two integers whose sum is positive 6 and whose product is negative 135. so if i think about 135 pretty easy it's what 27 times 5 27 is 9 times 3 9 is 3 times 3 and so if i think about possible factors the one that jumps out right away is what

15 and not right 15 and 9. so if i did positive 15 and negative 9 so plus fifteen and negative nine the sum is going to be positive six the product is going to be negative 135. so let's just erase this and this we'll just kind of drag this up here and what you're going to see is that Nothing is actually going to cancel here i don't have any common factors a plus 3 that quantity a plus 3 that quantity and 9 are my factors in the numerator negative 1 a plus 15 and a minus

9 are my factors in the denominator there's nothing i can cancel so for most teachers you would just report your answer in this format for others they want you to go through and kind of multiply everything together so we can do that real quick So 9 is going to be multiplied by the result of this so we know that this is a squared plus 6a plus 9. then if i do the multiplication let me kind of just do this down here that would be 9a squared plus 54a plus 81. okay so that's going to be

my numerator then this is over Down here we just factored this so we know what it is it's negative 1 times again this would be a squared plus 6a and then minus 135 and then a few things you can do you can write the denominator as a squared plus 6a minus 135 and just put this negative out in front you'll see that in a lot of Answers or you could just make each term the opposite of what it was so you could say okay this is negative a squared and then minus 6a and then plus

135. there's quite a few different ways you could write this it's just a matter of what your teacher wants from you all right let's take a look at another one we have 36 over x squared minus x squared over 6 and then this is over 6. now don't get confused with this one this is your numerator for the complex fraction This is your denominator okay the denominator does not have to have a fraction involved it can just be a number six but in order to not confuse yourself here go ahead and write it as 6

over 1. so now it's clear when i'm thinking about 6 over 1 the denominator is a 1. and then for 36 over x squared the denominator is x squared and then for this guy x squared over 6 the denominator is a 6. So what would the lcd be we'll just be 6x squared so then i can erase all this and just say okay i'm going to multiply this by 6x squared over 6x squared so if i do this 6x squared times 36 over x squared the x squareds would cancel this would cancel with this and

i'd have 6 times 36 which is 216. then minus we kind of erase this real quick You'd have 6x squared times x squared over 6. the sixes would cancel you'd have x to the fourth power so x to the fourth power and this is over if i have 6 times 6x squared that's just 36x squared so is there anything that i can do to factor this guy well if 216 was a perfect square i could use the difference of two squares there but it's Not so it's not really anything i can do a lot of

you will like to flip your answer around and say okay well this is negative x to the fourth power plus 216 over 36 x squared that's fine either way is acceptable all right let's take a look at one final problem i think that all of us can agree that this is a much faster method just have to be able to understand it again it doesn't take long to understand you have To be able to identify what denominator you're looking at to produce your lcd so for this one we have y plus one over x minus

two minus x plus two over x minus two and then this is over y plus one over y plus two minus x plus two over y plus two so what i'm looking at here is an lcd that's what the denominators here are the same x minus 2 In each case the denominators here are the same y plus 2 in each case so all i can do is just multiply the denominators together i can't do any better than that but again notice how in the numerator of the complex fraction i'm looking at this denominator this denominator

in the denominator of the complex fraction i'm looking at this denominator and this denominator so again you're looking at all the Denominators involved of that complex fraction so this one this one this one and this one all right so once we have our lcd going all we need to do is multiply the numerator denominator the complex fraction by that lcd so x minus 2 that quantity times y plus 2. now this is pretty simple to do because in this case when i multiply here in each case the x minus 2 is going to cancel and

i'm just going to be left with the y Plus 2 times the numerator in other words this is going to cancel with that in each case so y plus 2 times y plus 1. let's write that out and i'm going to leave it in factored form for now and then i'm subtracting away and you can put that inside of brackets or just remember that you're subtracting the whole thing away you have y plus two Times x plus two right those two quantities all right then this is over for the next one in a similar way

the y plus two is going to cancel here and here right when i do the multiplication to each so i'm going to be left with x minus 2 times y plus 1 and then minus In this case i'll have x plus 2 and then times x minus 2. okay so now once we get in this format a lot of you will stop and say well there's nothing i can cancel that would be wrong and then a lot of you will incorrectly say well i can cancel you know between the numerator denominator i can go this

with this again you can't do that because we have subtraction involved here If it's straight multiplication if all you have in your numerator and denominator is multiplication cancel away but if we have subtraction or addition involved you know and i know that you're seeing the subtraction here and here and the addition here and here and you're saying well how is that different it is different because these are factors right this whole thing is multiplied by This whole thing this subtraction sign is separating these two there's no multiplication involved in that right this is a factor

and this is a factor so that's how you got to look at it and again this is one of the most common mistakes that i see and so that's why i talk about it in each and every video so that at some point you would stop and say okay i can't cancel this with this no big no no Okay one of the things i want you to look at is that in the numerator here i have a common binomial factor of y plus 2. in the denominator i have a common binomial factor of x minus

2. so what happens if we pull that out if i pulled it out so in this case i have a y plus 2 out in front so inside of let's say brackets i would have what's left which is y plus one that quantity Minus the quantity x plus two now this is multiplication this is a factor and this is a factor this whole thing here let me just highlight it the whole thing is a factor now this is going to be over if i pull out the x minus 2 here inside of brackets i'm going

to have y plus 1 That quantity minus over here i'd have x plus 2. now we can cancel between the numerator denominator here because we are canceling common factors again let me kind of show this this is a factor and this is a factor okay so we don't have any subtraction here if i had this minus this and this plus this or this Minus this i can't cancel right because there's addition and subtraction involved i don't have that what i have is multiplication i have two factors here that are being multiplied i have two factors

here that are being multiplied so i'm going to cancel this entire factor that consists of the quantity y plus one minus the quantity x plus two with this factor again of the quantity y plus one Minus the quantity x plus two this over this is equal to one they can be cancelled so what i'm going to be left with here once i cancel that out it's just going to be a y plus 2 over an x minus 2. and again i know that when you first start doing this particularly in algebra 1 it's kind of

the first time you're introduced to these concepts it can be very confusing What you can cancel and what you can't again if you're only looking at multiplication so in other words here this quantity y plus two is multiplied by this quantity which inside has the quantity y plus one minus the quantity x plus two this is multiplication you just have factors here okay same thing down here you just have factors here now within those factors you do have addition and Subtraction i know that's what makes it confusing but again again again this thing right here

is a factor which can be cancelled with the same factor down here so that leaves me with what's left which is y plus 2 over x minus 2. hello and welcome to algebra 1 lesson 48 in this video we're going to learn about solving equations with rational expressions so before we kind of jump in and start talking about solving an equation with rational expressions i Just want to refresh everybody's memory with something that we learned at the very beginning of algebra one so i want you to recall that when we solve an equation with fractions

we can clear the fractions by multiplying both sides by the lcd so as a quick example let's say we saw something like four-fifths x plus seven thirds x and this is equal to ninety-four well if i didn't want to work with the Fractions of course i could work with the fractions if i wanted to but if i didn't want to work with them i could multiply both sides of the equation by the lcd and it would clear the fractions for me so in this particular case i have a denominator of 5 and a denominator of

3. neither one of those is going to factor so the lcd would be 5 times 3 or 15. so i would multiply this side of the Equation by 15 and to make it legal i would multiply this side of the equation by 15. remember your multiplication property of equality i can multiply both sides of the equation by the same non-zero value and i'm not going to change the solution so if i multiply 15 times four-fifths the 15 would cancel with the 5 and give me a 3 3 times 4 is 12. so this would be

12x and then let me erase this real quick we Would do that again over here so plus the 15 would cancel with 3 and give me a 5. 5 times 7 is 35 so it'd be 35x and this equals we'd have 94 times 15 which would be 1410. all right so the first thing i notice now is that i don't have any fractions anymore so now i just solve it as i normally would 12x plus 35x would give me 47x so this is 47x this is equal to 1410 divide both sides of the equation by

47 and i'll get that x is equal to 30. now in almost every occasion i tell you to check the work to make sure it's accurate i've already checked this i know that 30 is the correct solution but if you want to you can pause the video and go ahead and check it in the interest of time i'm going to skip that for now and just going to kind of get to the main content of the video all right So now we want to talk about solving an equation with rational expressions and the reason i

gave you that example is because we're going to use a similar technique here i want you to multiply each side of the equation by the lcd the lcd to clear the denominators okay so we're going to do that in this case just like we did in the previous example the next thing you're going to do is you're going to solve the equation and then we're going to check the Solution by substitution just like i always tell you to do but here's the key thing with a rational expression when you have variables in your denominator remember

we cannot divide by zero so i have this statement here we must reject any solution that results in a denominator of zero very very important that you understand that okay if you want to look at the denominators before you kind of get Going and say hey these are the restricted values do that that way if you get that as an answer you can just reject it immediately or if you want to go back and plug in see what results in zero after you've done the problem that's fine too all right so let's take a look

at the first example here we have one over three x and this is equal to x minus five over two x squared and then minus three x plus three over Two x squared so the first thing i'm going to do is i'm going to figure out what the lcd is okay what is my lcd so in this particular case i have a denominator that's 2x squared 2x squared and 3x so if i break this down i know this is 2 times x times x i know this is 2 times x times x and i know

this is 3 times x So again we know how to get an lcd at this point we've watched several videos where we've done this and essentially all i'm going to do is look for the largest number of repeats for any duplicate factor everything else just goes in so for x because it appears in everything i just go with the highest exponent which is x squared then for the number parts i have a 2 here and a 2 here it only occurs once in each so that's going to go in there And then i have a

3 here only occurs one time there so i'm going to throw that in there so i would have 3 times 2x squared or what we'd say as 6x squared for my lcd okay let me move this up here so the next thing that i want to do i want to multiply both sides of the equation by the lcd so i'm going to have 1 over 3x and i'm going to multiply this by 6x squared and this is equal to i'm going to put Parentheses around this x minus 5 over 2x squared minus 3x plus 3

again over 2x squared and i'm multiplying this guy by 6x squared so let me scroll down and get a little room we'll come back up to the original equation in a little while so for this one over here it's easy to see i can cancel this 6 with this 3 and get a 2. i can cancel one factor of x up here with this x here and so what i'm Going to be left with is just a 2x right here and this is going to be equal to i'm going to use my distributive property here

6x squared over 2x squared the 6 would cancel with the 2 and give me a 3 and the x squareds would cancel so essentially i would have 3 times this quantity x minus 5. so 3 times the quantity x minus 5 and then minus i'm going to have 6x squared times this Guy right here again it's going to be the same thing the x squareds are going to cancel and 6 over 2 is 3. so i'm going to have 3 times the quantity 3x plus 3. okay so moving on now i just want to do

some simplification i have 2x is equal to 3 times x is 3x 3 times negative 5 is minus 15. now i have a negative 3 here okay so pay close attention i'm going to distribute the negative and the 3 To everything so negative 3 times 3x is negative 9x negative 3 times 3 is negative 9. so let's simplify now i've got my 2x is equal to 3x minus 9x is negative 6x negative 15 minus 9 is going to give me negative 24. and now what i want to do is get all the x's on one

side and the numbers on the other right just like we've been doing since we started so i'm going to add 6x To both sides of the equation so i'm going to have 2x plus 6x which is 8x this is going to cancel and this is equal to negative 24. we can then divide both sides of the equation by 8 and we're going to see that x is equal to negative 3. so let me erase everything and i'm going to drag this answer back up to the top and i'm going to check to see that this

is a valid solution okay so what i'm Going to do is plug in a negative 3 for each x and so i would have 1 over 3 times negative 3 which would be negative 9. we can go ahead and just do that we'll put negative 1 9. this is equal to plugging in a negative 3 here so negative 3 minus 5 would be negative 8 over plug in a negative 3 there now it's going to be squared i'm plugging it in for x So negative 3 the negative and the 3 squared is 9. 9 times

2 is then 18. so i'm going to put 18 here and then minus 3 times negative 3 is negative 9 negative 9 plus 3 is negative 6. and then over again same thing here if i take negative 3 and i square it i get 9 then 9 times 2 is 18. so i have minus a negative that's plus a positive So that's the only thing i want to do there and so i want to just check this now i see that plugging in a negative 3 did not give me a denominator of 0. so we're

good to go there now we're just checking to see if the left and the right side are equal so negative 8 plus 6 is going to give me negative 2. so this would be negative 2 over 18 so i get negative 1 9 is equal to negative 2 over 18 and this would be negative 1 9. if i simplify and Divide numerator denominator by 2 i would have negative 1 over 9 is equal to negative 1 over 9. so our solution here which again was x equals negative 3 is correct all right for the next

one let's look at 2 over 5n plus 6 over n squared is equal to 2 over n squared so what is the lcd here what would that be well we look at the fact that i only have a number of 5 no other number in Any of the denominators so that goes in there and then each has an n this one is n squared n squared and n so we'll go with the highest exponent on the variable n and that's a 2 right so we just put n squared in there so now that's done we

want to multiply both sides of the equation by the lcd so let me scooch this up or actually i'm going to put it up here make it nice and out of the way and i would have 2 over 5n plus 6 over n squared this is going to be multiplied by 5n squared and this is equal to 2 over n squared times 5 n squared now if you wanted to take the route of seeing what couldn't my variable be equal to before i start i could do that i could say okay well i have an

n here and n squared and an n squared obviously n cannot be zero right it can be anything else other than zero If i get a value of n equals zero i've got to reject that solution but just as valid i could have not figured that out here figured out the problem and then came back up and plugged it in and say okay if i get an net equals 0 then i've got to reject that solution okay so let's go ahead and multiply so use my distributive property 5n squared times 2 over 5n the 5s

would cancel And one factor of n here would cancel with this n and essentially i'd have n times 2 which is just 2n then plus next i would do 5n squared times 6 over n squared and the n squareds are going to cancel and essentially i'm going to have 5 times 6 which is 30. now over here i can just cancel this n squared with this n squared 5 times 2 is 10. so this equals 10. and now we have a super simple equation To solve i subtract 30 away from each side of the equation

i get 2n is equal to negative 20. so i divide both sides of the equation by 2 and i get that n is equal to negative 10. okay so that is not 0 so i know we don't have to reject the solution but we still want to check it to make sure that we got the right answer so important that we do that when we start out okay so Again n is equal to negative 10 so i'm looking at 2 over 5 times negative 10 just plug in a negative 10 for n plus 6 over

if i square negative 10 i get 100 and this equals 2 over again if i square negative 10 i get 100. so over here if i multiply 5 times negative 10 i would have negative 50. now i need to get a common denominator going so i'm going to do two things i'm going to move this negative up into The numerator remember that's perfectly legal and then i'm going to multiply both numerator and denominator by 2. and so this is going to be negative 4 over 100 negative 4 over 100 so negative 4 plus 6 is

2 so this side would be 2 over 100 and this side is 2 over 100 you could Simplify each side of 1 over 50 if you wanted to but you're just trying to see that the left and the right side are equal which they are so again you can say that your solution here n equals negative 10 is correct all right for the next one i'm looking at 1 over v plus 6 minus 2 over v squared plus 3 v minus 18 this is equal to 2 over v minus 3. all right so this one's

going to be a little bit more time consuming but still Not very hard overall the lcd here is the first thing we want to find so the lcd what is that well for this denominator i have v plus six for this one i have v minus three it's pretty straightforward for this one i need to factor to see what's going on now when you first start getting these problems particularly in algebra 1 you're going to be able to guess that one of these at least is going to be a Factor of this in this particular

case it's going to factor perfectly to give you those two so give me two integers whose sum is 3 and whose product is negative 18. well of course that's going to be positive 6 and negative 3. so you'd have v plus 6 and v minus three so your lcd is just that it's v plus six which occurs here and here and v minus three which occurs here and here so once you have that the v plus six Times v minus 3 you want to multiply it by each side of the equation because of the length

of my screen i don't think i could possibly fit all of this in there so without having to scroll back and forth what i'm going to do i'm just going to say that i'm going to be multiplying this by this side of the equation so i'm going to multiply by this side and i'm also going to multiply it by that side but i'm just going to do the work down Here and then just report the simplified result so if i had v plus six times v minus three and i multiply this by one over v

plus six obviously the v plus six would cancel here and here and i just be left with v minus three so for the first one let me kind of scroll down i would have just a v minus three Okay let me scroll back up now and i'll just be going back and forth so this was v minus three here so now i'm going to multiply by this guy here now the denominator here is exactly this so we know that without even going through this the denominator here is going to cancel with this right here i'm

just going to be left with a minus 2 right so if i was to write this over here i would simply put minus 2 now then equals I would be on this side now and i would be multiplying this guy right here by 2 over v minus three so v minus three cancels with v minus three and i've got two times the quantity v plus six so two times the quantity v plus six okay so let's simplify each side and go ahead and solve so negative 3 minus 2 is minus 5. so v minus 5

Is equal to 2 times v is 2v and then 2 times 6 is 12. if i subtract 2v away from each side of the equation and add 5 to both sides of the equation what is that going to give me this is going to cancel and this is going to cancel on the left side i'm going to have negative v on the right side i'm going to have 17. so the opposite of v is 17 that means v is negative 17 right i can show that by Multiplying both sides by negative 1 or dividing both

sides by negative 1 whatever you want to do that gives me v is equal to negative 17. okay let's erase everything and then we're going to go back and check and i'm going to use the factored denominator to check it's just going to be easier for me so 1 over you'd have negative 17 plus 6 minus 2 over you have again V plus 6 so that's negative 17 plus 6 and that's multiplied by v minus 3. so negative 17 minus 3. these two quantities are multiplied together this is multiplication so make sure you don't just

write it as one long string because it's easy to just kind of substitute it in that way and not realize that it's multiplication and make a mistake so it's this times this then this equals 2 over You've got negative 17 minus 3. so let's crank this out real quick negative 17 plus 6 is negative 11. so i've got 1 over negative 11 or i can write negative 1 over 11 does not matter and then we're subtracting away we have 2 over negative 17 plus 6 is negative 11. negative 17 minus 3 is negative 20. so

if i multiply these two together i know i'm going to get a positive so what is 2 times 11 well that's 22. Then just put your 0 at the end the trailing zero that gives you 220. now this is going to be equal to 2 over negative 17 minus 3 so that's 2 over negative 20 or again i could write this as negative 2 over 20 doesn't matter so now what i want to do is just go through and simplify as much as i can so i need to multiply this by 20 over 20 so

i can have a common denominator so 20 times negative 1 is negative 20. So i have negative 20 minus 2 over a common denominator of 220. so this would be negative 22 over 220. and what would that reduce to well each is divisible by 22. 22 divided by 22 is 1. so this would be negative 1 and 220 divided by 22 is 10. so this is negative 1 10. so let me write that right there this is equal to if i divide this by 2 i get 1. if i divide this by 2 i get

10. So again negative 1 10 and let me write it down here for the sake of completeness we get negative 1 10 is equal to negative 1 10. so our solution which we have down here v equals negative 17 is correct all right for the next one we're going to take a look at negative 2 over x plus 5 plus 3 over x minus 5 this is equal to 20 over x squared minus 25. so for the lcd you can see here that x squared minus 25 Is what it's the difference of two squares it's

x plus 5 times x minus 5 and this matches what you have here and here so the lcd is just this the quantity x plus 5 times the quantity x minus 5. so if i multiplied let me just write that the lcd is that [Music] if i multiply both sides of the equation by this what would happen well if i multiply this part right here by the lcd The x plus 5 would cancel with the x plus 5 here right so what i'd have is a negative 2 a negative 2 times the quantity x minus

5 then plus in this particular case when i multiply this lcd by this the x minus 5 is going to cancel and it'll cancel here so i'd have 3 times the quantity x plus 5. in this particular case once i get past the equal sign over here I have x plus 5 times x minus 5. so this is going to cancel completely with this right this denominator would just cancel itself out and i'll simply just have 20. so now to solve this guy negative 2 times x is negative 2x negative 2 times negative 5 is

plus ten three times x is plus three x three times five is fifteen and this equals twenty now to simplify negative two x plus three x is x ten plus fifteen is twenty-five And this equals 20 subtract 25 away from each side of the equation and i get that x is equal to negative 5. so some of you will already see a problem let's go back up and substitute it in before we say anything let's go back up here let's erase if x is equal to negative 5 what's going to happen here when i plug

a negative 5 in for x here and here i'm going to have a problem Right if i plug in a negative 5 for x there i would have negative 5 plus 5 that equals 0. division by 0 is not allowed and if we get a result like that we have to reject the solution we reject x equals negative 5. it is not a valid solution if i plugged it in here negative 5 that quantity squared would be 25 25 minus 25 is 0. can't have that you could plug it in here negative 5 minus 5

is negative 10 but that doesn't Matter if it fails in one of the denominators you have to reject it you can never divide by zero so that's what you're looking for here you reject x equals negative five let me just put a big you know x through it and say it's not a solution and so we don't have a solution here we just say that there's no solution there's no solution [Music] All right let's take a look at one more problem and we're going to wrap this up again very very easy to do this you

just find the lcd multiply both sides of your equation by the lcd and then you just solve the equation like you normally would the only thing is you can't have division by zero so you've got to check that result so you don't get something like we did in the last problem so to start this 1 over 3v plus 15 minus 1 over v squared plus 5v is equal To 1 3. so again i'm looking for the lcd and for this guy i've got to factor it it can factor into three times the quantity v plus

five for this guy i can factor out a v and i'd have v plus five and for this guy i just have a three so it looks like i have a three here and here so that goes in the lcd i have a v and then i have a v plus five so my lcd is three v times the quantity V plus five let's erase all this i think i might be able to fit this on the screen so let's try that so i would have 3v times the quantity v plus 5 times 1 over

3v plus 15 minus 1 over v squared plus five v and then equals and let me just scroll down and get a little room going we have one third There so one third times three v times the quantity v plus five okay so we fit that on the screen now what i'm going to do here is i'm going to be using my distributive property and i'm going to distribute this to this so if i have 3v plus 15 again that's 3 times the quantity v plus 5 if i factor it the v plus 5 Would

cancel with this and the 3 would cancel with this so essentially i have v times 1 which is just v now let me kind of erase what i canceled there and do this again we're going to use our distributive property over here now in factored form this is v times the quantity v plus 5. so in this particular case v v is going to cancel and the v plus 5 is going to cancel So that's gone and what i'm left with is 3 times 1 or just minus 3. so then minus 3. and then over

here if i look only a 3 is going to cancel so i'm left with equals v times the quantity v plus 5. okay so we've got it in kind of a simple form won't take us very long to solve this now we have v minus 3 is equal to v times v is v squared plus v times 5 is 5v Okay so now what i want to do i want to subtract v away from each side of the equation and i want to add 3 to both sides of the equation so that's going to cancel

itself out so i'm going to have 0 is equal to v squared plus 4v plus 3. now we have this set up as a quadratic equation We cannot solve quadratic equations that we can't factor because i haven't taught you how to do it yet unless you know how to do it from outside of my course there is something called the quadratic formula and completing the square that we're going to get to at the very end of algebra 1. so until we get to that point unless we can factor this we cannot solve it but fortunately

i gave you a problem that you could do So this is going to factor into what two integers whose sum is four and whose product is three that's three and one so i would get that v plus three is equal to zero or v plus one is equal to zero so solving that i just subtract three away from each side here i get v is equal to negative 3 and then over here i subtract 1 away i get v is equal to negative 1. so here are my two Proposed solutions let's go back up and

check them okay so let's check each one separately and so i'm going to do 1 over let's start with negative 3. so 3 times negative 3 is negative 9 and negative 9 plus 15 is 6 then minus i've got 1 over v squared negative 3 squared would be 9 and then 5v 5 times negative 3 is negative 15. so 9 minus 15 would be negative 6. So this would be minus a negative 1 6 so this is plus positive 1 6. so essentially we could write this as what 2 over 6 which is one-third so

this is one-third and this says one-third over here so this one checks itself out this is good to go so now let's try negative 1. so 3 times negative 1 is negative 3. negative 3 plus 15 is 12. so this is 1 Over 12 and then minus i'm going to have 1 over v squared negative 1 squared is 1. so this would be 1. 5 times negative 1 is negative 5. so 1 minus 5 is negative 4. so i have minus a negative 1 4. now this is plus a positive and i could change this

to multiplying by 3 over 3. so this is going to be 3 12. So three twelfths now if i add one and three i get four so this is four twelfths and what does four twelfths reduce to it reduces to one third right four divided by four is one twelve divided by 4 is 3 so i get 1 3 equals 1 3 from over there so again we've found the big winner so v can equal negative 3 or v can equal negative 1. Hello and welcome to algebra 1 lesson 49. in this video we're going

to learn about applications of rational expressions so in this video basically what we're going to be doing is looking at some word problems that require equations with rational expressions now the two most common problems in this arena are going to be motion word problems which we saw back at the beginning of algebra 1 right use your distance formula there distance is equal To rate of speed times the amount of time traveled and then something that's new to us is going to be these work rate problems or some people say rate of work problems so these

are really really easy to solve we're going to get to them at the very end of the lesson all right so we're going to begin today with a motion word problem so i want you to recall that we use the distance formula let me just highlight that real quick The distance formula for this type of problem so we have d is equal to r times t remember this d stands for distance the r stands for rate of speed and t stands for time traveled so something that's going to be new for us here is that

we're going to manipulate the distance formula and we're going to be solving it for r or t so for example if i have Distance is equal to rate times time and i want to solve this for the rate let me highlight that how would i go about doing that well all we're going to do is realize that if t is multiplying r to get r by itself i just divide both sides of the equation by t and what's going to happen is the t is going to cancel here and here and on the right side

of the equation i'm just left with r it's isolated and so what i have Is i have distance over time is equal to rate now let's check that real quick and make sure that that makes sense we know from the traditional distance formula that if i am going let's say 40 miles per hour and i do this for 3 hours 40 times 3 is 120 i've traveled 120 miles okay so that makes sense If i plug in the same numbers here is that going to work so if i plugged in 120 here and i plugged

in a 3 here would i get 40 here well yes i would 120 divided by 3 is in fact 40. right so we're good to go there now the other thing we might have to solve for is time so if i have distance is equal to rate times time i'm going to use the same logic to solve for time i want to isolate this guy so i'm going to divide both sides of the equation by r So i can isolate the t let's cancel this with this and i'm going to be left with time is

equal to distance over rate and again you can say okay using this example if i plugged in 120 here and i plugged in a 40 there 120 divided by 40 is in fact three so we're good to go all right so let's go ahead and take a look at the first problem So the local river has a current of three miles per hour a boat takes the same amount of time to go 24 miles downstream okay that's going with the current in case you don't know that as it does to go 12 miles upstream upstream

is against the current what is the speed of the boat in still water so it might seem like this is a difficult problem but it's really not Let's go ahead and make a little table to kind of organize our information the two things we're going to think about is the scenario where we're going downstream again that's with the current and the scenario where we're going to go upstream and that's against the current let's scroll down here real quick so we'll kind of think about okay we have upstream and we have downstream okay so let me

just draw a little line Here now the components of the distance formula let's think about those so we have distance so i'm just going to abbreviate with a d we have the rate of speed which i'm going to use an r for and then we have the time so what i want to do here is just fill in this table and get the information that i'm going To need to create an equation that's going to allow me to solve and figure out an answer for this problem let's scroll back up to the top real quick

all right so it tells us in the problem that the boat is going 24 miles downstream in the same amount of time as it does to go 12 miles upstream so for the distance i'm going to put 24 miles under downstream and 12 miles under upstream okay so i'm going to Write 12 here and 24 here again the distance when i'm going upstream is 12 miles the distance when i'm going downstream is 24 miles now let's think about the rate so when we think about the rate of speed again this is how fast we're traveling

the only information we're really given is that the local river has a current of 30 miles per hour doesn't give us any other information Now what you're supposed to know here is that if you're going upstream you are fighting the current right so whatever the amount of speed you're going in miles per hour like if you're going in still water you've got to subtract three away from that because you're fighting the current so if i let a variable like x let's say let x equal the speed of the boat In stillwater what would happen is

the speed when i am going upstream would be x the speed of the boat in stillwater minus three because that's the speed of the current and that's what i'm fighting so let's scroll down and put that so again for the upstream speed it's going to be x the speed of the boat in stillwater minus three the speed of the current and then for the downstream speed it's just As easy right we think about the speed of the boat in still water is x and then now if i'm going downstream that means i'm going with the

current so the current is pushing on my boat it's making me go faster by three miles an hour so i would add three to that x so x plus three is the speed of the boat going downstream x minus three is the speed of the boat going upstream now how do i go about getting a time well reading back through The problem the only thing it really says about the time is that it takes the same amount of time to go 24 miles downstream as it does to go 12 miles upstream so what i can

do here as we talked about a little while ago is i can get a time for each one and then i can set those times equal to each other so let's take that one step at a time to get a time remember the d Equals r times t can be solved for t so if i divide both sides of the equation by r to isolate t as we did earlier this cancels with this and i've got t by itself so time is equal to distance over rate i have a distance here i have a right

here so i'm going to have a time the distance in this case is 12. the rate in this case is x minus 3. so my time for going upstream is 12 Over x minus 3. now for downstream i can do the same exact thing my time is going to be 24 which is my distance over x plus 3 which is my rate now that we have all this information figured out we want to go ahead and set up an equation and as i told you when we look at this problem it tells us that a

boat takes the same amount of time the same amount of time to go 24 miles downstream as it does to go 12 miles upstream so that means i could set the Two times equal to each other and i could solve for x and i'd figure out what the speed of the boat is in still water let me erase this real quick so i have the time for upstream which is 12 over x minus three and i'm going to set this equal to the time for downstream which is 24 over x plus 3. so in the

last section we learned how to solve equations of this form so one way we can Do this is we can find the lcd and multiply both sides of the equation by the lcd clear the equation of any denominators and kind of go from there but in this particular scenario the easiest thing to do would be to cross multiply remember that from proportions right i can multiply x minus 3 that quantity times 24 and i can multiply x plus 3 that quantity times 12. and why is that a shortcut well the lcd here Is going to

be what it's x minus 3 times x plus 3. so if i multiply both sides of the equation by this lcd well what's going to happen is this denominator here is going to cancel and i'm going to multiply 24 by x minus 3 which is exactly what i'm doing when i'm cross multiplying right this would be 24 times the quantity x minus 3 is equal to same thing over here the x minus 3 would cancel and i'd be multiplying 12 by x plus 3. that's what i'm doing when I'm cross multiplying so 12 times the

quantity x plus 3. so that's a little shortcut for you to kind of speed up the process if you get this scenario let's erase this real quick and so 24 times x is 24x and then 24 times negative 3 is going to give me negative 72 and this is equal to over here we have 12 times x that's 12x and then we have 12 times 3 that's 36. Okay so let's solve this equation very very easy to do i'm just going to add 72 to both sides of the equation and i'm going to subtract 12x

from both sides of the equation so this is going to cancel and this is going to cancel 24x minus 12x is 12x so this is 12x and this is equal to 36 plus 72 which is 108. okay as my last step here i'm going to divide both sides of the equation by 12 And this is going to cancel with this i'm going to have that x is equal to 9. so i've found the answer to the problem x again was the speed of the boat in still water and that's what i was trying to figure

out so x is equal to nine so the speed of the boat in still water is nine miles per hour so let's go back up to the top and write that so i'm just going to write that the boat travels Nine miles per hour in still water and that answers our question there what is the speed of the boat in stillwater so how would we check something like this well let's read back through the problem real quick so the local river has a current of three miles per hour a boat takes the same amount of

time to go 24 miles downstream again with the Current as it does to go 12 miles upstream again against the current so what is the speed of the boat in stillwater so i know at this point that upstream my rate of speed is going to be what well x is 9. so i plug in a 9 there 9 minus 3 is 6. so in other words i know that the time that it takes to go 12 miles upstream should be 12 over 9 minus 3 or 12 over 6 which is 2. And is that equal

to the amount of time that it takes to go downstream well my speed downstream if i replace this x with a 9 9 plus 3 is 12 is 12 miles an hour you basically have 24 over again 9 plus 3 is 12 so 24 over 12 would be two so it takes the same amount of time which is two hours to go upstream 12 miles as it does to go downstream 24 miles so we have the correct answer here right the speed of the boat in still water is nine Miles per hour so the next

type of problem we're going to deal with is known as a work rate problem or some of you will hear called a rate of work problem either way these type of problems involve how much time it takes to complete a job so in this scenario working alone paul can pour a large concrete driveway in five hours natalie can pour the same driveway in seven hours How long would it take to complete this job if they worked together so when you first start thinking about it if you don't get any direction on it it's kind of

hard to figure out how to solve this i know how long it would take for paul to pour a driveway it's five hours says it right there i know how long it would take for natalie to pour the same driveway at seven hours But if i combine their work rates and they work together how long is it going to take well the first thing i have to do is break this down into one unit of time in this particular case we're talking about hours now for these problems you might talk about days or weeks or

minutes whatever it is you're going to break it down into one of those units of time okay in one hour because i want one unit of whatever time I'm dealing with i'm dealing with hours so i want one of those what can paul do in one hour well paul completes the job in five hours so in one hour he's done one-fifth of the job one-fifth of you could say the driveway or of just the job just to be generic then natalie she's a little bit slower it takes her Seven hours to complete the job so

in one hour she's one seventh complete so one-seventh of the job or again you could say of the driveway so what i want to do next is sum the individual contributions from these two people if i do that i'm going to get the amount of the job that is completed by both of them in a one hour period again working together so what i want to do here is do one-fifth That's the contribution from paul plus one-seventh that's the contribution from natalie so i would get a common denominator here this would be times five over

five this would be times seven over seven and this would be what seven plus five over seven times five is thirty-five so seven plus five is twelve so this would be twelve Over 35 here and what 12 over 35 is that's the contribution again by both of them in a one hour period so let me just kind of scroll this down here and just say 12 35 is the amount of the job completed in one hour i'm going to let x here be equal to The number of hours to complete the job if they work

together we have 12 over 35 or 12 35 is the amount of the job completed in one hour if i multiply this by x which is the number of hours the again number of hours to complete the job I'm going to get one okay and the reason we use one is it's one completed job now to make this make sense for you because i know when you start dealing with this it's like why would you set it equal to one so on and so forth let me just scroll down real quick and give you a

very easy scenario let's say me and you paint a room in four hours working together so that's very very simple so in one hour we would paint a fourth of The room if i didn't know how many hours it would take let's say i multiplied that by x and then i filled in that with a four well i know that four times one fourth is equal to one it's because we're completing one room or one job however you want to think about that so that's our goal is to set this equal to 1 because we

are trying to see how long it's going to take to complete one job and now all we need to Do is solve this very simple equation if i have twelve thirty-fifths x is equal to one all i need to do is multiply both sides by the reciprocal multiply this by thirty-five twelfths and this by thirty-five twelfths and i'm gonna get one this is to cancel and i'll have x is equal to 35 12. now let's go back up to the top and make sense of this so i'm just going to answer it first so how

long would it Take for them to complete this job if they work together it would take it would take 35 12 hours to complete the job and you can do the division on that it's not going to give you an exact value but it will be 2.91 with a 6 that repeats forever so you could say about you know Approximately 2.92 hours if you wanted to just give an estimate but if you wanted to be precise which it's good to do on a test i would just write that fraction that you got which is 35

12 hours now let's make sure that this makes sense so again working alone paul can pour a large concrete driveway in five hours natalie can pour the same driveway in seven hours how long would it take for them to complete this job if They work together so what i'm going to do here is i'm going to take 35 12 35 12 which is the number of hours they work for i'm going to multiply this by the rate of speed that paul works at which is one-fifth and then plus i'm going to do 35 12 times

the rate of speed that natalie works at which is one-seventh and let's see if this equals one right one Completed job so this divided by this would give me seven so i'd have seven over twelve so that means that in this amount of time 35 12 hours paul completes seven twelfths of that driveway he completes seven twelfths of that job you might wanna say and then natalie's contribution is less because she works at a slower rate this divided by this would give me five She does five twelfths of the job so in this amount of

time paul contributes seven twelfths natalie contributes five twelfths which equals what seven plus five is twelve twelve over twelve is one so they do complete one job hello and welcome to algebra one lesson fifty in this video we're going to learn about direct variation so throughout your studies of algebra you're going to encounter many different types of variation Obviously as you move higher in algebra you get to algebra 2 and college algebra you know so on and so forth the types of variation that you're going to encounter are going to become more and more challenging

but for an algebra 1 course you're typically just going to deal with direct variation which we'll see in this lesson and then also inverse variation which we're going to cover in the next lesson so i want to just start out by saying y Varies directly with x if there is a constant k such that y is equal to k times x now this is generally the first thing you're going to read in your textbook when you're in a direct variation section so they have a special name you're going to see this 4k it's called the

constant of variation now We have seen this type of equation before i want you to think back many lessons ago when we learned about linear equations in two variables remember we were graphing these things we went through different forms of the line and if i had a linear equation of two variables in something known as slope intercept form we wrote this as y is equal to m the slope Times x plus b where b was the y-intercept right remember i could plug in a 0 for x and i would figure out what y would be

equal to by just looking at what b was so if i plugged in a 0 for b let's say i had a 0 for b and i had a y-intercept that occurred at the point zero comma zero so plug in a zero for x get a zero for y i have this exact equation here It looks different because this is an m and this is a k but it's the same equation right i just swapped out letters and i renamed it right in this case this is known as a slope and when we get into

the section on direct variation we change it and we say it's the constant of variation so same equation same properties one of the things that you might remember You know just to give you a little insight as x increases okay so as x increases if k is positive then y is going to increase so if k let me write if k is positive as x increases y increases but we can be more specific than that because as x As x increases by one unit we would see that y increases by k units right by this

constant of variation that is multiplying x or again you can think about it as the slope so to see an example of this let's just say we had something like y equals 7x well if i started out by making like a little table let's say this is my x value this is my Y value let's say i just start out by saying okay x is 1. so if x is 1 i plug in a 1 there 1 times 7 is 7. so y is 7. now as x increases by 1 unit so let me just

say x is going to increase by 1 unit so this goes to 2 now what's y going to go to without plugging anything in i should know that y would increase by 7 units right so 7 plus 7 is 14. so this should be 14. This increases by 7 units and does it plug in a 2 for x 2 times 7 is 14 so that is what you get again this is the case as you move up so if i go up by another 1 and this is x equals 3 now i would go up

by another 7. so this would be y equals 21 right so if i did 4 this would be 28. if i did 5 this would be 35. so on and so forth and this works in the opposite direction as well As i decrease x by 1 i'm decreasing y by 7. and to see that let's do another example let's say i had something like y equals 3x now so the constant of variation the k is 3 right or i could say m is 3. i could think of it as slope if that makes you feel

more comfortable for right now as i increase x by 1 unit y will increase by 3 units As i decrease x by 1 unit y will decrease by three units so in other words if i started out with let's say x equal to ten so plug in a ten for x ten times three is thirty y would be thirty if i decreased x by one unit so let's say i went down to nine i'm gonna put minus one there y is going to decrease by three units so 30 minus three is 27 and of course

if i plugged in a nine there Nine times three is 27 right so this decreases by three and you can keep going if i put an eight here this would be twenty four seven here this would be twenty one six here this would be eighteen so every time i decrease x by one i am decreasing y by 30 that constant of variation or again that slope so it's simple enough to understand i want you to think about a real world scenario now let's suppose you purchased gas for 5.25 cents per Gallon we could model this

purchase as y which is the total the total cost or kind of your ending transaction is equal to 5.25 this is going to be your k your constant of variation or your slope however you want to think about that And then x is going to be the number of gallons purchased so we've all done stuff like this let's say you go up to the gas pump and you have let's say twenty dollars in your wallet and it's two dollars a gallon make it really really simple you know that you can buy 10 gallons of gas

can't go over that because you only Have 20 bucks so it's the same thing here if we've modeled this purchases 5.25 per gallon of gasoline purchased well of course if i plug in a 1 there i'm going to pay 5 and 25 cents for that transaction if i plug in a 2 i just increase what i just had by 5.25 so now i'm paying 10.50 if i increase it to three i go by another 5.25 So now i'm paying 15.75 or 15.75 right so on and so forth all right so now that we have a

little bit of understanding under our belts let's look at a typical problem that you're going to see kind of when you get to the questions in the back of your book or for your homework or for a test and these these type of scenarios are very very easy to deal with so we have that y Varies directly with x and y equals 40 when x equals 5. so that's the first part they kind of give you an opening case to deal with and then they're going to give you a problem to solve it says find

y when x equals 3. so i'm not going to deal with this until last the first thing i'm going to always do is i'm going to find k the constant of variation in order to do that i need to do a little basic algebra So we say that again y varies directly with x and y equals 40. so i'm going to write this equation y equals k x that's my direct variation equation and then why they're saying is 40. so i'm going to plug in a 40 there and that says when x equals 5. so

i don't know what k is but i know x is 5. and to switch that around so it looks more traditional let's put 5k now how would i go about finding the value For k here well this is a linear equation in one variable and so all i need to do is divide both sides of the equation by 5 the coefficient for k and so this is going to cancel with this 40 divided by 5 is 8 so i get 8 is equal to k so once i have found my constant of variation all i

need to do is deal with this scenario that they're asking me to Provide an answer for so it says find y when x equals 3. well i know that k is 8 and i know x is 3. so that's pretty simple i'm just following this equation i'm just plugging things in right that's all i'm doing so i plugged in an 8 4k and i plugged in a 3 for x once i multiply these two together i'm going to find that y is equal to 24. so find y when x equals 3 again y is Going

to be equal to 24 and these problems are just that simple all right so here's the next one if q varies directly with m and q equals 5 when m equals 20 find q when m equals 24. so you might see this and go whoa what's with all these letters i'm very confused right now but just follow the traditional format of x and y so just start out with what you know y is equal to k times x okay So now it's q that's varying directly with m so this would be in this scenario q

would take the place of y and m would take the place of x and i know that could get extra confusing because we use m to denote slope and you have this k here which is you know the same thing as slope so just focus on what you're given and take it piece by piece i'm going to use m to represent x here and put my k there And then i'm given information q equals 5 so this is going to be a 5 when m equals 20. so m is going to be a 20. very

very easy to solve so let me drag this up here and what would we have 5 is equal to 20 times k divide both sides of the equation by 20 so i can isolate k and we're going to do some canceling this would cancel with this and you get k equals 5 over 20 5 over 20 Simplifies to 1 4. so 1 4 is equal to k so now that i've found this information my constant of variation i have this scenario that they want me to deal with which is to find q when m equals

24. so using this equation that i've already figured out i'm just going to plug in i'm going to find q so what is q when m is 24. so i know k is 1 4. and i know that m is 24 Multiply the two together 1 4 times 24 is 6 right because this would cancel with this and give me 6 6 times 1 is 6 so q equals 6. and again what i try to do in these scenarios where they change the letters around and typically you don't see this until you get to algebra

2 but i try to just replace it with what i know so i would say ok if y varies directly with x and Y equals 5 when x equals 20 find y when x equals 24. so i would plug in all the information to my original formula solve it and then just go back and say okay well y was representing q so y equals 6 is the answer i got so q equals 6 right so on and so forth that's how you deal with these letter changes when you first get them because this is something

you might see on the sat or the act and it's just them throwing a Wrench at you because they know that you've prepared for this but they want to see if you can adapt if you can go further than basically just regurgitating what you've learned in a textbook and kind of apply your knowledge when you get a minor scenario change so now that we've dealt with two basic scenarios with direct variation you're going to encounter something known as direct variation as a power so it's not more difficult you follow The same process to get your

solution but you're just going to have a power on your variable x now okay so y is equal to k again the constant of variation times x raised to the nth power so i have here y varies directly with the nth power of x if there exists a real number k such that again y equals k that constant of variation times x raised to the nth power so let's take a look at an example here We have if y varies directly with x squared and y equals 50 when x equals 2 find y when x

equals 10. so the equation with direct variation is a power we just write that y is equal to k times x to the nth power we're just going to plug in it's just that simple so if y varies directly with x squared so instead of an n i'm going to put a 2 there and then y equals 50. so i'm going to replace this with a 50. When x equals 2 find y when x equals 10. so let's find k first and then we'll deal with this scenario second okay so 2 squared is obviously 4

so you would get 50 is equal to 4k divide both sides of the equation by 4 and 50 is not divisible by 4 but it is divisible by 2. 50 divided by 2 is 25 so what you'd end up with is 25 halves is equal to k Let's drag this up here and we're going to apply that so we want to find y now when x equals 10. so we don't know what y is we know that k is 25 halves and we know that x is 10 and we know that it's squared so 10

squared is 100 so let's go ahead and write that out so we'd have times 100 here and essentially this is going to cancel with this And give me a 50 and we'd have 50 times 25. now to do this easily 5 times 25 is 125 put a zero at the end you get y is equal to 1 250. all right let's take a look at one more so if a varies directly with b to the fourth power and a equals 162 when b equals three find a when b equals one so again when They kind

of throw these different letters at you just start out with what you know don't panic just say okay well i know from my formulas that i memorized that y is equal to k times x to the nth power okay so now i have a that varies directly with b to the fourth power so a takes the place of y i still have my k and then b to the fourth power is going to take the place Of x to the nth power so it's just that simple just write what you know and then just substitute

based on the problem once you have a general understanding of this it's going to be very very easy to do all right so the next part is that a equals 162 162 when b equals three so b is going to be three Then find a when b equals one so we'll deal with this last the first thing we want to do is just find k so let's figure out what that's going to be let's drag this up here so i'd have 162 is equal to 3 to the fourth power which is what 3 times 3

is 9 9 times 3 is 27 27 times 3 is 81 so this would be 81k and if i divide both sides of the equation by 81 This is going to cancel with this and 162 over 81 is 2. so this is k is equal to 2. now that i have that information they want me to find a when b equals 1. so i want a is equal to i know that k is 2 so i'm going to plug in a 2 for that and i know that b here is 1. so i'd have 1

to the 4th power which i'm basically just multiplying by 1. so again as i just said 1 to the 4th power is 1 And then 1 times 2 is 2 so a here is going to equal 2. that's going to be your answer hello and welcome to algebra 1 lesson 51. in this video we're going to learn about inverse variation so in our last lesson we talked about something known as direct variation and we saw with that we had y varies directly with x if there is some constant k such that we had y Is

equal to k times x where again this k is called the constant the constant of variation we talked about a lot of things in that opening lesson we discussed the fact that in this scenario if k was positive so if k was greater than zero as x increases y increases And as x decreases y decreases and more specifically we could say as x increases by one unit y increases by k units or as x decreases by one unit y decreases by k units and so on and so forth now we're going to talk about another

scenario that you're going to encounter in algebra 1 and that's called inverse variation so the concept is the same so we have that y varies inversely with X if there is some constant k such that and here we have y equals k over x so again the k is called the constant the constant of variation now what's different here is that if k is greater than zero now as x increases what's going to happen to y Well just think about this for a second if x is in the denominator if x is getting bigger and

bigger what's happening to y well it's getting smaller right y is going to decrease and the thing about this let's think about some money for example let's say the four of us get together and do a job so let's say we're getting paid a hundred dollars for this job and we divide it by four people And this amount here tells us how much each person is going to get in this case that's 25. now if i increase the number of people in the group let's say i take this up to five okay so x goes

up to five what's going to happen to y well it's going to go down it's going to go to 20. so as i divide by a larger number i am getting a smaller result and again this is when k is a positive Value but the opposite is also true as x decreases y increases right so if i take this from 5 to let's say 2 now 100 divided by 2 is 50 so that's going to increase so again as x goes down y goes up and as x goes up y goes down so this is

the opposite of what we saw With direct variation where as x went up y went up and as x went down y went down again when k is positive all right so let's jump in and just look at a typical problem you solve these problems the same way that you approach them with direct variation so we have if y varies inversely with x and y equals 7 when x equals 5 find y when x equals 2. so just write your formula so y equals k over x And just plug in with the information they give

you so we have that y equals 7 we have that x equals 5. now the shortcut here is to realize that with this generic formula i can multiply both sides of the equation by x and what i'm going to have is x y is equal to k so that's kind of what most teachers will teach you as a shortcut to say okay if i want to find k it's just equal to x times y So if i plug in a 7 for y and i plug in a 5 for x i know that k is

35 right i know that this will be 35 35 divided by 5 is in fact 7. okay so once i know what the value is for k i can go through and figure out this scenario find y when x equals 2. well again i know that k is 35 And x is going to be 2 so i would get 35 halves which you could write a 17.5 if you wanted to or you could just keep it in fractional form as 35 halves all right let's take a look at the next one so if q varies

inversely with n and q equals one-half when n equals one-fourth find a q when n equals two-thirds so again this might be the type of Problem that really throws your for loop right you see this and you're like oh i'm used to working with y and x just think about what you normally get normally you get if y varies inversely with x right so we would write our equation to start as q is equal to k over n and once i have that set up i can just plug in and get my answer Now we

know that to find k we can multiply both sides by n so i can multiply this by n multiply this by n and i could say k is equal to what n times q so in the first scenario the scenario they give us so that we can find k we're told that q is one half so q is one half and we're told that n is one fourth so n is one fourth and so one-fourth times one-half would be one-eighth So k is equal to one-eighth okay so now that we know that we're told to

find q when n equals two-thirds so we have q is equal to we'll have one-eighth one-eighth divided by two-thirds so what we're gonna do here we're gonna have q is equal to 1 8 times the reciprocal of two-thirds which is three-halves This is going to be equal to 3 over 16. so q in this scenario is going to be 3 16. okay let's take a look at a pretty simple word problem that deals with ohm's law so this is something that you know as an algebra 1 student you might have come across this in your

science class you might not have for sure by the time you get out of high school you'll be very familiar with this Ohm's law so the current in a simple electrical circuit varies inversely with the resistance if the current is 30 amps when the resistance is 5 ohms find the current when the resistance is 15 ohms so pretty simple to solve one of these i'm just going to give you let me go down to another sheet of paper i'm just going to give you the generic Notation they use when you talk about ohm's law so

ohms ohm's law so you have the current okay the current which is represented with i so this is for the current then you have the resistance which is represented with r and then you have the voltage now in our word problem we didn't say anything about voltage but the voltage is going to be your Constant of variation it's not going to change so this is my k or my constant of variation and it represents the voltage okay so let's go back up real quick and let's get some numbers going so it says first off the

current in a simple electrical circuit varies Inversely with the resistance so current we represented with i resistance we represented with r so that means that our formula would look like this i is equal to some constant over r now we already said that the constant would be v the voltage so you could put v there you could put k there you could really put whatever you want there doesn't matter right you're just solving For that unknown so let's put v there to make sure that we're consistent with that formula all right so it tells us

in the starting scenario if the current is 30 amps so this is where my current goes so this would be a 30 when the resistance is 5 ohms so this would be a 5 and then we'll stop there we just want to find v we want to find v and then apply to that second scenario That gives us so let's solve for that so what i would do here i would multiply we'd have 30 equals v over 5 multiply both sides of the equation by 5 and i would get 150 is equal to v so

that's my voltage that's going to stay the same okay so now in the next scenario we want to find the current when the resistance is 15 ohms now the resistance goes up that means that the Current is going to go down right it's inverse variation so we're looking for a smaller value than 30 here right so if i plug in i have i is equal to i know that my voltage which is v that's on top is 150 and they're giving me this 15 for the resistance pretty easy to do this here 150 divided by

15 is 10. so i is equal to 10 in this scenario Because it's a word problem we probably want to put a nice little sentence there and say that the current is 10 amps when the resistance is 15 ohms okay one of the things you can notice about this again it's inverse variation so as r increased it went from 5 to 15 the i or the current decreased it went from 30 to 10. so that's consistent with What we know about inverse variation all right so to kind of wrap up the lesson we're also going

to see inverse variation as a power so this is no more difficult it's just plugging things in figuring out what k is and then following up with your scenario that they're giving you so y varies inversely with the nth power of x if there exists a real number k such that y equals k the constant of Variation over x to the nth power so y varies inversely with q squared so y is equal to your variable in the place of x as you normally get it is now q so you just put y equals k

over q squared and then we're given the scenario where y is equal to 4 so y is going to be 4 when q is going to be 5. so we're going to use this to solve for K and then we can figure out what this guy is so y is 4 when q is 5. 5 squared is 25 so k over 25 obviously k there is a 100 right if i multiply both sides by 25 i'm going to get that k equals 100. so now we want to do this scenario where we find y when

q equals 15. so i'd be plugging in A 15 there so let me erase this real quick let's get a little room going so i know k is a hundred so 100 is going to get plugged in there and q is 15 15 is going there and so what i'm going to have is y is equal to 100 over 15 squared now a lot of you don't know 15 squared is 225 yet but by the time you finish Algebra 2 you're going to know that for sure so 15 squared is 225. all right we can

reduce this fraction so each of these is going to be divisible by 25 100 divided by 25 is going to be 4. 225 divided by 25 is going to be 9. so this ends up being 4 9 as my answer all right so let's wrap up the lesson by looking at a word problem so we have that the amount of light measured in foot candles produced By a light source varies inversely with the square of the distance from the source so let's just label things i'm going to say that the amount of light measured in

foot candles produced by a light source let's just go ahead and just say that's l and this varies inversely with the square of the distance from the source so let's say the distance from the source let's say that's d And that's going to be squared so we'll put d squared down here and then for my constant of variation i'm just going to use k right nice and simple all right so let's copy this because we're going to need it okay so if the light produced 10 feet from a light source so that's the distance so

i'm going to plug in a 10 for d there is 12 foot candles so this is going to be 12. find the light produced 20 feet away From the same source so what we want to do again is just plug in this find k and then figure out what this second scenario is that they're giving us so if i plugged in a 12 for l i have 12 is equal to you'd have k over 10 squared which is a hundred to isolate k i'd multiply both sides by a hundred And so this would cancel with

this and you would get k is equal to twelve hundred now let me erase all this real quick and in this second scenario the one that we need to figure out it says find the light produced 20 feet away from the same source so now the distance the d part is going to be a 20. so i plug in a 20 there and i know that k is still 1200 so this is 1200 So the l is going to be equal to 1200 over 20 squared now 20 squared is 400 so 400 and i know

that i can cancel two zeros with two zeros right it's like cancelling out a common factor of 100 and then 12 divided by 4 is 3. so the l here or the light produced is going to be three and in terms of units it would be three foot candles so let's Go ahead and write that the light produced 20 feet away from the light source is three foot candles hello and welcome to algebra 1 lesson 52 in this video we're going to have an introduction to roots so i know for some of you you've learned

about square roots in a previous course but for others you've had no exposure to It whatsoever so we're going to just begin by talking about square roots i'm going to pretend like you have no prior knowledge of it and we're going to start from scratch all right so i want you to think back to when we first started talking about exponents and we learned what it meant to square a number so squaring a number means to multiply a number by itself basically it's when you have an exponent of two so for these four examples here

We have three squared so we know at this point that this is what it's three times three or nine if i have four squared this is 4 times 4 or 16. if i have a negative 7 squared with the negative and the 7 inside of parentheses this is negative 7 times negative 7 which is positive 49. and then if i have negative 12 squared again with the negative and the 12 Inside of parentheses this is negative 12 times negative 12 which is positive 144. now what if i gave you the following scenario i have here

that a squared so something squared is equal to 225. what is the value for a so you might be thinking how would i go about doing that well let's just go back to a simple example we saw that 3 squared was equal to 9 right why is it Equal to 9. well i can further break this down and say it's 3 times 3 which equals 9. now if i think about a number a number that when multiplied by itself so in other words this would translate to a squared is equal to a times a which

equals 225. so i've got to think about a number that when multiplied by itself is going to give me 225. Can you think of a number well if you can't you can just break 225 down using a factor tree so it ends in five so that means it's divisible by five 225 divided by five is 45 so let's write 45 times five and then five is going to be prime so let's circle that and then 45 can be further broken down into 9 times 5 5 is prime so let's circle that And then 9 is

3 times 3. so 3 times 3 and those are each prime so let's circle those and what can i see here i have 3 times 5 or 15 and then i have 3 times 5 again or 15. so 225 is 15 times 15. so we've found a number that when multiplied by itself is going to give us 225. so where we have a question mark here let me erase all this I can put a value of 15. now some of you will be screaming at this point going what about negative 15 if i took negative

15 and i multiplied it by negative 15 i would also get 225 so you'd be correct there so a could be 15 or it could also be negative 15. right if i had negative 15 and that was squared this would be negative 15 times negative 15 which would also give me a positive 225. So when we think about square roots okay we're going to think about a positive version and a negative version and i'm going to show you the notation for that in a minute let's take a look at another one so we have b

squared equals 100 so we want to know what is the value for b so if i took b squared and i said it's equal to what let's break this down into b times b Let's say that's equal to 100 so all i'm looking for here is i'm looking for a number that when multiplied by itself gives me 100. so again that's easy just make a factor tree for 100 and most of us know that 100 is what it's 10 times 10. right i don't need to go through and go all the way to 5 times

2 and then 5 times 2 because i know that 10 times 10 is what i'm looking for right so b here would be 10 but again it could also be negative 10 right negative 10 times negative 10 would also give me a positive 100 all right so let's think a little bit about the notation involved with what is known as a square root so in the previous two examples we looked at We said a squared is equal to 225 and then we said okay a is equal to 15 or negative 15 right because 15 squared

would give me 225 then also negative 15 squared would give me 225 as well and then we looked at a squared is equal to 100 and we said in that case a was equal to 10 or it could be negative 10. Okay so we want to think about kind of how this is asked to us in terms of our homework or a test just moving forward if i want what's known as the square root of a number i'm going to see this symbol right here so it looks kind of like this so if i wanted

to know what number squared would give me 225 i would put 225 underneath this symbol And i would say this is the square root of 225 or what number when multiplied by itself gives me 225 we know at this point that the answer is 15. okay we found that here now there's also a different way to kind of notate when i want negative 15. this right here is known as the principal square root or the positive square root so it produces a positive here and then we have what's known as the Negative square root so

if i want this one i put a negative out in front so i put a negative and then everything else is the same it's just asking me for the negative of this 15 which would be negative 15. okay same thing would go for this example here so if i want the square root of 100 this is the positive or principal square root of 100 so that's positive 10 and then the negative square root of 100 put a negative sign out in front would Be negative 10. okay so those are two different ways to kind of

notate that so let's look at kind of a general statement for this if a is greater than zero so we know that's a positive value then we can say that the square root of a is a positive number that when squared gives us a so in other words if i have the square root of again i go back to 225. so this represents a here so This is a positive number that when squared when multiplied by itself gives us this number back so in this case this is 15 because 15 times 15 would give me

225 or that number that's inside of that symbol back now we have that negative outside in front of the symbol here negative square root of a is a negative number that when squared gives us a so this is a the number is out of here so if i did negative Square root of 225 the answer here would be negative 15. so you want to know the two different ways to notate that because kind of moving forward on your test if your teacher says okay i want the square root of let's say the number not you

have to know that the answer is 3 just positive 3. if she was to say i want the negative square root of 9 you have to know that the answer is going to be negative 3 And then kind of moving forward as we get higher in math you might see them put plus or minus the square root of 9 and then in this case the answer is 3 or it could also be negative 3 right because i've accounted for the positive or principal square root of 9 and then also the negative square root of not

so in that case you get 3 or a negative 3. all right let's just go through some examples real quick If i have the square root of 0 okay the square root of 0 0 is not a positive or a negative number it's just 0. so in this particular case what's it going to be equal to what number multiplied by itself gives me 0 well 0 does so this is an automatic you see this on your task you have the square root of 0 it's just equal to 0. now the next one we have the

square root of 4 and then after that we have the negative square root of 4. so the Positive or principal square root of 4 is what it's 2 right because 2 squared is going to give me 4 right what number multiplied by itself gives me this well it's 2 right 2 squared is 4. then for this one i have the negative square root of 4. so what negative value when multiplied by itself is going to give me 4 well that's negative 2. so you can see the difference in the notation principle or Positive square root

of 4 negative square root of 4. so something you really need to pay attention to all right what about the square root of 16 can you think about a number that when multiplied by itself gives you 16. again if you don't know that off the top of your head always just build a factor tree so i can say okay 16 is what a lot of you right away know it's 4 times 4 but let's say you did let's say you start out with 8 times 2 You know that 2 is prime and for 8 you

know that's 4 times 2 and you know 2 is prime at that point you can stop because you say okay 2 times 2 is 4 you'd have 4 times 4 but let's say you just kept going okay we keep going this is 2 times 2. so now you see that you have what 2 times 2 or 4 this is 4 times 2 times 2 or 4 right so 4 times 4 Which is 16. so a number that is multiplied by itself that gives me this is my answer right so the square root of 16 is

4. and then what about the negative square root of 16 well again what negative number when multiplied by itself is going to give me 16. well negative 4 right negative 4 times negative 4 is positive 16. all right what about the square root of 400 well a lot of you are not going to know What number multiplied by itself gives you 400. so let's just start out with a factor tree so 400 i know it ends in zero so it's divisible by 10. so this would be what 40 40 times 10 and 10 i can

break down into 5 times 2. those are each going to be prime so let's circle those then for 40 i can break that down into 10 times 4 and 10 is 5 times 2 And 4 is 2 times 2. let's circle all these so what do i see here i've got one two fives so five times five and i've got one two three four twos so times two times two times two times two well what i can do is i can write this as what five Times two times two times five times two times 2

or 5 times 4 which is 20 times 20. 20 times 20 is 400 so that's your answer and sometimes when you don't know these off the top of your head you've got to do a little bit of work but pretty much in algebra one they're going to keep giving you kind of the same ones over and over again so you'll memorize them very very quickly you also have something on your calculator if You're allowed to use it known as a square root key right so you type in a number and then hit square root and

then it's going to give that to you automatically so the square root of 400 would be 20. now again we have the negative square root of 400 and so this is what this is negative 20. since negative 20 times negative 20 would give you 400. all right so we're also going to see this with fractions as well so we have the Square root of 9 16. so just think about what fraction times itself would give me 9 16. well just break up the numerator and denominator kind of separately in your head what number multiplied by

itself would give me 9. we know that would be 3. then what number multiplied by itself would give me 16 we know that would be 4 and you can check it Would 3 4 times 3 4 give me 9 16. 3 times 3 is 9 4 times 4 is 16 so yeah that does check out all right so let's talk about something that is going to come up kind of as we move forward when we square the square root symbol the two cancel each other out so squaring and square root are opposite operations just like

multiplication and division so in other words if i had something like 5 i start with that Let's say i divide 5 by 3 and then i multiply by 3. well what's going to happen i'm going to end up with 5 again right because this operation the division cancels with the multiplication right if i divide by three but then i multiply by three it's i just start out with that original number five right so this just equals five it's going to be the same thing if i started out with let's say 5 And let's say i

have 5 and i square 5 but then i take the square root i'm going to end up with 5 again right 5 squared is what it's 5 times 5 that's 25 and then if i took 25 and i took the square root of it i'm going to end up with 5 again right so it's just like multiplication division they're opposite operations so we have here the square root of 15 that's squared so in other words i start out with 15 i take the square root of it but Then i square it so it's like i

can cancel this and this and i'm just left with this right and another way to show that is just to say okay well what i have here is i have the square root of 15 times the square root of 15. so whatever number it is okay for 15 that multiplied by itself gives me 15 it's multiplied by that number so i would just get 15. right as an easier example let's say i Had something like the square root of 4 squared so writing this the same way this is square root of 4 times square root

of 4 does this equal 4. we know that the square root of 4 is 2. so what is 2 times 2 well 2 times 2 is 4 right so that's where this comes from kind of the shortcut to this is just to say okay well this is going to cancel with this and i just have this so what if i had the negative square root of 31 squared So if i square a negative right you think about that as just negative 1. negative 1 times negative 1 is 1. so this and the square root are

going to cancel and i'm going to be left with 31. again if you want to think about this in a different way think about if you had negative square root of 31 multiplied by negative square root of 31 negative times negative is positive of course Square root of 31 times the square root of 31 is just 31. all right so this one's a bit more complex but we follow the same logic so we have the square root of 8x minus 7 and this is squared so this is going to cancel with this and i'll just

have what's inside which is 8x minus 7. all right let's talk a little bit more about rational numbers and irrational numbers which we've kind of covered before but i want to make sure that you Have kind of a strong foundation in this before we get to algebra 2 and we really start knocking some numbers around so i want you to recall that a rational number let me just highlight that a rational number is one that can be formed from the quotient of two integers in decimal form it's going to terminate okay or it's going to

repeat the same pattern forever so you've seen examples where you get something like 21.3 and we put a bar over three because the three repeats forever that is a rational number or you've seen something like you know let's say you have point six eight seven six eight seven six eight seven you know where the six the eight and the seven repeat forever right and we could erase this and just put a bar over this so that pattern repeats forever or It terminates so you might just get point six eight two five nine one and it

stops those are rational numbers right you can always form these from the quotient of two integers so i should be able to say this is let's say 28 over three one integer over another so any number whose square root is a rational number is called a perfect square a perfect square you're going to hear me talk about that all the time you've also Heard me talk about this when we're talking about polynomials i refer to a perfect square trinomial so to go back to an example that we already know the square root of 225 is

what it's 15. 15 is a rational number so therefore 225 is a perfect square and then as another example that we already know the square root of nine is what it's three so three is a rational number so therefore 9 is a perfect square what about the square root of 1 4 Well this is what square root of 1 if i think about what times what would give me 1 it's 1. what times what would give me 4 it's 2. so the square root of 1 4 is one half and so therefore 1 4 is

a perfect square right now let's talk a little bit about irrational numbers so an irrational number is one that cannot be written as the ratio of two integers so i can't say okay this is you Know 15 over four for example i can't do that with an irrational number there are no two integers that i can put over each other to give me that irrational number so its decimal does not terminate okay it does not terminate nor does it repeat okay so something like the square root of 2. you cannot find two integers that you

can put over each other that would give You the square root of 2. if you type this into a calculator you're not going to get an exact value your calculator is either going to round or it's going to truncate and it's going to give you an estimate for it but not an exact value and usually we'll say this is approximately equal to 1.41 but if you type it up on a calculator And i just typed it up on mine i get 1.414213562 so after that final digit that you get realize that it continues forever and

ever and ever and it's not going to terminate or repeat the same pattern so that's the problem with an irrational number you can't really you can't really work with it that easily a famous example of this is pi we talked about this before when we talked About direct variation and i gave you an example where in the hollywood movie the life of pi this guy when he was a child is on a blackboard he's reciting all these digits of pi right and you can go on a computer and go into google and you can find

websites that give you all these digits to pi no matter how many you list it just keeps going forever an hour you can't list all of them so that's the problem again with an irrational number so when we get an Irrational square root generally we will approximate by rounding to the nearest hundredth and sometimes you won't approximate at all sometimes you're just going to leave it like this but in some cases it's necessary to kind of do an estimate in other cases you can kind of leave it like that just depends on the scenario you're

dealing with so square root of 11 for example would be Approximately punch that up on a calculator to get about three point three one you know six six two four seven and let me kind of stop there if i round it to the nearest hundredths that would be 3.32 so 3.32 what about square root of 7. so you punch that up on a calculator if we ran out to the nearest hundredths you'd have something like 2.65 all right so now that we understand kind Of the basics of finding the square root of a number and

we know the difference in notation between finding the principal or positive square root and kind of the negative square root we've got to move into some kind of harder material so the first thing that students struggle with would be the square root of a negative number okay so we also need to understand that not Every real number has a square root so if i ask you for the square root of negative 4 what am i asking you for the first thing students might do is they might get confused and say well i'm thinking back through

this lesson that we just had and he already covered this this is negative square root of 4 this equals negative 2. no that's not what i'm talking about in this case the number under the square root symbol is a 4 okay it is a positive Value the negative is outside this is outside okay that's a big deal now in this particular case i have the negative 4 inside this little symbol so what i'm saying is what number when multiplied by itself is going to give me negative four the answer to that is it does not

exist okay now when we get higher in math we're Thinking about these imaginary numbers but in our real number system does not exist right you can't square a number and get a negative number right you can't do that because a negative times another negative is going to give you a positive 100 of the time so there's no number that if i multiply by itself is going to give me a negative 4. so we would just put not a real Number okay it doesn't exist and you can try that on your calculator and depending on the

calculator you might get no real answer you might get not a real number you might just get error so on and so forth because the computer cannot calculate a number that when multiplied by itself is going to give you a negative 4. all right so now let's move on and talk about some higher level roots So we saw that the square root is the reverse of squaring so similarly the cube root is the reverse of cubing so cubing is nothing more than just raising something to the third power so in other words if i had

2 cubed this is what this is 2 times 2 times 2 or just the number 8. so if i reverse that process and i said here's my 8 here the symbol looks the same all i'm going to do is instead of having nothing here i'm going to put a 3. that tells me i want the cube root of 8 or i want a number that when multiplied by itself three times okay three times is going to give me eight and we know the answer to that is two all right so in general when we work

with roots okay not just square roots but roots in general you kind of think about okay a the number that is underneath the symbol is known as the radicand so this is the Radicand this is something that they might test you on when you first get the square root so this is the radicand n is known as the index okay this is the index and the index is important because it tells me what i'm looking for so for example when i just saw the cube root of 8 my index is three right my radicand is

eight so i'm saying what number again when multiplied by Itself three times would give me eight if i had something like let's say the fourth root of 16 okay i'm doing easy stuff the fourth root of 16 is asking what number when multiplied by itself four times is going to give me 16. well we would think about two you could also have negative two right but it would be a different notation but 2 times 2 is 4 times 2 again is 8 times 2 one more time is 16 and again you Could also say the

fourth root of 16 would be negative 2 but again there's that different notation for that now this symbol here is known as just a radical symbol a radical symbol and then the whole thing okay so the whole thing is known as a radical so if your teacher comes in and says We're studying radicals today or we're studying roots today okay they're basically talking about the same thing now one thing you would have noticed already is that when we have the square root we don't write the 2 right the index would be a 2 but we

don't write it it's left blank so that's one property of square roots if you write a 2 up there it's not incorrect it's just not traditional so If i put you know the square root of let's say 16 like that it's the same thing as if i put the square root of 16 like that right in each case it's 4. this is just a notational thing that we do so this one right here if i have an index of three this is the cube root if i have an index of four it's the fourth root

if i have an index of five It's the fifth root right so on and so forth you can put any number you want there but it starts getting very very complex if i'm looking for the fifth root of something i'm looking for a number that when multiplied by itself five times gives me that number the radicand so again let's look at some easy examples of this if i have the cube root of eight again what number when multiplied by itself three times gives me eight we Know at this point that that's two right two times

two is four 4 times 2 is 8. so now we have the 4th root of 16 which we already looked at again what number when multiplied by itself 4 times gives me 16. well that's going to be 2. and this is the principal 4th root of 16 so i want a positive 2. if i want a negative 2 then i would look at the negative 4th root of 16 in this case this is negative 2. this one is one you have to Pay attention to we know that in the case of a square root so

let's say i want the square root of negative 16. i already talked about the fact that that's not a real number right i can't square a number and get a negative result negative times negative is positive if i'm asking for an even root so Something to the second power let's say to the fourth power you know so on and so forth this is not going to be a real number right it's it's not going to be possible right because an even number of negative signs when multiplied together gives you a positive if i have an

odd index like i have here it is possible right because an odd number of negative signs would in fact give me a negative result so if i had Something like the cube root of negative 8 this is a real number it's negative two negative two times negative two is positive four positive four times negative two is negative eight in this case we have five five is an odd number so i can have the fifth root of negative 32 that would just be negative two to show you that just think about negative 2 times negative 2

times negative 2 times negative 2 times Negative 2. you have 1 2 3 4 5 negatives so that's going to be a negative and then 2 times 2 is 4 times 2 is 8 times 2 is 16 times 2 is 32 so that is negative 32 and so again the fifth root of negative 32 would be negative 2. so again a radicand with an odd index 3 5 7 9 11 you know so on and so forth can have a negative radicand so this part right here can be negative again if you have an odd

index if you have an Even index it cannot be and the reason for that again is you won't have a real number that will satisfy that so if i look at the cube root of negative 27 what number when multiplied by itself three times would give me negative 27. will it be negative three negative three times negative three is positive 9 positive 9 times negative 3 is negative 27. so let's look at the fifth root of negative 1024. So again if i have an odd index this can be negative and when i'm solving a problem

like this i would just automatically write a negative there right if i had five negatives i would get a negative so i can just think about the number part so 1024 kind of break this down with a factor tree i know it's divisible by 4 because 24 is if i divide 1024 by 4 I'd get 256. so 256 times 4 and 4 i know is 2 times 2. and let me circle those so what about 256 well i know that's divisible by 4 this is 4 times 64. and 64 is divisible by 4 this is

4 times 16 and 16 is divisible by 4 this is 4 times 4. so without continuing on this factor tree i've got 1 2 3 4 and this original up here 5 Factors of 4. so if i multiply 4 times itself 5 times i would get 1024 and that's what we're looking for so this would be negative 4 as your answer so let me erase this real quick and i can show that to you real quick so you'd have negative 4 this quantity would be raised to the fifth power and this would be equal to

negative four times negative four times negative four times negative four and negative four One two three four five negatives gives me a negative four times four is sixteen sixteen times four 4 64. 64 times 4 is 256 and if i multiply 256 times 4 i get 1024 right stick that next to the negative and i get my negative 1024 back which was my radicand so then what about the sixth root of negative 720 knot do we need to look to solve this no because i have a negative here and i have an even index so

this is not A real number hello and welcome to algebra 1 lesson 53 in this lesson we're going to learn about the distance formula so we just got done talking about square roots cube roots fourth roots basically radicals in general now one of the immediate applications of square roots would be the ability to find the distance between two points on the coordinate plane using the pythagorean theorem or your book might Say the pythagorean formula so the pythagorean formula basically looks like this you have this a squared plus b squared equals c squared it's a pretty

famous formula so even if you haven't taken geometry yet or you haven't seen it in a previous algebra course you've probably come across this at some point okay so what the pythagorean formula does is it tells us about the relationship between the sides and a right triangle So i alluded to geometry and i know that some of you probably haven't studied geometry yet because this is usually taken directly after an algebra and course so if you don't know anything about triangles it's not a big deal you can pick up what you need to know about

right triangles from our lesson today so let's focus on what we have here this triangle see a lot of stuff that's labeled and let's just kind of go Through that so the first thing is you'll notice that this symbol here is going to denote that we have a 90 degree angle or what is known as a right angle okay so when a triangle has a 90 degree angle or a right angle it is known as a right triangle okay a right triangle now with a right triangle you have two shorter sides that are known as

legs so here you have this leg a and this leg b okay so those are the two shorter sides of the right triangle Then the longest side is always going to be opposite of the 90 degree angle so opposite of that would be here that's labeled as c and it's known as the hypotenuse okay so the way the pythagorean formula works is it tells us that kind of the measure of leg a so from here to here if we square that so that's where you get the a squared then plus the measure of leg b

so from here to here so you take that and you square it so you're Summing those two amounts together so a squared plus b squared this is equal to c or the hypotenuse squared okay so this is true for any right triangle that you work with now what we're going to see is that we can use this kind of pythagorean formula to develop a distance formula that's going to tell us the distance between two points on a coordinate plane but before we jump into that let's look at a little sample problem just to get our

feet wet so when we're using our Pythagorean formula if two of the three sides are known we can solve for the third side that is unknown so in this case we're given that a is 3 so that's going to be from here to here okay that's my leg a b is 4 so that's going to be from here to here okay so that's my leg b we want to know what is c c is from here to here that's our hypotenuse okay so all i need to do is plug into my pythagorean formula and i

can crank out An answer so i have a squared plus b squared equals c squared okay plug in for a what is a it's 3 so just plug that in there what is b that's 4 so just plug that in there so 3 squared is 9 plus 4 squared is 16 this equals c squared i'm going to flip this around just so i have my variable on the left side so i'm going to say c squared equals 9 plus 16 is 25. okay so we haven't solved something like This yet and we're going to get

to this as we kind of progress through our course but essentially what i want to do if i want to isolate c since it's being squared i want to take the square root of each side right i'm always thinking about how can i undo what's being done to my variable right so since c the variable i'm trying to solve for is being squared to undo that i just take the square root okay and to make it legal i've got to take the Square root of each side so let me kind of scooch this down just

a little bit so i would take the square root of this side and the square root of this side now in most cases when you do this you want to go plus or minus over here okay in this case you're not going to and i'll explain why in a second first let's kind of solve just given the way that i've presented it here so this index of 2 right remember this Square root has an index of 2 is going to cancel with this exponent of 2 okay so i now have c by itself and it's

equal to plus or minus the square root of 25 which is 5. now does that make sense well remember if i erase all this and i bring this back to c squared equals 25 well c could be 5 right 5 squared is 25 and c could also be negative 5 right negative 5 squared is 25 so that's why you have that little plus or minus there we'll get to that Later on in the course as to why that's important but for now we just take it as a given but we need to understand that here

in this specific situation i don't need a negative 5 as an answer i only need the principal square root when i do this because 5 is my only valid solution why because i'm thinking about how far is it from here to here i'm basically thinking about the distance for the measure of c and that's not going to be negative Right negative 5 is a nonsensical answer it can only be positive 5. okay so we've seen an example of working with a pythagorean formula and now we're going to use it to find the distance between two

points on a coordinate plane essentially you're going to have a point that's here and here okay and what we're going to do is we're going to derive a formula that lets us solve for the unknown which is going to be the hypotenuse given the Fact that we're going to be able to find this horizontal distance here and this vertical distance here so what we have on the screen here is our distance formula again the distance formula is just allowing us to easily calculate the distance between two points in a coordinate plane it's a direct application

of what we're just looking at again the pythagorean formula so what we see here is that d which Stands for distance is equal to the square root of we have x sub 2 minus x sub 1 squared plus we have y sub 2 minus y sub 1 squared and again just like when we work with slope okay we can label one of our points as x sub 1 y sub 1 the other is x sub 2 y sub 2. it's not going to matter which is labeled as which you're going to get the same answer

either way so how does this relate to the pythagorean formula well let's look at An example and work through that completely and then we'll kind of bust through a bunch of examples all right so we want to find the distance between each pair of points and we're going to start out with 5 comma negative 2 and negative 7 comma negative 7. so let me go back up so we have 5 comma negative 2 and we have negative 7 comma negative 7. so the first thing i'm going to do is just plug into the formula we're

going To get an answer then we're going to go to the coordinate plane i'm going to show you where this comes from so d the distance is equal to the square root of so i'm going to label this guy as x sub 1 y sub 1 and i'll label this as x sub 2 y sub 2. again you can change that up it wouldn't matter so x sub 2 is going to be negative 7 just plugging that in for x sub 2 there then minus x sub 1 is going to be 5 just Plugging that

in there this is squared then plus we've got y sub 2 which is negative 7 just plugging that in for y sub 2 there then minus y sub 1 which is negative 2. be careful you're going to have minus a negative which is plus a positive this is going to be plus 2 and then this guy squared so all i did was i plugged it okay nothing nothing fancy there so my distance d is going to be equal to The square root of all i've got to do is just kind of crank this out negative

7 minus 5 is going to be negative 12 negative 12 square is 144 then plus negative 7 plus 2 is going to be negative 5. negative 5 squared is 25. so 144 plus 25 is 169. so d my distance is going to be equal to the square root of 169 which is basically 13. right so let me kind of erase this and i'll say the distance between the two points is going To be 13. okay so where did this come from i know most of you just want to be able to get through your homework

so once you have the formula you're basically good to go you can just plug in but i want you to understand where it came from so let's go to the coordinate plane and let's think about the distance between the two points kind of using a different approach so again we had 5 comma negative 2 And we also had negative 7 comma negative 7. so i've kind of pre-drawn everything to make this a little faster so we have 5 comma negative 2 that's right here okay that's your 5 negative 2 and you have negative 7 comma

negative 7 so that's going to be right here so you'll notice that i drew in an extra point to kind of form the right triangle so the way i did that was i took the x coordinate from here which was 5 and i took the y coordinate from here which Was negative seven that gave me my third vertex so i can again draw the right triangle so what i can see is that i can find the measure from here to here that's going to be a vertical leg for me that's going to be my leg

a i can find the measure from here to here that's going to be my horizontal leg or my leg b okay and all i need to solve for is this c which is the distance between these two points that's my Unknown okay so how do i find the measure for leg a well since this is a vertical line i'm just thinking about y values okay because it's parallel to the y axis so how far do i move from this y coordinate here of negative seven to this y coordinate here of negative 2. well most of

you would realize that you just need to subtract the y values right so you would just basically do inside of Absolute value bars negative 2 minus a negative 7 which would be equal to 1. it would be the absolute value of negative two plus seven which would be equal to the absolute value of five which is five you could also do it the other way because you use absolute value so i could reverse this completely and i could say that we had negative seven minus a negative two okay you get the Same answer so negative

seven minus a negative two inside of absolute value bars would be the absolute value of negative seven plus two which would be the absolute value of negative five which would be 5. okay now you might say well why are you using absolute value we're using absolute value because we're calculating a distance we don't want to end up with the negative value okay so notice how we just found the difference in y values now if we go back up you'll Notice that when we did our formula we have y sub 2 minus y sub 1 we're

just calculating the difference in y values so we're doing the same thing we don't have absolute value bars because we're squaring the result okay so if you square something that's negative if i end up with negative 5 and a squared i get 25 if i end up with 5 and a square i get 25 so i don't need absolute value bars so that's why they're not there the same thing is going to go when i find The kind of horizontal legs measure so now i'm dealing with a horizontal line which is going to be parallel

to the x axis so i'm thinking about x values now so i have an x value of negative 7 and i have an x value of 5. so i can either do the absolute value of 5 minus a negative 7 or i could do the absolute value of negative 7 minus 5 gives me the same answer either way so this would be equal to 1. this would be equal to the absolute value of 5 plus 7 which is 12. So the absolute value of 12 is 12. okay so again if i go back up i'm

looking for the difference in x values and again if it's negative 12 and a square to get 144 if it's positive 12 and i squared i get 144 so i don't need absolute value bars in that case okay so once we know that leg a's measure is 5 and leg b's measure is 12 well then i can plug it into my formula right so i can say that this is a so 5 is a and We're squaring that plus the b is going to be what it's 12 so we're squaring that this is equal to

c squared okay c squared and in this case we just re-labeled c with d for distance that's all we did so let's just label this as d it doesn't matter we're just changing the variables letter but it means the same thing okay so if i have d squared equals 5 squared plus 12 squared how do i solve for d Well at this point i could just take the square root of each side right i know i can simplify this further but all i'm doing up here if i go back to my formula once i have

this guy i can just take the square root of each side and i've solved for d okay let me go back down make that crystal clear so d squared equals you have 5 squared plus 12 squared if i take the square root of each side then this would go away and i have the square Root of this then i could just calculate this 5 squared is 25 and 12 square is 144 and then 25 plus 144 is 169. and we all know the answer from earlier d equals square root of 169 which is 13. okay

so just a quick insight into where this comes from it's a direct application of the pythagorean formula i know most of you you know you're more interested in just being able to kind of get the formula and crank out your homework but It's nice to know where it comes from in case you forget the formula you can kind of pull out a coordinate plane and kind of think your way through deriving this formula because it's very easy to get you just need the difference in x value squared plus the difference in and y value squared

that gives you your a squared plus b squared okay and you set this equal to c squared and from that you can kind of set up your distance formula all right so now let's kind of Wrap up the lesson by just kind of bursting through a few of these once you know the formula is very very easy so we have that the distance between two points is equal to the square root of you've got x sub 2 minus x sub 1 quantity squared plus you've got your y sub 2 minus your y sub 1 that

quantity squared okay and i'll just drag this down here and let me put that right there so this is going to be my x sub 1 y sub 1 this is going to be my x sub 2 y Sub 2. again you can change that around it doesn't matter so just plug in real quick we'll get our answer so x sub 2 in this case is going to be 6 and x sub 1 is going to be negative 3 minus the negative 3 is plus 3. be very careful when you work with these whenever you

have minus a negative you've got to make sure you do plus a positive okay very common mistake so then y sub 2 is going to be negative 6 and y sub 1 is going to be positive 6. so 6 plus 3 is 9 9 squared is 81 and negative 6 minus 6 is negative 12 negative 12 squared is 144. so you get 81 plus 144 which is going to be 225 and the principal squared of 225 is 15. so the distance between these two points is going to be 15. all right let's take a look

at a null so we have d the distance between the two points is equal to the square root Of you've got your x sub 2 minus your x sub 1 that quantity squared plus your y sub 2 minus your y sub 1 that quantity squared and let me kind of make that a little bit better okay so let's just change it up and label this as x sub 2 y sub 2. we'll label this as x sub 1 y sub 1. again it does not matter how they're labeled just like you saw with slope formula

so for x Sub 2 i'm going to erase this and put 8 for x sub 1 i'm going to erase this and put 7 and go ahead and do that 8 minus 7 is 1 1 squared is 1. okay so that's pretty simple for y sub 2 i've got a 1 for y sub 1 i've got a negative 3. so again be careful minus the negative 3 is plus 3 and 1 plus 3 is obviously 4. so if i have 4 squared that gives me 16 1 plus 16 is 17. so d the distance is

Equal to the square root of 17. now 17 is not a perfect square and 17 is a prime number so it's not going to factor right it's just 1 times 17. so when you get an irrational number for your answer then what you want to do is either leave it as the square root of 17 or you can approximate it if you want to but make sure you use the correct symbol because it's not precise so we could say d is approximately 4.12 okay it's approximately or about 4.12 again this is precise d is equal

to the square root of 17. i would always go with this answer but sometimes it makes sense to do this especially if you need to do some more calculations with it all right let's finish up the lesson and just look at one more so we've got d our distance between the two points is equal to the square root of again you've got your x sub 2 minus your x sub 1 squared Let me make that a little better a little clearer plus you've got your y sub 2 minus your y sub 1 squared and let

me make that a little better okay so i'll label this as x sub 1 y sub 1 and this is x sub 2 y sub 2 and again we're just going to plug in so for x sub 2 i want to plug in a 5. for x sub 1 i'm going to plug in a negative 5. so minus a negative 5 is Plus 5. so 5 plus 5 is 10 10 squared is 100. okay so this is 100 and then y sub 2 is going to be 4 and y sub 1 is negative 3. again

minus a negative 3 is plus 3 so 4 plus 3 is going to be 7 7 squared is 49 okay so now you've got 100 plus 49 which is 149 okay 149 and again 149 is a prime number so i'm just going to leave my answer like this i'm just going to say d is Equal to the square root of 149 just like that if you want to approximate it you can you can say d the distance is approximately going to be 12.21 okay just as an approximation but this guy right here is a precise

answer whereas this one is just an estimate hello and welcome to algebra 1 lesson 54. in this video we're going to learn about multiplying dividing and simplifying Radicals all right so let's begin today by talking about the product rule for radicals so this is the very first thing that's going to be in your textbook when you start talking about simplifying radicals so let me highlight that again the product rule for radicals so you're going to see something similar again in your textbook and it's going to start out by just saying that for two non-negative real

numbers let's just say A and b and they could be any two letters you want to use we have that the square root of a times the square root of b is equal to the square root of a times b now a lot of times when we see these kind of abstract concepts in our textbook it's hard to relate them to the problems we're going to see on our homework or on a test or you know wherever we might see them but it's very easy to kind of just work With things we already know and

prove these concepts so if i had something like the square root of 4 and i multiply this by the square root of 9. now we should be able to look at this rule and just kind of match this up and say okay well based on this rule i should be able to do the square root of 4 times 9 so the square root of 4 times 9 which would be equal to what the Square root of 4 times 9 is 36 and we know 36 is a perfect square right it's 6 times 6. so the

answer here would just be 6. now can we prove that this is mathematically valid yes we can we know that the square root of 4 is 2. so i could basically write that this is 2 times the square root of 9 we know that's 3 so 2 times 3 is what it's 6. So the end result here is the same i got 6 this way and i got 6 this way right now i could also have reversed that i could have started out with something like the square root of 36 and i could have factored

it i could have factored into the square root of 4 times 9 and then i could have broke it up and say okay this is the square root of 4 times the square root of 9. i could have said okay well the square root of 4 is 2 times the square root of 9 which is 3 and i could have got 6 that way so you're allowed to go back and forth so let's look at another just quick kind of example let's take two square roots that we know already we know that the square root

of 16 is four so let's do the square root of 16. and let's multiply this by the square root of 25 right we know that's 5. so by this rule again i can do this as the square root of 16 times 25 and 16 times 25 is 400 so this equals the square root of 400 and we should know from the videos we've done so far that the square root of 400 is 20 right so this would be 20. now again there's multiple ways to kind of look at this i could have started out with

the square root of 400 and i could have said well i don't know what that is so let me break that down i could use a factor tree and i could find Out that 16 and 25 are factors of 400 so i could say okay i can break this down into the square root of 16 times the square root of 25 and then these are two square roots that i know right square root of 16 is 4 times square root of 25 is 5. 4 times 5 i know is 20. so again another way to

kind of get that answer all right so before we kind of move on i want to address something that comes up pretty often It's a very very common mistake that you're going to see so in general if you have the square root of a plus b okay notice how there's an addition sign there this is addition this is generally not equal to the square root of a plus the square root of b and there are some exceptions to this if i did the square root of zero plus one this would be equal to the square

root of Zero plus the square root of one right this is going to be equal to one and so is this right so in some circumstances you can make it work but when we have a rule okay that we're going to follow it's got to work for everything it's got to be completely consistent i got to be able to rely on it and i can't rely on this because this isn't going to work in the majority of cases okay if i want to give you a quick example Let's say you had something like the square

root of four plus four this would be what if i simplify this correctly i would just perform the operation under the square root symbol so i would do 4 plus 4 is 8. so this is the square root of 8. and if i approximated this because this is an irrational number i would get about 2.83 Let's say and that's an estimate it's not an exact value but it's close enough to prove our point here if i was able to simplify the way that most students think that i can i could go okay this is the

square root of four plus four and i'll just separate this so this is the square root of 4 plus the square root of 4. well i know the square root of 4 is 2. so this would Be what it would be 2 plus 2 so this should equal 4 if this logic was correct but it's not right because this is the square root of 8 and it's about 2.83 so that's why this is let me put some big red through this this is wrong okay this is incorrect and again a lot of students make that

mistake please don't be one of them all right so let's use the product rule for radicals to do some basic multiplication here you'll see some Problems like this at the beginning of the section so you have something like the square root of 3 times the square root of 7. what i can do is write this as the square root of three times seven so three times seven so this is equal to the square root of 21. so just a way to combine things that make it simpler here's another one we have the square root of

five times the square root of two so i can say this is Equal to the square root of 5 times 2 5 times 2 is equal to the square root of 10. what about the square root of x times the square root of y well even though there are variables involved we're still going to follow the same process so i can write this as the square root of x times y x times y or just the square root of x y remember variables represent numbers so the properties that we use with numbers We use with

variables right because they're just representing those numbers we just don't know what they are what about the square root of 13x times the square root of 11y again just equals the square root of i'll have 13 times 11 times x times y so this equals 13 times 11 is 143 so this would be the square root of 143 x y x y so even if you had more than two of these you still follow the same process So if i had the square root of two times the square root of three times the square root

of five i can write this as the square root of two times 3 times 5. so 2 times 3 times 5 so this is equal to we'll have the square root of 2 times 3 is 6 6 times 5 is 30. so the square root of 30. so pretty easy overall to use the product rule for radicals that's kind of the first type of problem you're going To encounter in this section then you're going to kind of move on and talk about simplifying radicals so this is where you've got to take this rule that you

just learned and kind of apply it so a square root is considered simplified okay simplified when no perfect square exists under the radical sign okay so what do we mean when we say a perfect square remember a number is a perfect square if its square root is a rational Number so 4 is a perfect square because its square root 2 is a rational number 1 4 is a perfect square because its square root one half is a rational number so what i'm going to do here in each case is i'm going to break down the

number under the square root symbol and i'm going to be looking for perfect squares in its factorization so let's take a look at This kind of first one where we have the square root of 63. so you can go off to the side and make a factor tree if you want and say okay well 63 is what it's 9 times 7 i know 7 is prime and 9 is 3 times 3. so 9 is a perfect square so using my product rule for radicals i could break this down into the square root of nine times

seven and then i could further break it down Into the square root of nine times the square root of seven now we just showed that nine is a perfect square right we know it's 3 times 3 so the square root of 9 is 3. so i could put this is equal to this guy right here is just going to simplify to 3. so i'm just going to write a 3 instead of the square root of not and then it's multiplied by the square root of 7 so then times the square root Of 7. so this

3 times the square root of 7 is considered simplified if we look at what's underneath our square root symbol now 7 you can't factor 7 any further it's a prime number right so 7 is not a perfect square so no perfect square exists under that radical sign unless you were to consider something like 1 which we really can't do anything with that right so we would say 3 times the square root Of 7 is the simplified version of the square root of 63. so this is simplified now to prove that to yourself you could punch

up square root of 63 on a calculator and then also punch up three times the square to seven on a calculator and you're going to see that you're going to get the exact same value now before i move forward i want to show you kind of an alternative way to do this So we've seen this method let me kind of erase this and i want to show you something else you can do this is very very common so if i have the square root of 63 i can kind of factor it under my square root

symbol that i create so i could say okay well i know it's 9 times 7 9 is 3 times 3 so i would put 3 times 3 times 7. so without making a factor tree i just kind of factor it underneath the Symbol and then what i do is i look for pairs of numbers in this case i have a pair i have one pair of threes so for each pair that i have i can pull that outside of my square root symbol right because the square root of three times three would obviously be three

so if i have a pair of numbers i can pull one of those out so for each pair i pull One out so in other words i would just circle this and say i can pull one of those out and say i have three times the square root of seven so it's kind of an alternative way to look at that i know that you'll probably see that in some of your textbooks and it's a common strategy that a lot of teachers will show you when they're covering this in class all right let's take a look

at another one so let's say that you ran across something like the square root of 512. so if you're like myself you know two up to like the 20th power because you use it all the time but in an algebra 1 course you probably only know stuff like 2 up to the 6th or 7th power 512 happens to be 2 to the 9th power but you might not know that so if you didn't know it again you've got to have some method like a factor tree to kind of break the number down so let's say

i said 512 Is divisible by 2 it's also divisible by 4 right because it ends in 12. so i could say this is 128 times 4 then 4 is what it's 2 times 2 and 128 is 64 times 2. now once you start getting into some numbers where you can recognize that it's 2 to a certain power like we know 64 is 2 to the 6th power even if you didn't know that once you break down 64 into what 32 Times 2 you should know that this is 2 to the fifth power so you should

be able to stop and say okay well if this is two to the fifth power and i have one two three four more well then really the whole thing would be two to the power of five plus four or two to the ninth power so let me erase this and once you realize that there's a few different ways you can kind of tackle something like this So kind of the traditional way or the way you just saw a minute ago you can under the square root symbol write out nine factors of two so two times

two times two times two times two times two times two times two and then times two so nine factors of two then for each pair of twos so here's one here's a second Here's a third here's a fourth for each pair you're gonna pull out one two right because the square root of two times two would be two right square root of four is two so this is equal to again for each one you're pulling out one so this is 2 times 2 times 2 times 2. then what's left inside is a lonely two So

then all this times the square root of two we know that two to the fourth power or two times two times two times two is sixteen so this would simplify to sixteen times the square root of two is there anything else i can do to simplify this no there's not the square root of 2 is not going to be able to be simplified any further because under that square root symbol there's no more perfect squares right 2 is a prime number it's Not going to factor any further and 2 is just simply not a perfect

square so there's nothing else we can do now another way to kind of tackle this problem you're going to get the same result either way but it's just kind of a quicker way to do it if i saw something like the square root of 512 and i recognize very quickly that this is 2 to the ninth power you could write this as the square root Of 2 to the ninth power now later on when we get to fractional exponents you're gonna really be able to simplify this quickly because i'm gonna teach you that the square

root symbol is like raising something to the one-half power okay and we'll get to this later but i'm not i'm not trying to go too far ahead but for right now we're stuck with kind of this symbol so we have the square root Of 2 to the ninth power now if i had an even exponent on that 2 something like 2 to the 8th power so let's rewrite this as square root of 2 to the 8th power times square root of 2 to the first power or just 2. right everything is equal here i haven't

changed the value of anything so the square root of 2 to the eighth power would be what well think about what i just taught you a minute ago if i had two to the eighth Power and this was raised to the power of one half well i would multiply exponents keep the base the same this is 2 raised to the power of 8 divided by 2 or 2 to the fourth power 2 to the fourth power we already know is 16. and again it's getting a little far ahead of where we are at this point

but i just want you to realize that there's going to be some strategies coming up That are going to really reduce the amount of work you have to do with these problems you know another thing you can do if you didn't have kind of the exponent trick under your belt yet you could just really think about okay well how could i break 2 to the 8th power down into 2 to some power squared that would get me to 2 to the 8th power right you could realize that this is 2 to the 4th power that

amount squared and then we take the Square root of that well then we learned in the last lesson that this cancels with this and i'm left with 2 to the fourth power so kind of any way you think about this this would be 2 to the fourth power times the square root of 2 or again 16 times the square root of 2 as your simplified answer all right what about something like the square root of 48 well 48 i can kind of do that in my head so let me kind of just write this as

the Square root of i know i could do 4 times 12. but one of the things about 12 is it's also got a 4. so i've already got a 4 times a 4 there so i could do what i could do 16 times 3. so i could do square root of 16 times 3. i know 16 is a perfect square this is a perfect square square root of 16 is 4. so i could write this as the square root of 16 times the square root of 3 And then bam i've got my answer i know

square root of 16 is 4 times the square root of 3. now 3 is a prime number can't be factored any further and it's not a perfect square so this is as simple as we can make this and again if you wanted to do this the other way you could say okay well the square root of 48 is equal to what the square root of you could do 4 times 4 times 3 and then Again if you have a pair of fours one of those is going to go outside of the square root symbol and

you'll just have the square root of 3 that's left so kind of either way you want to look at this you're going to get the same answer what about the square root of 12 well twelve again i can break that down into the square root of four times three the square root of four is two so this Would be equal to the square root i'm gonna do of four times three and then i'll write the square root of 4 times the square root of 3. again we know that the square root of 4 is 2

so i can just simply write that so this would be 2 times the square root of 3. and again if you wanted to you could simply say okay well the square root of 12 the square root of 12 is equal to the square root Of 2 times 2 times 3. so i have a pair of twos okay let me circle that so a pair of twos so one of those is going to come out in front of the square root symbol and what's left inside is that 3. so 2 times the square root of 3

kind of either way you do this all right so let's take a look at one more problem before we move on and talk about the quotient rule for radicals which again is kind of the same Thing so we have the square root of 108. so again if i don't know anything about 108 some of you might know its prime factorization off the top of your head others don't right so if i look at it i know it's divisible by 2 i know it's also divisible by four it's what four times 27 so i could put

four times 27. i already know that four is a perfect square so there's something that can be Pulled out here and four i can just factor in a two times two and 27 i know is 9 times 3. so 9 is another perfect square and 3 is prime and then 9 is 3 times 3. okay so once i've kind of done that i can eyeball this at this point and say okay well i know that 4 is a perfect square so 2 would come out 9 is a perfect square so 3 would come out 2

times 3 is 6 so what i'd have here is 6 Times the square root of 3 and i did that just from a factor tree and you're going to get that good at as well but for right now let's kind of take it slow and say this is the square root of let's do 4 times 9 times 3 then we'll break this up into the square root of 4 times the square root of 9 times the square root of 3 And then we can break this up into what square root of 4 is 2 times

square root of 9 is 3 times square root of 3. and then one more time we have 2 times 3 that's 6 times the square root of 3. and again this can't be simplified any further because 3 is a prime number i can't break that down anymore and it's not a perfect square so let me erase this and one last time let me just show you the Alternative way to do it so if i have the square root of 108 this is equal to 1. the square root of let's do 2 times 2 times 3

times 3 times 3. so then again for each pair i'm going to pull one out so this is a pair so i'm going to pull that out this is a pair i'm going to pull that out and then nothing to really pair with that so let me not circle it let me put A little box around it so we can identify that it's different so we'll put the square root of 3 for that so this equals 2 times 3 is 6 then again times the square root of 3. so again either way you do it

you're going to get 6 times the square root of 3. so overall it's not too difficult to simplify radicals once you kind of understand the basics and you have a general idea of what's going on it's Just basic arithmetic so now we want to talk about the quotient rule for radicals this is no more difficult particularly in algebra 1 they're not going to give you anything that's too challenging and we're just going to start by saying if a and b are non-negative again non-negative real numbers and b is not equal to 0 then we have

this the square root of a over b is equal to the square root of a over the square Root of b so all we're doing is we're separating this into the numerator having a square root and the denominator having a square root so in other words if i had something like the square root of 1 4 we know at this point this is equal to one-half but we can break it down and say the square root of 1 4 is equal to the square root of 1 over The square root of 4 this is equal

to what square root of 1 is 1 square root of 4 is 2. so we can see that that is mathematically valid all right so what about something like the square root of 15 over 12 well when you get a problem like this what they want you to do is just go through the steps and in most cases they're going to give you a bunch of stuff that's going to cancel out so you can simplify So this would be the square root of 15 over the square root of 12. but don't just do that once

you've separated it look for ways to simplify it based on what you did in the previous section so i know the square root of 15 is what it's the square root of five times the square root of three okay i can't do anything with either of those But if you look at the square root of 12 i know i can break that up into one i know i can do the square root of four times the square root of 3. now square root of 3 over square root of 3 is what if i gave you

the square root of 3 over the square root of 3. this is the same non-zero number over itself so this is going to cancel and be equal to one okay so these same rules apply so This cancels with this that's a one so now what i'm left with is the square root of five which i can't do anything with over the square root of 4 which we know is 2. so i end up with the square root of 5 over 2 for my simplified answer all right let's take a look at another one let's say

we have the square root of 20 over 64. all right so again what we're meant to Do here is just break this up into the square root of 20 over the square root of 64. now 64 is a perfect square so all i really need to do in that is just take the square root and that's going to be 8. that's simple enough now for the square root of 20 i could break that up as the square root of 4 times the square root of 5. i know the square root of 4 is 2 so

i could just write this as 2 times square root of five over eight so because this is straight multiplication i can cancel common factors so this two could cancel with this eight and give me a four so this would simply be the square root of five over four all right so for the next one we can kind of simplify this two different ways kind of the way we've been doing it so far we could set this up as what the square root of five Over the square root of a hundred twenty-five obviously i can't do

anything with the square root of five but what i can do is something with the square root of 125 i can write this as the square root of 5 times the square root of 25. now we already know that this will cancel with this right square root of 5 over square root of 5 is 1. so i'm going to write a 1 for the numerator There and then the square root of 25 is 5. so let me just write that this is equal to 1 over 5. now in this particular situation i could have gotten

that by just simplifying the fraction to start right if i divide 125 by 5 i get 25 if i divide 5 by 5 i get 1. so if i have the square root of 1 over 25 basically if i think about that at my head the square root of 1 is 1 And the square root of 25 is 5. so that's kind of a quicker way to do that in this scenario and you get the same answer either way what about something like 3 times the square root of 9 over the square root of 25

we know the square root of 9 is 3 so this would be 3 times 3 over the square root of 25 is 5. so this would be 9 5. what about something like 3 times the square root of 2 over 3 times the square root of 18. Well right away again because this is all multiplication we could just cancel this with this to start and just i have the square root of 2 over the square root of 18 which i could say is the square root of 9 times the square root of 2 and i

can break this down further because this is 3 and then this would cancel with this so what i'm left with is a 1 for my numerator and then a 3 for my Denominator so this is going to be equal to 1 3. so all in all simplifying radicals is a very very easy topic we're going to start to get into some stuff that's a little bit more challenging now we're going to start talking more about variables and higher level roots so when radicals appear with variables in the radicand we have to be very careful why

do we have to be careful why do we have all these restrictions well i want You to remember that if you have something like the square root of let's say negative 4 this is not a real number so if i gave you the square root of x can i assign x to be negative 4 no i cannot right because that's not a real number later on at the end of algebra 1 and as we get into algebra 2 and college Algebra we're going to have these things called imaginary numbers that allow us to deal with

negative square roots but we haven't gotten there yet so for right now we're just going to say that the variables that appear in the radicand are going to be non-negative so let's take a look at one that's a little bit more challenging so we have negative 9 times the square root of 112 x to the fourth power so just put your negative 9 out in front And then times the square root of let's try to break this down inside and let's just start with the number part so 112 i know is divisible by 2. it's

also divisible by four right the last two digits would be a twelve so obviously that's divisible by four it would be four times twenty-eight so let's start with that twenty-eight times four four is a perfect square it's two times two 28 is 4 times 7. again 4 is a perfect square it's 2 times 2. so at this point i know the number part the square root of 112 would simplify to what 2 times 2 or 4 times the square root of 7. right so once you start getting good at this you can kind of eyeball

that and say okay i would have 4 outside of the symbol okay and then that would be multiplied by negative 9 and then inside the symbol for the number part and have a 7. but Again let's kind of slow this down a little bit and let's write the square root of 4 times 4 is 16 so let's write 16 times 7 and then we have this x to the fourth power to deal with which we'll think about in a minute now if i look at x to the fourth power x to the fourth power this

is what this is x squared squared or i could think about it as X squared times x squared x to the fourth power is a perfect square it's x squared squared so if i took the square root if i took the square root of x to the fourth power this would be equal to x squared right because again i'm looking for what squared gives me x to the fourth power and the answer to that is x squared and i know it's a little bit more challenging to understand this when you start dealing with variables and

Especially with the exponents getting larger and larger but really it's it's not too difficult if you just kind of break it down you know completely into pairs if i was to say okay i have x to the fourth power and this is x times x times x times x again i have one pair of x's then a second pair of x's so two of those would kind of come outside the symbol right if we did it That way so this would end up being what x times x or x squared so let me erase all

of this and i'm going to just completely break this down now so this is equal to we have negative 9 times the square root of let's put 16 times the square root of let's put x to the fourth power times the square root of 7. we know that we'd have negative 9 times the square root of 16 is 4 The square root of x to the fourth power is x squared and then the square root of 7 we're not going to be able to do anything with so now we just multiply negative 9 times 4

that's going to give me negative 36 and then we have x squared and then we have times the square root of 7. and if you don't completely understand this example don't feel too bad it's It's one of the more complex ones you can pause the video go back restart the problem and you can do it in a different way if it's more comfortable for you you can again let me erase all this you can completely break this thing down and say okay this is negative 9 times the square root of 4 times 4 times 7

times let's say you did x squared times x squared or you can even further break this down All the way and say i have 2 times 2 times 2 times 2 times x times x times x times x times 7. any way you do this you're going to get the same answer so if this is what makes you feel more comfortable this is what helps you to understand it then spend the extra time to do it so this is equal to you have negative 9 times remember for each pair one's going to come out so

this would be a 2 coming out this would be another 2 coming out This would be an x coming out this would be another x coming out and then the 7 would stay in there so let me put a box around that and so this would be times the square root of 7 so this equals what now it's just straight multiplication negative 9 times 2 is negative 18 negative 18 times 2 is negative 36 and then for the variables x times x is x squared and then times the square root Of seven so our answers

match up right this is the exact same as this and this was a little easier to understand right when you first start out again if you need to spend the extra time to break it down completely go ahead and do it and then as you get better as you get a better understanding you're going to be able to speed up the process and go very quickly on these problems all right let's look at another kind of hard example so we have eight Times the square root of 400 x to the fifth power y to the

fourth power so we have eight times the square root i know the square root of 400 is 20. so let's just write the square root of 400 times the square root of x to the fifth power times the square root of y to the fourth power now again i just said the square root of 400 is 20. so let's write equals 8 times 20. so we've got that part figured out what about x to the fifth Power well again there's multiple ways to kind of think about this i can say that i have x to

the fifth power is equal to x times x times x times x times x and then for each pair i'm pulling one out so that means i'd pull out two of these guys that would be x squared that's the first kind of way you can think about it the second way is just thinking about exponents in general so i Know that if i have a square root i'm looking for kind of an even value here right it's got to be something that's going to be divisible by 2. so if i wrote this as x to

the fourth power times x to the first power i know that i could take four and divide it by two right so i know i can get two x squareds out of this guy right so this can be x squared squared times x to the first power And again when you think about it this way and it's very very simple at this point if i have x squared and that's squared and then i take the square root of this whole thing again this is going to cancel with this and i'm left with this just the

x squared so kind of either way you want to look at that you would have x squared here and then times the square root of you'd have x to the first bump Now i still have y to the fourth power to deal with but again when we start thinking about exponents very very quickly you're going to memorize the square root of something raised to the fourth power is that variable squared so the square root of y to the fourth power is y squared so let me kind of scooch this down a little bit and put

that this is times y squared and again why is that the case well if i have y to the fourth power is equal to y Times y times y times y for each pair one's going to come out so two of these come out so y times y is y squared again if you wanted to you could write this as y squared squared and just think about this being under the square root symbol well of course if i had y squared and that was squared if i took the square root of this guy this would

cancel with this and i'd be Left with y squared so you do whatever you have to do to understand this at first again when we get to kind of fractional exponents we're going to have some methods that are going to really speed our work up here but for right now again we do what we have to do and so 8 times 20 is 160 and then times x squared times y squared and then times the square root of x all right so the last thing i want to do and This might not come up in

algebra one you might have to wait to get to algebra 2 to kind of see these problems but it might come up in your course so i want to cover it you have higher roots that you can apply these techniques to so if i have the nth root of a times the nth root of b i'm able to say this is equal to the nth root of a times b so in other words if i had the cube root of 27 Times the cube root of let's say 8 this is equal to the cube root

of 27 times 8 which would be the cube root of 216. now again this is something we can check the cube root of 27 is 3. the cube root of 8 is 2. so 3 times 2 should give me 6. the cube root of 216 is 6. 6 times 6 is 36 36 times 6 is 216. same thing for the Quotient rule the nth root of a over b is equal to the nth root of a over the nth root of b so for example if i had something like let's say the fourth root of

1 16 is equal to the fourth root of 1 over the fourth root of 16 and the 4th root of 1 is 1 the fourth root of 16 is two so this would be one half and of course you can prove this one-half times one-half Is one-fourth one-fourth times one-half is one-eighth and one-eighth times one-half is one sixteenth all right so let's take a look at a few examples here and these are going to be pretty challenging so don't feel overwhelmed if you can't get them right away so we have 5 times the cube root

of 375 times x cubed times y to the 8th power so the first thing i'm going to do is break down 375 and i'm going to be looking for perfect Cubes now and when we talked about a perfect square it was a number whose square root was a rational number so for a perfect cube it's a number whose cube root is a rational number so for example 8 is a perfect cube right it's cube root is 2 that's a rational number or 1 8 is a perfect cube right it's cube root is one-half which is

a rational number so for 375 i notice it ends in five so i can go ahead and divide by five and I find this would be five times seventy-five so five is prime let's circle that and then 75 is what it's 25 times 3 and 25 is 5 times 5. so these are all going to be prime and i have a perfect cube here i have 5 times 5 times 5 so 5 cubed or 125 times 3 is how i could break down 375 so let's start with that so let's say this is 5 times

the cube root Of and i'm just going to write 125 and then times the cube root of 3 and then times if i have x cubed and i take the cube root of that that's just going to be x if i have an exponent of 3 and have an index of 3 those are going to cancel each other out and again we're going to learn that when we get to fractional exponents if i have the cube root of something it's like Raising it to the power of 1 3. so if i had x cubed and

it was raised to the power of one-third this would be equal to x base stays the same multiply exponents three times one-third is one okay but again we'll learn that in a future lesson but right now you just think about it as i have x times x times x that gives me x cubed so x cubed is a perfect cube so let's write the cube root of x cubed And then times now i have y to the eighth power how can we break that down to kind of understand what's going to be able to come

out so now what i'm looking for is a number that's going to be divisible by three so the closest one would be six right so if i wrote this as y to the sixth power times y squared i can think about y to the sixth power is a perfect cube right because if i took the cube root of that i could think About this as what y squared times y squared times y squared that would give me y to the sixth power so i could write this as the cube root of y squared cubed times

the cube root of y squared i know i got a lot of stuff going on in the screen i'm trying to fit it on there but so i'm going to put my equals over here and then 5 times we Know the cube root of 125 is 5. the cube root of 3 i can't really do anything with that let me write that way down here that's just going to stay then i have the cube root of x cubed we know that's x at this point then i have the cube root of y squared cubed so

again this you can about this would cancel with this and i'm left with y squared so this would just be times y squared And then i have this times the cube root of y squared so in the end what this is going to be equal to 5 times 5 is 25 times x times y squared so x y squared and then times the cube root of 3 y squared and all i did was just combine those two that were under the radical symbols so for the next one let me look at four times the cube

root of 108 x to the fourth power y to the Fourth power so if i broke down 108 what would i get so 108 is 27 times 4 4 is 2 times 2 and then 27 is what that's a perfect cube it's 3 times 3 times 3. so let me just kind of stop there and let me write this as 4 times the cube root of so 108 i'm going to write 27 here the Cube root of 27 times the cube root of 4 okay and we can stop there all right now i have x

to the fourth power y to the fourth power again what can i do there to make this easy think about the closest number to four that's going to be divisible by three well that's going to be three right so if i had x to the third power times x to the first power this right here the cube root of that is x right we talked about that in the last problem so This right here would say the cube root of x cubed times the cube root of x and then i can do the same thing

for y to the fourth power let me kind of scooch all this down a little bit so i put times the cube root of y cubed times the cube root of y so you're just breaking these things down to where you're putting in a format where you have a perfect cube and then something that's not going to simplify so i'm just Going to go through now i have 4 times the cube root of 27 i know is 3. the cube root of 4. let's leave that for a second the cube root of x cubed i

know is x right because this would cancel with this and give me this then the cube root of x let's leave that for a minute the cube root of y cubed is y again this would cancel with this and leave me This and then the cube root of y let's write this right now so 4 times 3 is 12 times xy so let's just replace this with 12xy and then times the cube root of you're going to have a 4 that can't be simplified and x that can be simplified and a y that can't be

simplified so i just write them all under one radical symbol right you could spread it out and put cube root of 4 Times cube root of x times cube root of y but really you don't need to do that right this is just a more compact way to write it so 12xy times the cube root of 4xy all right let's take a look at one final problem and then we'll wrap up this lesson so a lot of stuff to take in and before i kind of do this problem i want to just make a little

side note i didn't say this but another error that students make is they might think something like Square root of let's say 4 times cube root of let's say 8 can be combined right i can't the index in each case has to be the same this is a 2 and this is a 3. so this is a no go here i can't do anything with that i have to evaluate those separately and say well the square root of 4 is 2 the cube root of 8 is 2 so it's 2 times 2 or 4. but

i can't you know make one Radical symbol and say you know i've seen students go okay well this is the sixth root of 32 right that's not gonna work that's that's wrong okay so another common mistake all right so let's tackle this last problem here so we have negative eight times the fourth root of 32x cubed y so right off the bat we know none of the variables are going to be able to be simplified how do we know that well if i Have the fourth root of something this would have to at least be

the fourth power for me to pull it outside of the radical symbol so in my simplified version i know i'm going to have x cubed y so i can just kind of box those off and forget about it what can i do with 32 to simplify 32 is what 4 times 8 8 is 4 times 2 and we can already see that what this is 2 to the fifth power which i know most Of you know that at this point so it's 2 to the 4th power times another 2 right or 16 times 2. so

what i can do here is i can write this as negative 8 times the fourth root of 16 times the fourth root of 2 times x cubed times y so everything else that's not going to simplify and then it's very very simple right the fourth root of 16 is 2. so i put equals negative eight Times the fourth root of 16 is two we can go ahead and just simplify that now negative eight times two is negative sixteen and then times the fourth root the fourth root of 2x cubed y 2 x cubed y and

there's nothing else to simplify here right so that's just your answer negative 16 times the fourth root of 2x cubed y hello and welcome to algebra 1 lesson 55. in this video we're going to learn about adding and subtracting radicals so before we discuss adding or subtracting radicals we need to learn about something known as like radicals and before i even tell you about like radicals i want to kind of just remind you that we saw something like this already so i want you to recall when we first started working with terms we talked about

something known as like terms Right so in the case of a single variable like terms would be the same variable raised to the same power so in other words if i had 3x and i had 5x those were like terms but if i had something like let's say 4x squared and i had 5x cubed those were not like terms those were not like terms And why because you had to have the same variable which in this case you have the same variable in this case you have the same variable but the kicker to that is

you also have to have the same exponential power and in this particular case you have a two here and a three here so they're not like terms so again similar to this we have like radicals so like radicals have exactly the same index so exactly the same index And exactly the same radicand so in case you forgot the terminology this right here is your index so this is your index and i know a lot of times we work with square roots and when you see a square root let's say you had square root of four

the index is left blank right but really the index is a two just with square roots because they're so common we just leave it blank and it's understood to be A 2. so this is your index i'm just going to write a 2 in right there now the number or it could be the variable that's underneath your radical symbol is known as a radicand so in this case the a would be called the radicand so this would be the radicand so for us again to have like radicals we Have to have the same index and

the same radicand so that's what we're going to be looking for all right so in this first example here i'm going to look at the index in this case it's going to be a 2 here and a 2 here right if you don't see anything it's understood to be a 2 because we recognize this as a square root then i'm looking at the radicand in this case it's the 6 and in this case it's a Six so we have the same right the same radicand and then we also have the same the same index so

these would be like radicals these are like radicals what about something like 7 times square root of 9 and 2 times the cube root of 9. well these are not like radicals although the radicand is the same in each case this is a 9 and this is a 9. you'll notice that your index here is a 3 and then your index here is a two right if you don't have an index you can write in a two or just understand that it's a two it's for a square root right so the indexes are not the

same in this case so these are not like radicals so not Like radicals all right so pretty easy overall to understand if you have a like radical or not a like radical again i always explain this with that concept of like terms because essentially it's the same level of understanding once you figured out what like terms were it was very very easy to combine them right using the distributive property so when we talk about like radicals it's going to be the Same thing it's going to be very very easy to combine them using your distributive

property so when we have like radicals we add or subtract using the distributive property now before i even do this example let me just show you real quick with something that you already know if i had something like 2x plus 3x i know that that's 5x but at this point we might have Forgotten that we started out by saying that we could factor an x out and write this as x times the quantity 2 plus 3 right because x was common to each so if i factored it out i'd have a 2 left here and

a 3 left here so that's where i got this 2 plus 3. now if i did that operation 2 plus 3 i'd get 5 and i'd be multiplying by x so i get 5 x so we kind of shorten this process up by Saying okay i know that the x is common to each of these so really all i need to do is add the coefficients and then slap an x at the end so i would say 2 plus 3 is 5. slap an x at the end i get 5x or if i was adding

something like let's say 5x cubed plus 7x cubed i know these are like terms the x cubed is just going to be there right same variable raised to the same Power that's going to be there i just add the coefficients so 5 plus 7 is 12 so i have 12x cubed so we already know how to execute that process and it's basically going to be the exact same thing with radicals so if i look at this first example here a lot of you can kind of guess what to do so i have 5 times the

square root of 10 plus 2 times the square root of 10. so the first thing you do is check to make sure that you have Like radicals so we have the same index they're both square roots and then we have the same radicand this is a 10 and this is a 10. so we're good to go there so essentially all i'm going to do is add the numbers outside here so 5 plus 2 would give me 7 and then it's times that common square root of 10. so 7 times the square root of 10. you

just think of it as i have 5 times some quantity plus 2 of the same Quantity that gives me 7 of that quantity and that quantity is the square root of 10. right if i had 5x plus 2x this would be 7x right it's no different what we're doing here and if you wanted to use your factoring that you've learned to this point to completely break it down you could do that as well i have a square root of 10 that's common to both so i could factor that out i would have the Square root

of 10 and then inside of parentheses that i have a 5 plus a 2. now what does this equal well i have square root of 10 times 5 plus 2 is 7. so times 7 and i could rearrange that and write 7 times square root of 10 like that so whatever you need to do to make this make sense for you again the easiest way to do it is just To think about what's outside of your radical symbol you're adding those together or subtracting them depending on your operation that you're performing and then what's common

here in this case is square root of 10 is just coming along for the ride so what about something like 3 times the square root of 8 minus the square root of 8. well the first thing is to realize remember if we saw something like 3x minus x we would say okay there's There's no coefficient for x so it's understood to be 1. well it's kind of the same thing here it's we have an understood 1 that's multiplying that square root of 8. so if i go to perform this operation here all i need to

do is think about the numbers in front so 3 and then we can think about that as negative 1. so 3 minus 1 would be 2 and then times the square root of 8. now are we good to go if we leave our answer like this No we're not right we learned in the last lesson that we can simplify our radicals so in the previous example we ended up with an answer of 7 times the square root of 10. i can't simplify that any further because what's underneath that radical symbol a 10 is 5 times

2. there's no perfect square underneath there that i can kind of work with so i've got to leave it in that format in this example if i factored 8 i could make it 4 times 2 4 Is a perfect square so i got to get that out of there so let's rewrite this as 2 times square root of 4 times square root of 2. i know the square root of 4 is 2. so essentially what i'm going to have is what 2 times 2 or 4 times the square root of 2. so you've got to

put that extra step in there we already know how to simplify so once we learn how to add and subtract we're going to throw that extra step in of Simplification that we've already learned all right what about something like negative 2 times the square root of 3 minus 5 times the square root of 3. so i'm just looking at this and really this right and the negative 5 there you can think about it as plus negative 5 or minus 5 again whatever makes you feel more comfortable but essentially i just think about negative 2 plus

negative 5 and that's going to give Me what negative 7 and then i have square root of 3 that's just coming along for the ride so this equals negative 7 times the square root of 3 and there's nothing i can do to really simplify that any further all right so the previous examples were kind of the easy ones the ones you get at the very start of this section then you start getting to stuff that's a little bit more complicated so i have here that in some cases we will Need to simplify one or more

radicals to get like radicals so you might get some problems that involve addition and subtraction where you look at them and it looks like you don't have like radicals but then you do a little simplifying and it turns out that you do have like radicals so for example here i have 3 times the square root of 8 minus 3 times the square root of 18 plus 3 times the square root of 8. so obviously these are like radicals it's the exact same thing so they can be added together but this is not a like radical

with anything i have the square root of 18 that's that's not going to work right because the radicand is different but if i go ahead and perform the operation that i can to start i know 3 plus 3 is 6 so i'd have 6 times The square root of 8. so 6 times square root of 8 and then minus 3 times the square root of 18. so what i want to do now is just think about simplifying each one of these so if i think about the square root of 8 i know i could write

that as the square root of 4 times the square root of 2. and this is 6 out here so square root of 4 is 2. so i have 6 times 2 times the square root of 2 or 12 times the square root of 2. so let's write that this equals 12 times the square root of 2. now for this guy right here i've got what i've got 3 times square root of 18 i could write the square root of 9 times the square root of 2. Now 9 is a perfect square so the square root

of 9 is 3. so i could write this as a 3 here 3 times 3 is 9. so if i replace all this i really have minus 9 times the square root of 2 and voila we can see that we have the same radicand now a 2 in each case and the same index also a 2 in each case because this is square roots we're dealing with so i can do this operation now i would just do 12 minus 9 that would give me 3 And then times the square root of 2 right and i

can't simplify that any further so that would be my answer all right let's take a look at another so we have negative 3 times the square root of 5 minus 2 times the square root of 5 plus 2 times the square root of 45. so as we look at it right now we don't have like radicals all the way across but i can perform operations with these two radicals I could do negative 3 minus 2 that's negative 5 times the square root of 5 and then i have plus 2 times the square root of 45.

now i can simplify this guy and let's see what would we get i know the square root of 45 can be written as the square root of 9 times the square root of 5. so let's write that so negative 5 times square root of 5 plus 2 times the square root of 9 times the Square root of 5. we know the square root of 9 is 3. so let's write this as negative 5 times the square root of 5 plus 2 times again square root of 9 is 3 times the square root of 5. so

we're going to find is that we do have like radicals we're going to have negative 5 times the square root of 5 plus 2 times three is six so this is six times square root of five And now i can just do negative five plus six that's going to be one so this would end up being one times the square root of five or you could really just write the square root of five right one times anything is just itself all right let's take a look at another one so we have 2 times the square

root of 54 plus 2 times the square root of 24 minus 3 times the square root of 18. now in this particular case we don't have any like radicals to start So i want to break down everything completely and see what i got so i know the square root of 54 if i think about 54 i know it's 9 times 6 so let me do 2 times the square root of 9 times the square root of 6 and then plus i've got 2 times square root of 24. i know 24 is 4 times 6 so

let me do square root of 4 times square root of 6. and then minus i have 3 times square Root of 18 so i could do square root of 9 times square root of 2. okay so put equals here square root of 9 is 3. so 2 times 3 would be 6. so this would be 6 times the square root of 6 and then plus square root of 4 is 2 so 2 times 2 is 4 so 4 times square root of 6 and then minus we have 3 here square root of 9 is 3

so 3 times 3 would be 9. so this would be minus 9 and then times the square root of 2. now i have like radicals for these two this one is not i have square root of 2 here i have square root of 6 here and here so you've got to have like radicals to combine these okay you can't just combine them if they're not like radicals i see that mistake all the time don't do that So what we're going to do is just combine what we can so in other words i would say 6

plus 4 is 10 so this would be 10 times the square root of 6. can't make that any simpler than minus 9 times the square root of 2. this is all that i can do i can do no more here okay now this is kind of similar to if you had something like let's say you simplified as much as you Could and you had 5x squared minus 3x let's say these aren't like terms so i can't go any further that's my simplified answer it's the same thing here they're not like radicals in this case so

i can't simplify any further so let's take a look at one more and if you understood the main lesson on radicals and you understood how to simplify i think that this particular topic within the topic of radicals is pretty simple overall So if i look at four times the fifth root of 96 plus two times the fifth root of 128 minus four times the fifth root of negative eight and then minus two times the fifth root of four i don't have any like radicals to start so i've gotta look and see what i can do

so let's start with just 96. if i think about 96 i know that would factor into 32 times 3. now 32 is 2 to the fifth power So i already can see that i can do what four times the fifth root of 32 times the fifth root of three and just to kind of save myself a little time i could just kind of simplify this as i go again 2 to the fifth power is 32 so i know the fifth root of 32 is 2. so i can put that this is going to be 2

and of course If i multiply 4 times 2 i get 8. so let's just again save ourselves a little bit of time and put eight times the fifth root of three okay so let's try to simplify this one now i think at this point that we all know that 128 is two to the seventh power so if i have 2 to the 7th power and i have something that's an index of 5. again you got to kind of think about these things to speed yourself up I would think about this as 2 to the 5th

power times two squared to the fifth power is 32 so i'm gonna break this up and say i have two times the fifth root of 32 times the fifth root two squared would be what that's four so i already know that the fifth root of 32 is two two times two is four so this is going to be four times four times the fifth root of four So now i want to think about negative four times the fifth root of negative eight is there anything i can do to simplify there well it looks like i

can't but actually i can do a little trick with that now if you think about eight it's what it's it's two cubed now 3 isn't high enough to match up with this index of 5 to be able to simplify In that way i can't do anything with the 8 but i can do something with the negative right if you think about that as what negative 1 times 8 okay and then the negative 1 raised to the fifth power would give me negative 1. so i can think about it as saying that i can actually pull

out a negative 1 from this and let me show you how we go about doing that so let's write Plus negative four and then times i'm going to put the fifth root of negative one and then times the fifth root of eight okay and again i'm legally allowed to do that per the product rule for radicals i haven't done anything illegal now what is the fifth root of negative one it's negative one right if i took negative one And i multiplied by itself five times i'd get negative one so i can erase this and just

put negative one here now i can simplify now by saying negative four times negative one is positive four so this would be positive four times the fifth root of positive eight now okay so not a simplification in the normal way that you'd expect but a way to kind of clean it up a little bit Then we have minus two times the fifth root of four and nothing i can do with that right if i think about four it's just two times two can't really do anything with that so i only have like radicals here and

here so all i'm going to be able to do in this scenario is combine those so if i look at this i'm gonna have what eight times the fifth root of three and then plus i can do four minus two that's two Times the fifth root of four and then plus 4 times the fifth root of 8. and again these are not like radicals so there's nothing else i can do to combine or simplify any further hello and welcome to algebra 1 lesson 56. in this video we're going to learn about rationalizing the denominator all

right so before we kind of move on in algebra 1 we need to learn what a simplified radical is just like when we work with fractions we Always had to simplify it before we reported our answer we're going to be expected to do the same thing when we work with radicals so there's three main things you have to do when you work with radicals so i have here that the simplified form of a radical and this is kind of the first thing the radicand remember that's your number or your variable or could be your expression

that's under your radical sign This contains no factor except for one that is a we have perfect square in the case of a square root perfect cube in the case of a cube root perfect fourth in the case of a fourth root and then so on and so forth so what we're basically saying is that if you saw something like the square root of 8 8 has a factor of 4 right 4 is a perfect square so because this is a square root this is not simplified because one of The factors of 8 is a

perfect square i could write this as the square root of 4 times the square root of 2 and then the square root of 4 is 2 so i can write this as 2 times the square root of 2. so we've been doing this for a few lessons now so we already understand this concept but this goes for higher powers as well if i had something like let's say the cube root of 16 i know that 8 is a factor of 16 and 8 is a Perfect cube so i could write this as the cube root

of 8 times the cube root of 2 and then the cube root of 8 is 2 so this is 2 times the cube root of 2. so that's one of the things you're looking for when you're simplifying a radical all right the next thing and we've already learned how to kind of deal with this scenario the radicand cannot contain fractions okay so if you have something like Let's say the square root of 1 4 you've got to split that up and say that this is equal to the square root of 1 over the square root

of 4 and of course this is something you can simplify square root of 1 is 1 square root of 4 is 2 so that becomes one half but you can't leave it in this format the radicand again this is the number or the variable or the expression that is underneath the radical symbol cannot Contain fractions when you report your simplified answer then there's the last thing here that is there is no radical present in any denominator so we can't have something like let's say 2 over the square root of 5 or something like 6 over

the square root of 7. they don't want that that's considered not simplified so basically the purpose of this lesson is to teach you how to deal With these scenarios right here we've already learned how to simplify this and then also kind of the first thing we talked about which is you know pulling out a perfect square or pulling out a perfect cube so we're focusing on this in this lesson so what do i do when i come across something like three over the square root of two again we cannot have a radical in the denominator

So this right now is called an irrational number this is an irrational number we don't want this so what could we possibly do to get rid of it well i didn't say that i couldn't have a radical in the numerator if i just don't want one in the denominator all i need to do is perform a little trick that we learned back when we studied fractions i'm going to multiply the denominator by The square root of 2 and i'm going to do the same thing to the numerator remember the square root of 2 over the

square root of 2 is a complicated form of 1. same non-zero number over itself it's one okay now what happens when i multiply the square root of two times the square root of two that gives me two so the denominator is now a rational number Okay it's a rational number henceforth the term rationalizing the denominator we're taking it from an irrational number and we're making it into a rational number so again that's rationalizing the denominator that's where the term comes from then i just multiply 3 times square root of 2. can't make that any simpler

i just write 3 times square root of 2. now the value of this and this Is the same bunch up on a calculator you're going to get the same irrational number but it's just considered simplified to have it in this format versus this one all right let's jump through and look at some examples here and there's some tricks that you need to know when you're dealing with this it's a very simple process overall so let's say i saw something like 9 over the square root of 12. do i want to just say okay I'm going

to multiply this by square root of 12 over square root of 12. yes i can do that but i'm going to show you in a minute that there's an easier way to do it so this times this would give me 12. right square root of 12 times square root of 12 is 12 and then i have 9 times square root of 12. the problem is here i can simplify the Square root of 12 right one of its factors is a perfect square right 4 is so i could really write this as 9 times the square

root of 4 times the square root of 3 over 12. i know that the square root of 4 is 2. so let's erase this and just put a 2 here now between the numerator denominator i could do some cancelling right if i Think about 9 and i think about 12 they each share a factor of 3. so let's cancel this with this and put a 3 here and a 4 here and then let's cancel between the 2 and the 4 and leave a 2 and basically i'm going to have a 3 times the square root

of 3 in the numerator over a 2 in the denominator so 3 times the square root of 3 over 2. now you can do it that way we've Rationalized the denominator we've reported a simplified radical but there's an easier way to kind of look at this let's say i erase this and start the problem over what if before i begun i simply said okay this is the square root of 12. now i know that 12 has a perfect square involved in it square root of 4. if i simplify that before i begin it's Going to

lessen my work so let's put equals 9 over i can say square root of 4 times square root of 3 and i know the square root of 4 is 2. so let's just put 2 times the square root of 3. now because this right here is a rational number i don't really need to worry about it all i need to worry about is this irrational number this square root of 3. so when i multiply now i'm just Going to multiply the numerator and denominator by the square root of 3. and what's going to happen is

square root of 3 times square root of 3 is 3. so that's 3 and then you still have your 2 there so 2 times 3 let me just write it like this would give you 6. i'm not going to write that out because there's some things we can cancel obviously and then we have 9 times square root of 3. so 9 times square root of 3. and then this will cancel with this and Give me a three so kind of either way i do that i end up with three times square root of three over

two but i feel like if you simplify before you begin if you just attack the problem and simplify before you know you just start jumping in there you end up with less work in the end all right let's take a look at another one so we have 2 times the square root of 64 over the Square root of 24. so let's simplify so we have 2 times the square root of 64 is 8 right we should know that at this point so 2 times 8 over the square root of 24 i'm going to write is

the square root of 4 times the square root of 6. now the square root of 4 is 2. so without even doing that i know i can cancel this with this square root of 4 and 2 are the same so that's gone So really what i have here is 8 over the square root of 6 and i'm simply going to multiply the numerator denominator by the square root of 6 and that's going to give me what i'll have 8 times the square root of 6 over square root of 6 times square root of 6 is

6. and then i can cancel one factor of 2 between numerator and Denominator that's going to give me a 4 here and a 3 here and so i'm going to end up with 4 times the square root of 6 over 3 as my simplified answer all right let's take a look at square root of 5 over 3 times the square root of 8. again we want to simplify before we begin so i know square root of 5 i can't do anything with that 3 times the square root of 8 i know the square root of

8 at this point is square Root of 4 times square root of 2. so square root of 4 times square root of 2 then i know this is 2 right here so let's write this as the square root of 5 over 3 times 2 would be 6. so 6 times square root of 2 and now all i need to do is multiply the numerator and denominator here by the square root of 2. so the square root of 5 times the square root of 2 is square root of 10 Over we have 6 times square root

of 2 times square root of 2 square root of 2 times square root of 2 is 2. 2 times 6 would be 12. so i get square root of 10 over 12. and again that's as simple as you can make that what about something like the square root of 3 over the square root of 7. so i can't simplify either of these so basically all i'm going to do is look at My denominator if i have the square root of 7 there i'm going to multiply by the square root of 7 over the square root

of 7. square root of 7 times square root of 7 is 7. square root of 3 times square root of 7 is square root of 21 so i get square root of 21 over 7. what about something like the square root of 48 over the square root of 42 again i want to simplify before i kind Of start rationalizing the denominator a lot of students just jump right in and go okay i want to do square root of 42 over square root of 42 that's going to give you more work in the end right you

don't want to do that so let's let's see if we can simplify this so square root of 48 i know 48 is 16 times 3 so 16 is a perfect square so we do square root of 16 times square root of 3 and for 42 I know 42 is 7 times 2 times 3 so no perfect squares in there right i got a 7 i got a 2 and i got a 3. so really it's just square root of 42 right just square root of 42. so this right here i can simplify i know that

would be 4 so this becomes 4 times square root of 3 over the square root of 42. so now we're just going to multiply the numerator and denominator by that square root of 42. And i know we end up doing that so you might say well why did you go through all the process of you know trying to simplify it it just saves yourself a lot of work in the end right you just want to take that extra time and try to simplify because even reducing this a little bit is going to knock some of

your workload down so if i have the square root of 427 square root of 42 i know that's going to be equal to 42. so that's my denominator For my numerator i'm going to have 4 times the square root of 3 times the square root of 42. now 3 times 42 is 126. but if i think about 126. i want you to remember that 42 was what 42 was 7 times 2 times 3. so you're gonna have a perfect square involved in this it's going to be 9. so i can think about this as 42

times 3 42 again is 7 times 6 6 is 3 times 2. so i could really think about this as 14 Times 9. so let's write this as the square root of 14 times the square root of 9 and then the square root of 9 is 3. so 3 times 4 would be 12. i'll have 12 times the square root of 14 over 42 and then i've still got some more simplifying to do 12 and 42 Share a common factor of 6 right so 12 divided by 6 would be 2 42 divided by 6 would

be 7. so in the end my simplified form is going to be 2 times the square root of 14 over 7. and again lots of different ways to kind of attack these problems if you wanted to start out by saying okay times square root of 42 over square root of 42 that's fine you can do that but i really think if you spend the extra time to try To simplify in the beginning it saves you work in the end all right so now that we've kind of gone through the basic scenarios and this isn't a

very difficult topic right we've kind of blown through those we may be faced now with multiple terms in the numerator or denominator and we'll cover more complex scenarios in the next lesson we're going to do a lesson called further operations with radicals and we're going To cover some more complex scenarios there so what if you saw something like 7 plus 2 times the square root of 2 over 2 times square root of 13. well this is already pretty well simplified right i can't do anything with square root of 13. can't do anything with square root

of 2 can't add here because i don't have like radicals so what am i going to do well all i need to do is multiply the denominator here by the square root Of 13. if i do that i'm going to rationalize the denominator but i've got to do it to the numerator as well now pay close attention here the main mistake that students make is they'll just multiply this by one of these terms you've got to use the distributive property and multiply this by both terms that are in that numerator so in other words i've

got to have the square root Of 13 that's multiplied by 7 and then plus the square root of 13 that's multiplied by 2 times the square root of 2. so square root of 13 times square root of 2 is square root of 26 so my numerator becomes 7 times square root of 13 plus 2 times the square root of 26. now in the denominator square root of 13 Times square root of 13 is 13 so you're going to have 2 times 13 or 26 simplifying is a little bit more challenging here because remember we can't

just go through and cancel stuff we've got addition the biggest mistake students make in algebra 1 is they start going through and saying oh i can cancel this with this this is 13 now no you can't do that remember you cancel common factors so In order for me to be able to cancel this with something up here i would have to factor something out for it to cancel and there's nothing that's common that i'm going to be able to factor out i can't factor out a 2. this is a 7 this is a 2. so

there's not really anything i'm going to be able to do to simplify this further i just report my answer at this point and say i have 7 times the square root of 13 plus 2 times square root of 26 and this is over 26. what about if we come across negative 2 plus the square root of 6 and this is over the square root of 22. again if i look at my problem and i don't see anywhere where i need to simplify right nothing can be simplified at this point well i'm just thinking about rationalizing

the denominator so i have the square root of 22 down there so i'm going to multiply the numerator by square root of 22 And the denominator by the square root of 22 and remember i'm going to multiply this by both terms so my numerator becomes what negative 2 times the square root of 22 plus we'll have square root of 6 times square root of 22 that'll be the square root of 132 and of course we can simplify that and we'll do that in a minute and then this is over We have the square root of

22 times the square root of 22 that's going to give me 22. now this right here you should know can be simplified i can't do anything with this one because 22 is what it's 11 times 2 so there's no perfect square as a factor of 22. but for 132 a lot of you know 132 is 12 times 11. so if i have 132 and that's 12 times 11 12 is what is 4 times 3 so 4 is a perfect square so i could really write this as 4 times 33. so let's go ahead and do

that so let's say this is equal to negative 2 times the square root of 22 plus 33 times 4 so i'll put the square root of 33 times the square root of 4 over 22. okay so now this equals i've got negative 2 times the square root of 22 Plus i know the square root of 4 is 2 so i'm going to put 2 times the square root of 33 over 22. now is there anything i can do to cancel further yes there is because in this case i have a 2 here and i have

a negative 2 here but i also can think about that as negative 1 times 2. so if i was to factor a two out okay From each term in that numerator so in other words if i said i have two and then inside the parentheses i'd have that negative is still there so negative square root of 22 and then plus factor the two out from there so just square root of 33 and this is over 22. now this is multiplication here 2 is a factor and this whole thing inside the parentheses here negative square root

of 22 plus square Root of 33 is a factor so now i can cancel common factors so this can cancel with a factor of 2 down here and so this is 11 okay very important that you understand the difference between when you can cancel and when you can't again that's a very very common mistake in algebra one so then this equals what's left in the numerator after i cancel is a negative square root of 22 plus square root of 33 Over 11. all right so let's take a look at a few now with some higher

roots involved not much more difficult just a little bit more thinking involved so if i have something like 10 over the cube root of 4 what do i do a lot of people just come through and they go okay well i'm going to do cube root of 4 over cube root of 4 and okay 10 times cube root of 4. over 4 times 4 is 16. so this would be the cube root the cube root of 16. now here's the problem with doing that particularly when you get into higher roots this is not simplified the

cube root of 16 16 is not a perfect cube this gets broken up into what the cube root of eight times the cube root of two so this would be two times the cube root Of two i'd still have a radical in the denominator so this is wrong this is wrong okay what you need to think about here is what do i need to do to get to a perfect cube so in other words if i have 4 here and 4 is 2 squared well i need another 2 another 2 to get to a perfect

cube so what i want to multiply by here in this case is the cube root Of 2 over the cube root of 2. and what's going to happen is i'm going to have 10 times the cube root of 2 in my numerator over 4 times 2 is 8. so this would be the cube root of 8. 8 is a perfect cube so i would get 10 times the cube root of 2 Over cube root of eight is two and then i can simplify by cancelling a common factor of two between here and here so this

is a five so my final answer there will be five times the cube root of two now another thing you could have done and i don't i don't advise this but a lot of students will end up doing this you could have just taken your opening scenario and done it twice so in other words i Can say well i've got cube root of 4 there so i can just multiply this by cube root of four times cube root of four and i see a lot of students doing this and i don't know why you wouldn't

just do the couple extra seconds to get the you know easier calculation but if you want to do this you can it's going to give you the right answer it's just a little bit more work So if i multiply the cube root of 4 times the cube root of 4 times the cube root of 4 i would end up with what the cube root of 64. that's my denominator and then in the numerator i would have 10 times the cube root of 16. now a lot more work to simplify when you do this The cube

root of 16 is the cube root of 8 times the cube root of 2. so let me write 10 times the cube root of 8 times the cube root of 2. now we know this is 2 so we'll simplify that in a minute the cube root of 64 is 4. so let's write that as 4. okay so now i have 2 times 10 which is 20 times the cube root of 2 over 4. so now i've done all that extra work and I'm going to get the same answer 20 divided by 4 is 5. so

either way i'm going to end up with what 5 times the cube root of 2. so again if you want to do this you can but it's going to just take you longer to get the final simplified result in the end so let me just say this as a rule i would like for you as a student to try to get in the habit of multiplying by the smallest radical possible because This is going to cut down on the amount of simplification you have to do that's what you always want to think about how can

i multiply by the smallest radical possible so let's take a look at one final problem and i want you to just think about it just pause the video and think about what's the smallest radical i can multiply by to accomplish this all right so if you've tried this you would see okay you have 6 over the 4th root of 27. what is 27 just start with that well it's what it's 9 times 3 and then 9 is 3 times 3 so really it's 3 cubed now if it's 3 cubed if it's 3 cubed and i

have an index of 4. what does that tell you i need to multiply by one more 3. 3 cubed times 3 to the first power will give me 3 to the 4th power so that's a Perfect 4th right 3 to the fourth power is 81. that's what we'd be looking for here as the smallest radical which would be the fourth root of three to multiply by to simplify this thing so i want to multiply the numerator by the fourth root of three and the denominator by the fourth root of three and this will equal what

in the numerator just six times the fourth root Of three in the denominator the fourth root of 27 times the fourth root of 3 is the fourth root of 81 and i know the fourth root of 81 is 3. so this equals 6 times the fourth root of 3 over again this is 3. now i can cancel here 6 and 3 6 divided by 3 is 2. so this would be 2 and i end up with what 2 times the 4th root of 3. Now again if you wanted to do this a lot of students

will look at that right there and say oh what am i going to do so they'll multiply by the fourth root of 27 four times okay please don't do stuff like that it just it's way too much work it's too complicated take five seconds and build a factor tree and figure out what is the smallest radical that i can multiply by to accomplish this goal that's going to save you so Much work in the end hello and welcome to algebra 1 lesson 57 in this video we're going to learn about further operations with radicals so

before i begin this lesson i want to revisit the simplified form of a radical that we learned in the last section so kind of the first rule that we encountered was that the radicand remember that's your number or your variable or your expression that's under The radical symbol okay so this contains no factor other than one which is and then we have a perfect square in the case of a square root a perfect cube in the case of a cube root a perfect fourth in the case of a fourth root and then so on and

so forth so as an example of this i typically use the square root of eight so if i have something like the square root of eight at this point in algebra we all know That eight is four times two and we should know that four is a perfect square right it consists of two times two so this is not considered simplified because it's a square root and one of its factors is a perfect square and one of its factors again is 4. so i can put that this is equal to and i can split this

up as the square root of 4 times the square root of 2 and i can say that the square root of 4 represents a rational number 2. right It's a perfect square so what they want me to do in order to simplify this is just to replace this with a 2 right replace it with its rational number so i'd have 2 times the square root of 2 which i can't simplify so there's no difference between this guy right here 2 times square root of 2 and the square root of eight in terms of its value

right punching both in on a calculator you get the same thing but if You turn this in on a test you're probably not going to get full credit because it's not considered simplified okay and let me do an example with a perfect cube let's say we had the cube root of 16 okay that's going to keep it nice and simple i know that this consists of what 16 i could do is 8 times 2 and 8 is a perfect cube so again one of the factors Of 16 is a perfect cube so i would break

this up and say this is the cube root of 8 times the cube root of 2 and again this represents a rational number the cube root of eight is two so i'm simply going to write this as two in the simplified version and then i'm gonna keep what doesn't simplify so times the cube root the cube root of two so 2 times the cube root of 2 is simplified the cube root of 16 is just unsimplified again they represent the same value and you could go on and on when you talk about a you know

a fourth root and look for a perfect fourth and one of the factors of the radicand right or you could do that with a fifth root or sixth you know so on and so forth all right so kind of the next thing we talked about was that the radicand cannot contain fractions so if i saw something like let's say the Square root of 2 over 25 something like that this would not be considered simplified so what they want you to do is break this up using your quotient rule for radicals so this would be what

the square root of 2 for the numerator over the square root of 25 for the denominator and the square root of 25 happens to be a rational number right it's 5. so let's just replace that with a five so we'd have The square root of two over five and again this would just be the simplified version of that then kind of the last thing that we talked about was that there is no radical present in any denominator so you'll recall that this was basically the focus of the last section where we learned about rationalizing the

denominator so if i had something like 2 over the square root of 7 we said that this right here was an Irrational number right square root of 7. so we want to rationalize this denominator in order to do that i'd multiply both the numerator and denominator by the square root of 7. and so square root of 7 times square root of 7 would be 7 and i went from an irrational number square root of 7 to a rational number which is 7. so that's where you get Rationalizing the denominator from then in the numerator you

just have 2 times the square root of 7. so again these two 2 over the square root of 7 and 2 times square root of 7 over 7 they represent the same value it's just that this one is simplified and this one is not so if you report something like this on your test as an answer you might get a wrong answer you might get points taken away your teacher might Forgive you and you know give you full credit but in a lot of scenarios you're going to at least lose some points for not simplifying

and reporting your answer in this way it's just like if you reported a fraction something like two-fourths right you don't want to do that you want to simplify and say it's one-half so what we're going to do now is we're going to use all the knowledge we've gained so far and we're just going to look at some typical problems that You will encounter with radicals i'm keeping it very simple here for algebra 1. when we get to algebra 2 college algebra we're going to look at some very very difficult problems so for right now let's

just keep it simple and let's start out with the square root of 5v times the quantity the square root of 10 plus 3v so what i'm going to do here because these are not able to be combined inside the parentheses here i'm going to use my Distributive property to simplify so i'm going to multiply this times this and also times this and we're going to see what we can simplify so we'll have the square root of 5v multiplied by the square root of 10 and then plus the square root of 5v multiplied by 3v now

can i do anything to simplify here well yes i can the square root of 5v times the square Root of 10 i could write that as what the square root of 5 times 10. think about breaking this down completely square root of 5 times square root of 5 times square root of 2 and then times square root of v then plus over here the square root of 5v i could break that down but really i'm not going to be able to do anything with that 5 is prime v is to the first power Nothing i

can really do so then times 3v i can do stuff here because the square root of 5 times the square root of 5 is 5. so i can simply write that and say i have 5 times the square root of what's left square root of 2 times the square root of v we combine that and say we have the square root of 2v then plus i have the square root of 5v times 3v Nothing i can do so i just kind of change the order and i put 3v out in front that's more traditional times

the square root of 5v now is there anything else i can do to simplify i want you to think back to when we added and subtracted radicals we had to have like radicals okay these are not like radicals if i look at this guy right here and i look at this guy right here they're not like Radicals why well they have to have the same index which in this case they're each square root so that's a check but then they have to have the same radicand this is a 2v and this is a 5v those

are not the same okay so they're not like radicals where students get confused on this is they're used to looking at like terms so if i had 2v plus 5v Those are like terms i can combine that and get 7v but that's not the case here it doesn't it doesn't work that way with radicals for like radicals you have to have the exact same radicand so if this was 5v and this was 5v you're good to go but in this case we have 2v and 5v not good to go so we don't have like radicals

so our simplified answer here is just 5 times the square root of 2v plus 3v times the square root of 5v all Right let's take a look at another one so we have the square root of 10n times the quantity 4 plus the square root of 2n so again i'm going to use my distributive property here and i'm going to start with the square root of 10n times 4 then plus the square root of 10 n times the square root of 2n so in this first case there's not going to be anything i can do

to simplify Why because i have a 4 out in front i'm just going to switch the order of that to make it more traditional underneath the square root symbol i have a 10 that factors in a 5 times 2 there's no perfect square in there right so there's nothing i can do then i just have n that's n to the first power nothing i can do there so this first part here or the part on the left will just be 4 times the square root of 10 n hey nothing else i can do then plus

here there's going to be something i can do let's let's completely break this down let's say i have the square root of 10 is 5 times 2. so square root of 5 times square root of 2 times square root of i have that n there then times square root of 2n i'm going to write square root of 2 times square root of n Now what i have here is i have square root of 2 times square root of 2 and i also have square root of n and square root of n so what i want

to do let me report this first part so 4 times square root of 10 n plus square root of 2 times square root of 2 is 2. square root of n times square root of n is going to be n so i'd have 2n there And then what's left just the square root of 5. let me kind of extend this a little bit and these are not like radicals and i can't simplify any further so i'm just going to report my answer as 4 times the square root of 10 n plus 2n times the square

root of 5. all right so for the next one i have the square root of 10x times the quantity negative 4 times the square root of 2x plus 2x So again i'm going to use my distributive property here so this times this and then plus this times this all right so we're going to have that negative 4 out in front times the square root of 10x times the square root of 2x then plus i'm going to have i'm going to put 2x out in front times the square root of 10x Now if i think about

10 again it's 5 times 2. so i know that i'm going to have what i'm going to have a 2 here and a 2 here so a pair of 2s 1 2 comes out right i can kind of write it in a different way remember we looked at that in the previous lesson where we first started learning how to simplify so if i did negative 4 out in front Times the square root of i'm going to go ahead and write this as 2 times 2. i've got a 2 here and i've got a 2 when

i factor 10 and then times 5 and then times x times x so i can think about it like that and then plus 2x times the square root of 10 factors into 5 times 2 but i don't really want to do it there because it's not going to help me let's just leave it as 10. and then x i'm not going to Factor that i can't i can't do anything with that it's just x okay so then each pair here i'm going to pull one out right so 2 times 2 square root of 2 times

square root of 2 is 2. so for each pair 1 comes out and then if i have x times x here 1x will come out square root of x m squared of x is x so i'd have negative 4 times Again one of these is going to come out one of these is going to come out so negative 4 times 2 times x and then times what's left under that square root symbol which is 5 so times square root of 5 and then plus 2x times square root of 10x let me scroll down here so

negative 4 times 2 is negative 8 so this would be negative 8 times x so negative 8x Then times square root of 5 then plus 2x times square root of 10x so that's going to be my simplified answer there i don't have like radicals so there's nothing else i can do right negative 8x times square root of 5 plus 2x times the square root of 10x all right let's get into something that's a little bit more complex so i have the quantity 2 times the square root of 3 plus 6 times the quantity negative 4

times Square root of 3 minus 2. so the way we're going to do an operation like this we're going to use foil just like when we multiply two binomials together right we have two terms here and two terms here so let's use foil so i'm going to do my first terms so this guy times this guy 2 times negative 4 is negative 8. square root of 3 times square root of 30 is 3. so i'd have negative 8 times 3 Which we know is negative 24. so let's go ahead and write that okay then the

next thing i'm going to do is i'm going to multiply my outer terms so this one times this one so 2 times negative 2 would be negative 4 and then times that square root of 3. so now i'm going to multiply my inside terms so this one times this one 6 times negative 4 is negative 24 and then times square root of 3. and then lastly I'm going to do my last terms so 6 times negative 2 is negative 12. okay so once i have it in this format pretty easy to simplify and in this

case we do have like radicals we have the same index we have the same index we have the same radicand so we can combine what's in the middle so if i look at negative 4 times square root of 3 minus 24 times square root of 3 Negative 4 minus 24 is negative 28 so this would be negative 28 times that common radical which is the square root of three now negative 24 and negative 12 can also be combined negative 24 minus 12 is negative 36 so i can write negative 36 out in front or behind

doesn't matter so we'd have negative 36 minus 28 times the square root of 3 as our answer All right for the next one i have a similar situation i have the quantity 4 times square root of 2 minus 5 times square root of 7k and then this is multiplied by the quantity negative 6 times square root of 5 plus 3 times the square root of 7 k so again i'm going to use foil so first terms 4 times square root of 2 multiplied by negative 6 times square root of 5. so 4 times negative 6

is negative 24. The square root of 2 times the square root of 5 is the square root of 10. now i want to do my outside terms so i'd have 4 times square root of 2 times 3 times square root of 7k so 4 times 3 would be 12 so plus 12 then the square root of 2 times the square root of 7k would be the square root of 14k then my inner terms so i have negative 5 times the square root of 7k multiplied by negative 6 times the square root of 5. so negative

5 times negative 6 is positive 30 and the square root of 7k times the square root of 5 would be the square root of 35k then i want to end by multiplying my last terms together so negative 5 times the square root of 7k times 3 times the square root of 7k so let's do negative times positive is negative 5 times 3 is 15 okay 15 but then i've got square root of 7 k times square root of 7k that's going to give me 7k so really i'd have negative 15 times 7 which is negative

105 times k so what can i simplify here well the answer to that is nothing this is 14k this is 35k that's not the same these are not Not like radicals okay very important that you understand that so my simplified answer will just be negative 24 times square root of 10 plus 12 times the square root of 14k plus 30 times the square root of 35k minus 105 k all right for the next one i want you to think back to when we saw the difference of two squares and we know the difference of two

squares looks Like this it's x squared minus y squared and when we factor it we end up with x plus y times x minus y right those two quantities now what you'll see in this case is you have this right here and this right here as the same term and then this right here and this right here as the same term what's different Would be the signs right you have a plus and a minus now look at what you have here you have a 6 and a six and then look over here you have square

root of two and square root of two then your signs are different you have a positive and a negative so they throw things at you especially on timed examinations to kind of say okay well you should recognize this pattern and you should Know that all i need to do is square this and then subtract away this squared right so really the answer i can just do this in my head 6 squared is 36 and then minus the square root of 2 squared would be 2. so 36 minus 2 would be 34 right so that should

be your answer again you can do that in your head once you're familiar with this pattern so let's go ahead and crank this out the long way and then again we'll do It the short way so if i was to use foil here so let's start with 6 times 6 that's 36 and then 6 times negative square root of 2 would be minus 6 times square root of 2 then next i have the square root of 2 times 6 so that's plus 6 times square root of 2. and then last i have square root of

2 times negative square root of 2 that's minus 2 right square root of 2 times Square root of 2 is 2. so now again we realize that the two terms in the middle here the negative 6 times square root of 2 and the positive 6 m squared of 2 those are going to cancel out and all i'm left with is the first term that we saw squared minus the second term that we saw squared right so it's 36 minus 2 which again is 34 and you saw me do that in my head and the reason

i can Do that in my head and you can too is because i recognize the pattern right away right so this ends up being 34 and again the quick way to do that is square the first thing square 6 that's 36 minus the second thing squared square root of 2 squared would just be 2 36 minus 2 is 34. all right so here we have that same scenario 8 minus the square root of 5 that quantity times the quantity 8 plus the square Root of 5. very easy to do this if you recognize it you

just square the first guy so if i square 8 i'd get 64 then minus square the second guy square root of 5 squared is 5. so 64 minus 5 would be right we can do this the long way just to verify that we get that answer but i'm going to promise you 100 of the time you will so 8 times 8 would be 64. the outer would be Positive eight times square root of five then the inner we'd have negative eight times square root of five then the last negative square root of five times square

root of five we know it's negative square root of 5 times square root of 5 is 5. so again the two terms in the middle are going to drop out right those are just going to cancel and i'm left with 64 minus 5 which again is 59. All right for the next one again we're going to utilize our special products formulas so do you recall something like the quantity x minus y squared right you should be able to regurgitate this formula at this point it's x squared minus right because we have a minus there 2xy

plus y squared so following that format here i would Have the first chi squared so 3x squared 3 squared is 9 x squared is just x squared i would have minus 2 times the first guy which is 3x times the second guy which is the square root of 2y and then plus the second guy squared so the square root of 2y squared the squaring would just undo the square root i'm left with just 2y so plus 2y So if i simplify this i get what 9x squared minus 2 times 3 is 6 i'd have 6x

times the square root of 2y and then plus 2y so that's as simple as i can get that i'm i'm not going to be able to further simplify because there's nothing to combine there so i would just erase this here and show you i got the same answer if i did it the long way so remember i need To expand this if i want to do it the long way i would say this is 3x minus the square root of 2y and then if this is squared i would say this is 3x minus the square

root of 2y times again 3x minus the square root of 2y so the first terms 3x times 3x would be 9x squared the outer 3x times negative square root of 2y negative 3x times the square root Of 2y the inner would be the same thing right negative square root of 2y times 3x so negative 3x times the square root of 2y and then the last would be negative times negative is positive square root of 2y times square root of 2y is 2y so if i combine the terms in the middle i know that i have

like radicals there so i can do that because i have the same index and the Same radicand right i have 2y in each case so really i just think about negative 3x plus negative 3x that's negative 6x so this would end up being 9x squared minus 6x times the square root of 2y and then plus 2y and again it's exactly what we got right there i just used a quicker method when we first did it and that's the method that you should use right memorize those special product formulas and again They're going to help you

for the rest of your life right it's just something that's just going to tremendously speed up your processes all right let's take a look at another one and for this one i'm going to match that formula of the quantity x plus y squared so this is x squared plus 2xy plus y squared so exactly what we just looked at the only difference is going to be your sine is plus in this case for that second Term right if i have the quantity x minus y squared everything is the same except for that minus sign right

there all the rest is the same so if i was to do this and i'm not going to do this one the long way just to save a little time i would say what the first chi squared so if i square 5 i get 25 if i square x squared i get x to the fourth power right x squared squared keep x the same Multiply two times two that's four okay so now the next thing i want to do is i want to put plus two times this guy right here so 2 times 5 is

10. i'm just going to write that so 10 times x squared times this guy the square root of 7z and then the last thing i want to do is put plus this guy squared so if i square the square root of something i just get The radicand back right if i square the square root of 7z i'm just going to get 7z now looking at this there's nothing i can combine here so my simplified answer would be 25x to the fourth power plus 10x squared times the square root of 7z plus 7z all right so

now what we're going to talk about for the remainder of this section is what to do when you have a more complex Denominator that's involved and you need to rationalize the denominator right you don't want any radicals in the denominator but sometimes it's not so simple you might get something that's more complex and before we start jumping into that we have to learn about something known as a conjugate okay so a conjugate is formed when we change the sine of something like this so if i have 3x plus 7 i would have 3x and i

would Have 7 but my sign would change okay my sign would change so the conjugate of 3x plus 7 is just 3x minus 7. as another example if i had something like 2x squared minus 5 the conjugate would be change the sign so 2x squared this would become plus and then this would stay the same so the terms are the same i'm just changing the sign so if i had let's say x plus y as an example the Conjugate would be x minus y if i saw a plus b the conjugate would be a minus

b if i saw the square root of seven plus three the conjugate would be square root of seven minus three i'm keeping the two terms the same all i'm doing is just changing the sign it's very very simple all right so now that we know what a conjugate is let me just read this to you so when we simplify a radical expression with two terms in the Denominator and at least one is a radical remember you can't have any radicals in your denominator what you're going to do to get that radical out of the denominator

we're going to multiply the numerator and denominator by the conjugate of the denominator okay and the reason that's going to work is you're setting up a scenario of something like this a plus b That quantity times a minus b okay you have the same terms okay in the first position of each you have the same terms in the second position of each your difference is the sign so what's going to happen is the middle two terms when you do foil there drop out you know we saw this earlier in the lesson where we talked about

you know our special products formulas this ends up being what a squared the first guy squared minus b squared right the second Guy squared very very easy to kind of calculate that the two terms from the middle are going to drop out and you're going to end up with a denominator that is radical free so let's see an example so we'll start out with this kind of easy one we have 8 over 6 minus the square root of 5. so i'd multiply this by the conjugate of the denominator right we're going to do that to

the numerator denominator so If i think about the conjugate it's what keep the terms the same so 6 and square root of 5 6 and square root of 5 but the sign is going to change right so instead of a minus it's going to be a plus now am i doing anything illegal if i'm multiplying this by this no this is a complex form of one right six plus square root of five over six plus Square root of five complex form of one i'll multiply by one i don't change the value right so this is

legal now in the numerator if i multiply i just use my distributive property right so i'm just going to say well 8 times 6 is 48 then plus 8 times square root of 5 is just 8 times square root of 5. that's simple enough right i can't i can't do anything else there and then for the denominator again Realize that you have that pattern where you have six and six same term square root of five square root of five same term different signs negative positive okay so i can use that little formula to speed up

the process and just say okay well i have the first guy squared 6 squared is 36 minus the second guy squared square root of 5 squared is just 5. So my denominator is going to be 31 so i would have 48 plus 8 times square root of 5 over 36 minus 5 is 31 so my denominator now is radical free and again that's our goal here because if we report our answer with a radical in the denominator we're going to get points taken away at minimum right it's not considered simplified all right let's look at

one that's a little bit Harder not harder but more tedious we have the square root of 2a squared over negative 3 minus 5 times square root of 5a so the first thing i do is if i see something i can simplify just do it right this this really should be what this is a squared underneath a square root so this would be a times the square root of two so i should really just line this out And write it like that right and i can let me kind of just write it over here as a

times square root of 2 over negative 3 minus 5 times square root of 5a so let me just kind of move this down here to get us some more room again if i want to remove the radical from the denominator in this case i'm going to multiply the numerator denominator by the conjugate of that Denominator so to get my conjugate the terms stay the same so i'd have negative 3 and i'd have 5 times square root of 5a that stays the same so let's do that in each case what's going to change is the sign

so in this case it's negative so this would be positive and this would be positive okay so now that we've worked that out a times square root of 2 times negative 3 is what This is negative 3a times the square root of 2. can't do anything else with that then plus i've got 5 times a that's 5a times the square root of 2 times the square root of 5a so i'm going to write that as the square root of 5 times 2 or 10 times a so the numerator nothing else i can do there it's

just going to stay as Negative 3a times square root of 2 plus 5a times square root of 10a okay what about my denominator what can i do here well again we're going to use that formula i've got the same term here and here same term here and here signs are different right so square the first guy square negative 3 you get 9 then minus square the second guy 5 squared is 25 and then the square root of 5a squared Would be 5a so you get 25 times 5 or 125 and then a now can i

simplify that any further no these are not like terms right 9 is not a like term with negative 125 a can't do anything further so my simplified answer here is negative 3a times the square root of 2 plus 5a times the square root of 10a over 9 minus 125 a all right for the next one i have negative 2 over 5b to the fourth power minus 2 times the square root of v cubed now what i can do before i even start i can simplify so negative 2 over 5 v to the fourth power minus

2 the square root of v cubed i can think about the square root of v cubed as the square root of v squared times the square root of v square root of v squared is v so i can write this as minus 2v times the square root of v So i like to simplify before i even think about anything else because it just makes my calculations easier in the end so let me kind of drag this down so we have a little room to work okay let me put this over here and so now i

want to multiply the numerator denominator by the conjugate of the denominator so the terms are the same 5 v to the fourth power and 2v times square root of v Over we have 5v the fourth power and 2v times square root of e the signs are going to be different right so this is a negative this would be plus and plus all right so in the numerator let's start out negative 2 times 5 is negative 10. so that's negative 10 and then we have v to the fourth power then negative 2 times 2 would be

negative 4 And then times v and then times the square root of v then over now again if i'm using that formula i don't need to go through and do foil right i have the same thing here and here same thing here and here the signs are different so the first guy squared 5 v to the fourth power squared 5 squared we know is 25 v to the fourth power squared so v to the fourth power squared V stays the same 4 times 2 is 8. then minus the next guy is going to be squared

so 2v squared is what that's 4v squared square root of v squared is v so i'd have v squared times v or v cubed v cubed so now my answer here negative 10 v to the fourth power minus 4 v times the square root of v over 25 v to the 8th power minus 4 v cubed so is there anything else i can do to simplify Here well yes there is notice that you have that variable v here here here and here so that means i could factor a v out from the numerator and the

denominator and i could cancel now you're going to only be able to think about the lowest exponent on any of them in this case that's going to be a 1 right so if i factored out a v from the numerator let me kind of scroll down get a little Room going i would have inside of parentheses negative 10 times v cubed minus from here i'd have a 4 the v is getting factored out so that's gone it's just v to the first power so that's gone i'm square root of v and then over for this

guy if i factor out a v i'll have 25 v to the seventh power minus four v squared now i can cancel this v with this v and now my simplified Answer is going to be negative 10 v cubed minus 4 times square root of v over 25 v to the seventh power minus 4 v squared all right for the last problem we're going to look at i have 3a plus 2 times square root of 3a over negative 3 plus square root of a cubed so again if there's something i can simplify before i start

i want to go ahead and do that i know that this right Here if i have the square root of a cubed really that's a times squared of a right i think about the square root of a cubed as square root of a squared times square root of a this is really a so it's a times squared of a so let's let's just simplify this as 3a plus 2 times the square root of 3a over negative 3 Plus a times the square root of a let's drag this down here to get a little room going

and we're going to multiply the numerator denominator by the conjugate of the denominator so at this point we know that would be negative 3 minus right we're changing the sign a times the square root of a so negative 3 minus a times squared of a all right so now that we've got that figured out let's multiply So for the numerator here i'm going to use foil it's going to make it easy 3a times negative 3 is negative 9a for the outer i've got 3a times negative a times square root of a so that's going to

be negative 3a times a is 3a squared and then times square root of a for the inner i've got 2 times square root of 3a times negative 3. so that's negative 6 times square root of 3a and then for the last i'm going to have 2 times negative a or negative 2a times the square root of 3a times the square root of a so that would be the square root of 3a squared so i know the square root of a squared is a so i can pull that out and make this a squared times the

square root of 3. now is there anything i can do to Simplify here no there's not right there's nothing else i can do i know it's long but there's there's nothing i can do to simplify so then this is over now i'm going to use my special products formula here and that tells me to take the first guy in square negative 3 squared is 9 and then minus the second guy squared so then that would be a squared times square of a times square root of a Would be a so really a squared times a

or a cubed so my denominator becomes nine minus a cubed nothing i can do with that so i have negative nine a minus three a squared times square root of a minus six times square root of three a minus two a squared times square root of 3 over 9 minus a cubed as my simplified answer hello and welcome to algebra 1 lesson 58 in this video we're going to learn about Solving equations with radicals so a radical equation is an equation where there is a variable in the radicand so let me repeat that a radical

equation is an equation where there is a variable in the radicand so i know most of you know this but in case you don't because you might be saying what's a radicand again if we had something like a square root Or cube root or fourth root or whatever it is let's just go with the square root because that's the most common the radicand is the number or the variable or the you could say expression that's underneath that symbol so something like square root of 3x this would be your radicand so again a radical equation one

more time is an equation where there is a variable in the radicand so as an example We see something like the square root of x plus two equals five in your radicand you have that variable x as another example we have 4 times the square root of x is equal to 2x minus 3. here the radicand just consists of x but that's a variable and so this is a radical equation all right so in order to solve a radical equation we need to introduce a new property of equality known as the squaring property of Equality

okay we've also learned you know the addition property of equality and the multiplication property of equality so this is just another tool in our chest you know to solve equations moving forward so the squaring property of equality there's one kind of little hiccup to this thing when both sides of an equation are squared all solutions to the original equation are among okay among the solutions to The squared equation now that seems like it's not a big deal but it really kind of is if i use something like the addition property of equality as we recall

that looks like this let me kind of scroll down we have x plus 2 equals 4 as an example i know the solution to this equation is x equals 2. If i added 3 to both sides of the equation so let's say i add 3 over here that would make this 5 i have 3 over here that would make the 7. the solution is still x equals 2 still works out if i have the multiplication property of equality so something like let's say 3x equals 12 i know this is x equals 4. if i multiply

both sides by negative 2 this would be negative 6x And this would be negative 24. x equals 4 still a solution plug in a 4 there you get a true statement with the squaring property of equality this is not always going to be true so let's suppose we had x equals 3 very very simple i know that if i replace x with 3 i get a true statement but let's say i square both sides of this so x becomes squared And 3 becomes squared so now i have x squared equals 9. now we haven't covered

how to solve an equation like this yet but let's let's just think about it it's a very very simple equation so we can kind of you know mentally figure out what this is what squared would give me 9 well i know 3 does right 3 squared would be 9. 3 times 3 is 9. but i also know that negative 3 Squared would give me 9 as well so the solution for this equation for this equation would be x equals 3 or x equals negative 3. so there's two solutions there but in the original equation if

i took this negative 3 and i plugged it in for x here we'd have a big problem we'd have negative 3 is equal to 3 that's false that does not work so let me kind of line this out and say wrong Okay that's wrong so you can see that when we square that equation we came up with an equation that had solutions that did not work in the original one so that's why it's very very important that when you use the squaring property of equality you have to check each and every solution that you generate

in order to make sure it works in the original equation all right so now let's Just take a look at the procedure to solve a radical equation so i have solving a radical equation we want to isolate the radical so that's the first thing you're going to do now in the most complicated situations you're going to have more than one radical involved and in that case you just want to isolate one radical at a time now the next step is going to depend on what type of root you're looking at do you have a square

root do You have a cube root do you have a fourth root in algebra one we're just going to deal with square roots so i just put square both sides so we would square both sides in the case of a square root then we would combine any like terms now at this point if you had any radicals that remained you're going to go back up to this first step and you're going to come back down you're going to keep doing that Until you're radical free then lastly you're going to solve the equation and check all

potential solutions in the original equation okay again this is very very important check all potential solutions in the original equation it's not going to happen to you every time but at least sometimes the solutions you're going to get from that transformed equation is going to give you problems or not Work in that original one and so it's not going to be a solution that you can use as an answer for your problem all right let's start out with something very very easy so we have the square root of n is equal to 3. so the

radical is already isolated here okay so i don't need to do that step and i would just start out by just squaring both sides so if i have the square root of n is equal to 3 And i square this side and i square this side what's going to happen is we know that this cancels with this and i just have n and 3 squared is 9. so n equals 9. i get one solution there so plug this back in for n if i plugged in a 9 there what would i have i would have

the principal square root of 9 is equal to 3 and that is true right square root of 9 is 3 you get 3 Equals 3. so this checks out as a solution let's take a look at another easy one so we have that the square root of x is equal to 5. again the radical is already isolated for me i don't need to do that and so all i need to do is just square both sides of the equation [Music] and so what's that going to give me well this is going to cancel with this

and i'll just have x 5 squared is 25. I get one solution only and so again if i plug in a 25 for x there the square root of 25 we all know that's 5 so this is a valid solution let's look at one that's a little bit more challenging so we have the square root of 15 p minus 56 is equal to p so the radical again is isolated for me so what i'm going to do is i'm going to again square both sides so this equals p so i'm going to square This side

and i'm going to square this side okay so if i square this this cancels with this and i'm left with the radicand so i'm left with 15 p minus 56 and this equals p squared so what we have now is a quadratic equation and we've worked with these in a previous lesson and i taught you how to solve them when they're factorable We have not gotten to the point to where you have a quadratic equation it's not factorable you know how to solve it we're going to get to that when we start talking about completing

the square and then the quadratic formula that's going to be the very last lessons of algebra 1. so for this lesson i gave you stuff that you could factor all right so for this i'm just going to subtract 15p away from both sides [Music] i'm going to add 56 to both sides and that's going to make this side equal to 0. so what i'm going to have let me scroll down here a little bit and i like 0 to be on the right side so let me move everything over so i would have p squared

minus 15 p plus 56 is equal to 0 and i'm going to factor my left side Okay so i need two integers whose sum is going to be negative 15 and whose product is 56 it's gonna be negative seven and negative eight so minus seven minus eight so we know how to solve this we take each factor okay that had a variable involved we'd set it equal to zero it's called the zero product property okay so we'd have p minus seven equals zero And then also we could put or p minus eight equals zero add

seven of both sides here add eight to both sides here and we get p is equal to seven or p is equal to 8. now we're not done because we need to check this in the original equation so if i plug in a 7 4p there i would have the square root of 15 times 7 which is 105 minus 56 and this should be equal to 7. so 105 minus 56 is 49 we know that the square root of 49 is 7 right this would give me 7 equals 7. so that's a valid solution for

the next one we have eight okay so let's try that out now this stays the same and 15 times eight now would give me 120. so if i do the square root of 120 minus 56 that's the square root of 64. 120 minus 56 is 64. so this would be the square root of 64 is equal to 8 and we know that's true right this would be 8 equals 8. so this is a valid solution as well so here p equals seven or p equals eight well let's take a look at another one so we

have the square root of negative one hundred plus twenty x is equal to x This is already isolated for us so let's go ahead and just go into the step where we square both sides so negative 100 plus 20x is equal to x i'm going to square this side and i'm going to square this side so this is going to cancel with this and i'll just have this i'll have negative 100 plus 20x is equal to x squared [Music] Again i'm going to set this up by just putting everything on one side and setting it

equal to zero so let me add let me add 100 to both sides and let me subtract 20x away from each side that's going to be gone and be equal to zero and i like my zero to be on the right and i like everything else to be on the left so let's just rewrite it as x squared Minus 20 x plus 100 is equal to zero okay so now that that's set up again i gave you stuff that you could factor so let's go ahead and factor this guy we'll have x here and here

and then give me two integers whose sum is negative 20 and whose product is 100. so we know that's going to be a negative and a negative right negative times negative is positive negative plus Negative is negative so if i think about this i could do negative 10 and negative 10. negative 10 and negative 10. i know that this would only have one solution right because it's basically x minus 10 as one factor and x minus 10 is the other factor so this would be one solution of x equals 10. so let's check this out

in the original equation okay so if i have square root Of negative 100 plus 20 times 10 20 times 10 is what that's 200 this should equal 10. let me kind of scooch this out of the way that over there okay so if i simplify this negative 100 plus 200 is 100 so i'd have the square root of 100 equals 10 and that's true right you get 10 equals 10. so yes this is a valid solution x equals 10. all Right let's take a look at another one so now we have negative b plus the

square root of 6b minus 2 and this equals 1. so notice how this is not isolated okay the radical is not isolated so we're getting into some more challenging problems really just more tedious so i can't just go ahead and start squaring things right i can't square both sides that's going to be really really sloppy if i do that so what i Want to do again i want to isolate the radical so i'm just going to add b to both sides of the equation and i'm going to end up with the square root of 6b

minus 2 is equal to 1 plus b now if i square both sides of the equation i'm going to get rid of the radical so this is going to cancel with this and i'm going to have 6b minus 2. And remember this has to be expanded i can't do that common mistake where i just go okay this is 1 squared plus b squared we've covered that many many times and at this point in algebra 1 you should realize this needs to be expanded now fortunately for us we know the formula for this to do it

very quickly it's this guy squared plus 2 times this guy times this guy plus this last guy squared so really it's b squared plus 2b plus 1. Let's rewrite that b squared b squared plus 2b plus 1. let's scroll down a little bit and some room going and let's add 2 to both sides of the equation so that's gone and let's subtract 6b away from each side of the equation so that's gone so i would have zero over here on the left is equal to We have b squared minus four b plus three and let

me just kind of switch this around and say okay i have b squared minus 4b plus 3 equals 0. again that's just personal preference you don't have to do that it's just up to you okay so now what i want to do is factor the left side so this is going To be factored and equal to zero and i know the first terms would be b and b so now i need to find two integers whose sum is negative four and whose product is 3. so i know this has to be a negative and a

negative right because negative times negative is positive negative plus negative is negative it's easy because 3 is a prime number right it's only 1 times 3. so it has to be 3 and 1. right so b minus 3 that Quantity times b minus 1 that quantity so to solve this again use the zero product property we're going to set each factor with a variable equal to 0. so we'd have b minus 3 equals 0 or b minus 1 equals 0. we'd add 3 to each side of the equation here we'd add 1 to each side

of the equation here and what we're going to end up with is b is equal to 3 or b is equal to 1. all right so let's check This and we had b is equal to 3 or b is equal to 1. all right so let's start out with b equals 3. so let's plug in a 3 here and here and see if we get a true statement so we would have negative 3 plus the square root of 6 times 3 is 18 minus 2 so we know we can just replace that with 16 And

this should be equal to 1. square root of 16 is 4 so i'd have negative 3 plus 4 that is 1. you get 1 equals 1. so yes this is a valid solution all right for the next one i'd replace my b with a 1. so i'd have negative 1 plus the square root of 6 times 1 which is 6 minus 2 which would give me 4. so the Square root of 4 and this should be equal to 1. now the square root of 4 is 2. so i'd have negative 1 plus 2 that gives

me 1 and that equals 1. so we get 1 equals 1. again this solution is valid b equals 1. so we have b equals three here or b equals one okay for the last problem that we're going to look at in this lesson we're going to dive into something that's much more tedious this is a type of problem where you want to focus as much as you can because it's Very very easy in these longer problems to make a silly mistake that costs you the right answer in the end right the worst thing is you

do all this work you made some silly mistake and then you go to check your answer and you're like ah you know i made a mistake and you have to start the problem over so just pay close attention to the ones like this that are very very tedious okay so we have negative 1 is equal to the square root of 2 minus a Minus the square root of negative 1 minus 5a now up into this point we've seen that we would square both sides after we isolated the radical and we were radical free in this

case we have two radicals involved so that's not going to happen what we have to do is isolate one radical at a time okay and work our way through the problem so that's what makes this much more tedious so let's just copy the problem so negative 1 Is equal to the square root of 2 minus a minus the square root of negative 1 minus 5a now i want to isolate one of these guys doesn't matter which one i choose i'm going to put the same answer in the end i think it would be easier for

me to isolate this guy right here and so i'm going to add the square root of negative 1 minus 5a to both sides So over here it's going to cancel itself and go away and i'm just going to add it over here so plus the square root of negative 1 minus 5a now what i can do at this point is i can square both sides of the equation and i'll get rid of one and only one of the radicals so i have negative one plus The square root of negative one minus five a and i'm

going to square this guy and then this is equal to we have the square root of 2 minus a and i'm going to square this guy now we know at this point that this would cancel with this and i'm left with this so i'm going to put equals 2 minus a over here for this side it looks like it would be kind of complex to do that but again we can use our special Products formulas to speed up the process here that's why we went through and memorized those so i know for this i would

square the first guy negative 1 squared is positive 1. so let's write that all the way down here and then the next thing would be we would have plus 2 times this guy times this guy 2 times negative 1 is negative 2 so i'll write minus 2 Then times the square root of negative 1 minus 5a then the last thing is i want this guy squared well if i squared the square root of negative 1 minus 5a we again know that this would cancel with this and i'd be left with this so i'd have minus

1 minus 5 a all right so now that that's worked out what i want to do at this point is Simplify so when i look across i see that i have a 1 and a negative 1. so those would cancel and then i see that i have a negative 5a and a negative a so i can add 5a to both sides of the equation that's going to cancel so i'm going to have this negative 2 times the square root of negative 1 minus 5a And this is equal to and i'm just going to reorder this

to 4a plus 2 okay negative a plus 5a is 4. now what i want to do is i want to square both sides of this equation one more time to get rid of this radical this is going to be a little bit tricky and i'm going to explain why so let me square this side and let me square this side Now when i square this guy i want you to think back to when i squared something like a times b squared this is equal to a squared times b squared right we all should agree on

that if i look at this more closely this is really what this is negative 2 times This quantity let me just put it in parentheses negative 1 minus 5 a okay so if i squared this if i squared this i would end up with what let me kind of erase this following that model i'd have negative 2 squared times this square root of negative one minus five a that guy squared now This will cancel with this and give me this where students make a mistake is you're still multiplying here so let me kind of go

through this let me let me move this down or i'll actually move it over to get some room and let me show you what i mean so when this comes down negative 2 squared which i have here would be positive 4. but this is still multiplying the result of this so the Result of this is the quantity negative one minus five a okay the common mistake that you see is let me just erase this students just go okay well this is 4 minus 1 minus 5a that's not correct again if we go back to the

example of a times b and this is squared we all should agree that this is a squared times b squared okay So it's the same thing here this is squared that's 4 times the result of this squared which is the quantity negative 1 minus 5a so you've got to make sure that you do that correctly all right so with that being said let's put this is equal to we have 4a plus 2 that quantity squared and again we can use our special products formulas for this square the first guy that's 16 a squared Plus 2

times this guy times this guy 2 times 4 is 8 8 times 2 is 16 then times a then plus the last guy squared all right so simplifying here on the left 4 times negative 1 is negative 4 and then 4 times negative 5a is negative 20a so this equals 16 a squared plus 16a plus 4. let me scroll down and so let me add 4 to both sides of the Equation here and let me add 20 a to both sides of the equation so this would be 0. so i'd have 0 is equal to

16 a squared 16 a plus 20 a is 36 a so plus 36 a and then plus 8. and again i like the zero on the right side it doesn't matter it's just personal preference so i'm going to rewrite this as 16 a squared plus 36 a Plus 8 equals 0. so before i move forward i notice that i can pull a 4 out of everything so if i pull a 4 out this would be 4a squared plus 9a plus 2 and this is equal to 0. so to factor the inside here i'm going to

use factoring by grouping so give me two integers whose product whose product is eight and whose sum is nine that's easy that's eight and one So let's rewrite that middle term we'll have four a squared plus eight a plus a plus two this is equal to zero okay so let's erase this and let's get our little room going so i'd write four in the first group i could pull out of 4a what would that leave me with a plus 2. in the second group i just pull out a 1 and that would leave me with

a plus 2. Okay so then this equals 0 and so what i'm going to do now i'm going to factor out the common binomial factor and that's this a plus 2 here so if i pull that out i'd have a plus 2 that quantity times i'd have 4a plus 1. [Music] okay this equals 0. so now in factored form i have 4 times the quantity a plus 2 times the quantity 4a plus 1. okay and then this equals 0. so using my zero product property i set each factor with a variable equal to zero so

i don't need to set four equal to zero that doesn't make any sense i'd set a plus two equal to zero and it also set four a plus one equal to zero so this is easy subtract two from each side you get a equals negative two for this one i subtract one away from Each side i get four a equals negative one i divide both sides by 4 and i get a equals negative 1 4. so a equals negative 2 or a equals negative 1 4. now let's go back and let's check each one and

make sure that they work in the original equation so we have a equals negative 1 4 And we have a is equal to negative 2. so let's start out with a equals negative 2. so if i plug that in let's have negative 1 is equal to the square root of 2 minus a negative 2 is like 2 plus 2. so the square root of 4 we might as well replace that with 2 we know what that is so we'll have a 2 there minus the square root of Negative 1 minus 5 times negative 2 is

negative 10. so i'd have minus a negative 10 which is plus 10. so really this would be 9. we know the square root of nine is three two minus three is negative one so that one works out just fine so this is a solution for our equation so let's take a look at a equals negative one fourth now so we have negative one is equal to The square root of two we have minus a negative one fourth that's like two plus one fourth okay so two plus one fourth now to deal with the fact that

we have fractions involved i could write 2 as 8 4 right 8 4. and really i could write this as 9 4. so the square root the square root of 9 4. and we know we can write that as the square root of nine over the square root of four square Root of nine is three square root of four is two so this is really three halves this is three halves then minus over here i have the square root of negative 1 minus 5 times a negative 1 4. so negative times negative is positive so

this would be positive 5 4. so plus 5 4. and of course to get a common denominator here I would simply write this as negative four over four negative four over four negative four plus five is one so this would be one-fourth this would be one-fourth and again i can break this down into the square root of one over the square root of four square root of one is one square root of four is two so this is one-half this is one-half so three minus one is two and this would Be over 2 2 over

2 is equal to 1. so what do i get here i get that negative 1 is equal to 1. that's not right negative 1 doesn't equal 1. and this is part of the problem again when you use that squaring property of equality you're going to end up with solutions that don't work in the original equation so this one does not work it's not a valid solution in the original equation Okay this is this is wrong not a true statement and so the solution here is a equals negative 2 and that's your only solution hello and

welcome to algebra 1 lesson 59. in this video we're going to learn about using rational numbers as exponents which is also known as working with fractional exponents all right so we're going to kind of just jump right in this is not a very hard topic to wrap your Head around and it's something that is going to dramatically increase your speed when you're simplifying radicals moving forward so let's suppose we saw something like three to the power of one-half times three to the power of one-half so at this point we have all the tools that we

need to simplify something like this now we know that from the rules of exponents we would keep our base three The same so that stays the same and then we would add the exponents right we would add the exponents because we're multiplying so this would be 3 to the power of one-half plus one-half now we already have a common denominator so this is pretty easy to do this would be 3 to the power of 1 plus 1 is 2 over the common denominator of 2 and then 2 over 2 is just 1. so this would

be 3 To the power of 1 which is just 3. now having seen me do this example what can you guess is going to be another way that we can write 3 to the power of one half well it's going to be the square root of 3 right because the square root of 3 times the square root of 3 is going to equal what at the end of the day it's going to equal 3. square root of 3 times square root of 3 is 3. Square root of 2 times square root of 2 is two

square root of let's say four times square root of four is what it's four and we know this one because square root of four is a rational number this would be two times 2 which we know is 4 right so we can prove the rule by looking at that so in each case i could rewrite these i could say this is really 2 to the power of 1 half times two to the power of one half which equals two right it's two to the power of one half plus one half which is two to the

power of one in this case i can say this is what it's four to the power of one one-half times four to the power of one-half which is four so having a number raised to the power of one-half Is the same thing as taking the square root of the numbers just how we write it with exponents now as something that's a little bit more complicated let's say we stay with the number three and we saw three raised to i don't know let's say the one third power okay the power of one third so then i

would multiply this by three to the power of one third again and then one more time so again using our rules for exponents What would this be equal to well i have the same base so that stays the same and then i would just add the exponents so one-third plus one-third plus one-third is equal to what it's equal to one right one plus 1 plus 1 is 3 3 over that common denominator of 3 would be 1. so this is 3 to the power of 3 3 which is equal to 3 to the power of

1 or just 3. so After having seen the last example what would you guess is another way that we could write 3 to the power of 1 3. we could do the cube root of 3 the cube root of 3 right because the cube root of 3 times the cube root of 3 times the cube root of 3 is equal to 3. so now that we've seen a few examples let's define our first rule So we want to let a be greater than or equal to zero so this just means that a which is right

here is going to be non-negative and then we're also going to say that n is going to be positive okay it's going to be a positive integer more specifically now so that means that n can't be zero and n is in the denominator so that's why it can't be zero and also n is not going to be negative okay Now when we have these conditions that are met we can say that a raised to the power of 1 over n is equal to the nth root of a so in other words this or looking at

it in exponential form is the base and when we transform it over to this format it's going to be the radicand then here n is your denominator this is the denominator and when we move over here it becomes The index now i suggest that you write this particular thing down on a flash card okay and what you want to do is as we go through the examples here keep your flash card out pause the video and then try to do it yourself okay you can even do this when you're doing your practice problems or your

homework okay after about 10 problems or so you're going to memorize this generic formula and you're going to do this without having to look At that flash card so let's start out with 216 raised to the power of one third so we know that this is going to end up being our radicand okay so let me write 216 and let me write the radical and we know that three is going to end up being the index okay that's going to end up being the index okay again just straight from your flash Card so we end

up with the cube root of 216 and most of you know at this point that that's going to be 6 right that's going to be 6. 6 times 6 is 36 36 times 6 once more is 216. what about something like 625 to the 1 4 power so again this part right here would be the radicand and it would be the fourth root right because this would be your index So again this would be the index and then this would end up being the radicand so i have the fourth root of 625 and that's equal

to 5. right 5 times 5 is 25 25 times 5 is 125 and then 125 times 5 is 625. what about 169 to the power of one-half again we know this is the same as just taking the square root of the number raise something to the power of one-half Like taking the square root so i'll put equals the square root of 169. and again if you didn't remember that you could use the flash card and say okay well this is going to be my radicand and then this the denominator is going to be my index

now when we work with square roots we know that we don't display the two right we have an understood index of a two When there's no index that's displayed so we all know that the square root of a higher 69 would be 13. what about 243 to the one-fifth power well again this part right here this 243 would be my radicand okay this would be my radicand and then i'm taking the fifth root of that because this right here my denominator would be the index Let me rewrite that make it a little cleaner so what

is the fifth root of 243 well again it's three three times three is nine nine temperatures 27 27 times three is 81 and then 81 times three is 243. all right for the next one let's look at negative 512 raised to the power of one-third now when i gave you the rule i specifically said that this value here Which is a when we talked about our rule is non-negative so what gives why is this negative well i want to show you that you can also incorporate this rule with a negative we just have to keep

in mind that when you work with roots remember you don't want to end up with a situation where you have an even index and you have a negative radicand right that's going to produce A not real solution okay so we want to kind of stay away from that for right now until we get into algebra 2 when we can kind of deal with those scenarios so i'm going to write the cube root the cube root of 512 and it's negative so i'm going to put that in there and this is okay because the negative there

when i think about a cube root is fine right negative 1 times negative 1 times Negative 1 is negative 1. so i can just put a negative out in front here and then just think about 512. what is the cube root of 512 well in case you don't know that and i know up to this point i just kind of gave you the answers let's get a little practice in figuring this out this ends in 12 so i know it's divisible by 4 and this would be 128 times and i know that 128 is 2

to the 7th Power some of you don't know that so you could continue and say okay well this is 64 times 2 and at this point you should recognize 64 is 2 to the 6th power so really if i wrote this as 2 times 2 i could say i have 2 to the sixth power times 2 times 2 times 2. so really this is 2 to the sixth power times 2 cubed or 2 to the ninth power two to the ninth power now here's where we can really kind of Get things going okay if i

have two to the ninth power because that's what 512 is forget about that negative for a second and i took the cube root of that how can i use what i just did to simplify that well if i realize if i realize that i'm taking the cube root and that's like raising something to the one third power I can keep this as two to the ninth power and then i can raise this whole thing to the power of one third now by the power rules for exponents i know that i would multiply this is like

saying i have two to the power of what nine times one third is nine thirds which is really equal to two to the power of three or 8. so this is going to equal negative 8. And the reason i did this this way is because it's going to lead into what we're going to do next okay i wanted to give you a brief example so negative 512 raised to the third power is negative eight so what would i do in a scenario where i have nine to the power of three halves well just like in

the last example i'm able to kind of break this up using my power rules for exponents so what i can do i'm just going to look At this for a second i can rewrite this as 9 to the power of one half and then this is going to be raised to the third power and again if you use the power rules for exponents you'll see that you end up with the same thing i'd multiply the exponents here and keep the base the same this is equal to 9 raised to the power of 3 over 2

right so i can go back and Forth but the reason i would want to write it like this is because it makes it obvious what i need to do right i know that 9 to the power of 1 half at this point in your mind might still not register as the square root of not it does for me but i've been doing this for a long time so you might want to say okay this is the square root of 9 just following the processes we've been Doing right 9 is the radicand 2 is the index

then once i've written it like that i can raise this to the third power and then it's completely crystal clear that this is 3 right square root of minus 3. that's raised to the third power or 3 cubed which gives us a final answer here of 27. all right let's try another one we have 625 raised to the 3 4 power again i'm going to break this up And i'm going to think about this as 625 raised to the power of 1 4 and then i'm going to cube this and again i'm going to write

this thinking about this as the fourth root of 625 again this is the radicand so that comes down here 4 is the index and then i'm going to cube this guy so what is the fourth root of 625 talked About that earlier it's 5. so this would basically just be what it'd be 5 cubed which is 125. all right so now let's look at the more tedious rule you definitely want to write this down because this is one that takes a little bit longer to get down so if a is greater than or equal to

zero so again if a is non-negative and m and n are integers with n greater than zero okay so n is greater than zero because we want it to be positive So then we have this scenario a raised to the power of m over n is equal to the nth root of a and this is raised to the power of m this is what we just did don't let these letters confuse you if i was to write this as 9 to the power of 3 halves just like we just did all i did was 3

in this case is m so 3 is going to go on the outside n is 2 in this case so it's going to be The square root of right that's my index and the a in this case is just the radicand it's 9. okay it's the same thing so again if you struggle with generic formulas like this write it down and then before you do anything on each problem just say okay my a here is 9. write a 9 okay write a 9 underneath my m here is 3. i had a 3 there my n

here is 2 write a 2 there and then just translate it into this formula here Okay well i have a a 2 here i have a 9 here and i have a 3 here once you start doing that over and over and over again this becomes really really easy okay so especially if you struggle with these formulas please do that every time now for completeness sake we know the answer to this square root of 9 is 3 3 cubed is 27. now here's one you probably have not seen yet So we have a raised to

the power of a negative m over n so we have this is equal to 1 over we have the nth root of a and this is raised to the power of m there's no difference between this formula and this formula other than we just take the reciprocal we take the reciprocal of this guy okay and we raise it to this guy being positive it's just like when we saw 2 to The power of negative 4 we took the reciprocal of the base and we made the exponent positive so really to make this crystal clear i

could say if i have a to the power of negative m over n i could write this as 1 over a to the power of m over n right all i did was i took the reciprocal of the base so i have 1 over a and then the exponent just became positive instead of Negative m to positive n as an example of this let's say i saw something like 27 raised to the power of negative 2 thirds so following this 27 represents a the negative two the whole thing is representing the negative m or you

could say two represents m and you have a negative out in front however you want to think about that then n represents 3 okay So if i translate this over here i would basically have what i'd have 1 over 27 raised to the power of 2 3. and then if i said i had one over the cube root of 27 and then this guy would be squared i'd end up with a final answer of one over the cube root of 27 is three so it'd be three squared which is 9 so i'd end up with

1 9 in that scenario the key to this is just to write down the generic Formulas match things up and just go slowly right just take it step by step all right what if we saw 9 to the power of 5 halves again i'm going to break this up i'm going to say this is 9 to the power of one half and this is raised to the fifth power and again this would be the square root of nine raised to the fifth power which is what three to the fifth power Which is 243. what about

144 raised to the power of negative three halves again take the reciprocal of the base so take the reciprocal of 144 that's just one over 144 and then raise it to the power of three halves exponents just gonna become positive okay just that simple so then i break this up i have 1 over this is 144 Raised to the one-half power and then it's cubed so i'm going to write this as 1 over the square root of 144 and then that's going to be cubed and so the square root of 144 is 12. so this

ends up being 1 over 12 cubed which is one over 1728. what about negative 128 raised to the power of eight sevenths so again I can break this up and say this is negative 128. now this is inside of parentheses so this guy is going to be raised to the power of 1 7th and then i'm going to raise the whole thing let me just use brackets here to the eighth power and then let me write it in a better way for you to where it's more familiar this would be The radicand the negative 128

and i would have the seventh root because that's my denominator and this whole thing would be raised to the eighth power now we know that the seventh root of negative 128 is negative two so again negative 2 is the result there and then this is still inside of parentheses here and it's raised to the Eighth power so when we raise it to the eighth power we're going to end up with 256 and again because this negative is inside of parentheses this would end up being positive 256. all right now let's look at a few with

some variables involved the same rules apply okay just because you're working with variables doesn't mean it got more difficult all right so i have 2x to the power of two-thirds Times three x to the power four thirds times two x squared now we know what to do here we're going to multiply the coefficients together 2 times 3 is 6 6 times 2 is 12. then we have x in each case so it's the same base we're going to add the exponents so i have an exponent of two-thirds an exponent of four-thirds and an exponent of

two which to get a common denominator i'm going to write as Six-thirds two plus four is six six plus six is twelve so this is twelve thirds which equals four so my exponent on x would end up being a four so i end up with what 12x to the fourth power as my answer what about something like m to the five thirds power times three m squared Three times three is nine and then m to the five thirds power times m squared so we know m stays the same we just have five thirds plus two

so i could write two is six thirds so this ends up being eleven thirds so eleven thirds all right what about this one we have x to the power of negative two times y to the power of negative one Fourth and this is raised to the one-half power then we're multiplying by x to the power of negative two times x to the power of negative four times y to the five thirds power so the first thing i want to do is use my power rule here so x stays the same in this scenario and we

have negative 2 raised to the one-half power so negative 2 times one-half we know that this cancels with this i'd Have negative 1. so this is x to the power of negative 1 and then for y i have the power of negative 1 4 times one-half so 4 times 2 is 8 so this would be negative 1 8. okay then times over here if i have x and i have x that stays the same and then negative 2 plus negative 4 is negative 6. and then lastly i just have y to the 5 3 Power

so if i think about these two here i can combine them using the product rule for exponents i would have x to the power of negative 1 plus negative 6 which is negative 7 and then as far as y goes i can combine these so i'd have y to the power of if i have negative 1 8 plus 5 3 what's that going to be equal To well we've got to have a common denominator so let me multiply this by 8 over 8. let me multiply this by 3 over 3 i would get negative 3

over 24 plus 40 over 24 so this is 37 over 24 37 over 24. so you can leave it like this or you could write it as Y to the power of 37 over 24 over x to the power of 7. remember you can drag this into the denominator and change the sign of the exponent so if there's a negative 7 when it's in the numerator when it gets dragged below into the denominator it becomes positive hello and welcome to algebra 1 lesson 60. in this video we're going to learn about the square root property

so many lessons ago we gave you a basic definition for a Quadratic equation so here's an example of one we have x squared minus 9x plus 14 equals zero so what makes this a quadratic equation well a quadratic equation has a squared variable like here we have x squared and it's got no variable involved with an exponent that's larger than 2. so in other words if i added x cubed to this it would no longer be a quadratic equation Now another thing is that this quadratic equation is already in standard form for us and this

is just a convenient way to look at things and it's displayed such that we have our squared variable all the way to the left then we have the variable that's to the first power and then we don't have a variable here now some people in some textbooks will say well this is your spot for x to the power of Zero right x to the power of zero is one if i multiply 14 by 1 i just have 14 right so just kind of a mathy way of looking at that but we've seen quadratic equations before

up to this point we've only looked at ones that we could solve by factoring the purpose of this lesson is to start thinking about techniques that we can use when we can't factor our quadratic equation So if i have something like x squared minus 9x plus 14 equals 0 i can actually solve this by factoring we're going to do so to get a little bit of a review so if i factor the left side here just like factoring a trinomial so what would i need the first position is going to be x and then the

first position here is going to be x as well and then i want two integers whose sum Whose sum is negative nine and whose product okay whose product is 14. well again we think about the factors of 14 you've got 1 and 14 that's not going to work and then you've got 2 and 7. if i play with the signs i can make that work notice that you have a negative here and a positive here negative plus negative would give me a negative and negative times negative Would give me a positive so i want a

negative 2 and a negative 7. so now that i've factored this into the quantity x minus two times the quantity x minus seven and set this equal to zero what do i do next again remember i use my zero product property or some of you called the zero factor property doesn't matter so if i have something like a times b is equal to zero Remember that at least one of these has to be equal to zero for this to be true a could be equal to zero b could be equal to 0 or they both

could be equal to 0. well following the same logic here remember these are each factors that's multiplication so this is a factor and then this is a factor so just like i did there i would say that either x minus 2 is equal to 0 Or x minus 7 is equal to 0 or they're both equal to 0. so i would just solve each resulting equation add 2 to both sides over here and i'll get x is equal to 2 and or add 7 to both sides of the equation here i get x is equal

to 7. so this equation has two solutions x could be equal to 2 or x could be equal to 7. In the interest of time i'm not going to go through and plug things in and show you that the left and the right side are equal in value you can pause the video at this point and do that for yourself i can assure you that both solutions are valid so now that we've reviewed how to solve a quadratic equation by factoring what happens if we get something that we can't factor and i'm going to start

out with something that we can factor just so i Can prove it to you but let's just say for example i give you something really really easy so we have x squared is equal to 16. now this by definition is a quadratic equation i have a squared variable here and i have no variable with any higher power and i realize that some of you will say well you're missing the x to the first power that doesn't matter it's still a quadratic equation So how would i go about solving this for x i want you to

use what you've learned up to this point to think about how i would isolate how i would isolate x isolate x well think about when we isolate the variable we're usually trying to undo what's being done to it if i have an equation like x plus 3 equals 7 3 is being added to x so i do the opposite and i subtract 3 away from each side and i get x is equal to 4. if i have multiplication something like 6x equals 24 since 6 is multiplying x i do the opposite which is division to

both sides of the equation and this becomes x is equal to 4. so just like with these basic examples i want to again isolate x and if i'm squaring x the opposite of that is to Take the square root so if i take the square root of this side what's going to happen is this will cancel with this and i'll be left with x so i have isolated x i have successfully done that but again just like when we use the addition property of equality and we use the multiplication property of equality we've got to

do the same thing to each side so i've got to take the square root Over here now the square root of 16 is 4. so let's put x equals 4. now are we done is this our correct answer well it's partially correct x does equal 4 but x here can also be equal to negative 4 and let me show you why we missed this when we first start doing it recall a few lessons ago that we defined This symbol right here to mean the principal square root and we defined this symbol to be the negative

the negative square root and why did we show two different things well it's kind of a notational thing now when we first started talking about square roots if i said what is the square root of 64 I'm not even going to put the symbol i first defined it by saying give me a number that when multiplied by itself would give you 64. well you know that 8 would do that but also negative 8 would do that so to account for the two possibilities if i wanted the principal or positive square root of 64 i would

do that and if i wanted the negative square root of 64 i would do that so this would produce negative 8. Now what we didn't do here is account for the fact that negative 4 squared would also give us 16. so what's going to happen is when we take the square root of both sides let me kind of clean this up go back to what we had originally when we take the square root of each side we've got to put a symbol out in front over here that says plus or minus and this is just

notation that refers to These two different scenarios i want to take the positive and the negative square root of 16 when i do this so this will cancel with this and i have x equals plus or minus square root of 16 is 4. so this is actually two different solutions two solutions this is x is equal to 4 or x is equal to negative 4. And there's a lot of ways to prove that to yourself if you just go back to the original equation if i plug in a 4 4 squared is 16 if i

plug in a negative 4 where the negative and the 4 inside of parentheses i square that i would also get 16. and one last quick way to prove that to yourself just use what you know let's say that i looked at this equation In a different way let's say that we go over here and we say that okay we have x squared is equal to 16. i want you to subtract 16 away from each side of the equation so we know this would cancel and become 0. now can i factor the left side of this

equation yes i can this is the difference of two squares i have x squared and really i could write 16 if i wanted to as four squared Okay so the difference of two squares which factors into x plus four that quantity times x minus four that quantity and this equals 0. if i set each of these factors equal to 0 and i solve we know that we have x plus 4 equals 0 [Music] or x minus 4 equals 0 which gives us two solutions this one subtract four away from each Side you get x equals

negative four this one add four to both sides and you get x is equal to four so it's exactly what we got here so you can see that using this square root property which i'm going to define for you in a minute gives you the same answer as when you solve it using factoring all right so let's go over the official definition of the square root property so when we have a quadratic equation of the form X squared equals k so something squared it doesn't have to absolutely be x we'll see some more complicated examples

at the end of this lesson but let's just take this slow so let's just say we have x squared is equal to some number k and k is greater than zero so that means k is positive well then we can use the square root property to get a solution and the square root property looks like this again if k is greater than 0 Meaning k is positive and x squared is equal to k then x is equal to the principal square root of k so the principle or x is equal to the negative the negative

square root of k and generally what we do to save space is we'll write x is equal to the positive or principal square root of k and then we put the negative Underneath and then we write square root of k so x equals plus or minus that's how it's read plus or minus the square root of k this is just a compact way to write this okay so let's take a look at an easy one so we have x squared equals 25. so again if x squared is equal to k and k is greater than

0 Then x is equal to plus or minus the square root of k so i'm going to do that over here i'm going to take the square root of each side but again i've got to account for the positive and the negative version so let's write that down here so we'll have x squared is equal to 25 and i'm going to take the square root of this side and then over here i'm going To put plus or minus the square root of this side so we know at this point that this cancels with this and

we'll have just x then it equals plus or minus the square root of 25 we know at this point is 5. so what this notation is telling me is i have again two solutions x can be equal to 5 or x could be equal to negative 5. And you can easily see if i plugged in a 5 here 5 square would be 25 if i plugged in a negative 5 negative 5 inside of parentheses squared would also be 25. all right let's take a look at another easy one so again if x squared equals k

and k is positive then x equals plus or minus the square root of k so x squared equals four so we have this same scenario here So x squared equals four i'm going to take the square root of this side and then over here i'm going to put plus or minus the square root of this so this cancels with this and i just have x and it's equal to plus or minus the square root of 4. square root of 4 is 2. so again this tells me i have two solutions x is equal to 2

Or x is equal to negative 2. and again if you plug this in 2 squared is obviously 4 and then negative 2 negative 2 times negative 2 or negative 2 squared is also 4. all right so let's look at one that's a little bit more complex so let's say you're in this section on the square root property and you turn to a problem that looks like this 6k squared equals 54 and you're thinking okay well i have a Square here so let me take the square root of 6k squared equals 54 plus or minus square

root of that is that gonna get you where you need to go no it is not so by the formula what you want is something squared equals to some number now we don't have a perfect square because 6 is not a perfect square So what you want to do is think about basic things that you could do to get a perfect square over here now we're going to get into some very complicated examples of this in the next section where we talk about completing the square but for right now i want you just to think

about what can i do to get k squared by itself well i can divide both sides of the Equation by six this would yield k squared is equal to nine so now i have a perfect square on the left and what's going to happen is if i take the square root of k squared and set this equal to plus or minus the square root of 9 i get my solution right this is going to give me k Is equal to plus or minus square root of nine is three so this yields k equals three or

k equals negative three so does this work as a solution well let's erase all of this over here and let's plug in so if i had 6 times 3 squared would that equal 54 3 squared is 9 9 times 6 is 54 so that checks out What if i had 6 times negative 3 inside of parentheses squared well it's the same thing i end up with 6 times 9 6 times 9 is again 54 so this checks out as well now additionally we can use this rule when we have a squared binomial so i just

want you to pretend that you're substituting in for that x you're substituting this whole thing so if x minus 5 that quantity squared is equal to 169. again to get x by itself i've got to first get rid of this squared here so let me take the square root of this side and then plus or minus the square root of this side now it's going to be a little bit more work in terms of simplification because now this will cancel with only this and i'm left with this x minus 5 here and this equals plus

or minus the square root of 169 Which is plus or minus 13. now i have two possible scenarios here so that's where it becomes more work i have to separate this now into x minus 5 is equal to positive 13. or x minus 5 is equal to negative 13. so on these more complex examples you're going to have more things that you have to do so let me add 5 to both sides of the Equation here and i get x is equal to 18 then or let me add 5 to both sides of the equation

here and i'm going to get x is equal to negative 8. all right so let's copy this and we'll go back up to the top and we're going to check our solutions so again we found that x was equal to 18 or x was equal to negative 8. so let's start by plugging in an 18. so if i had 18 Minus 5 and that amount was squared would this equal 169. 18 minus 5 is 13 so we'd have 13 squared equals 169 so that checks out for the next one we have negative 8. so if

i plug in a negative 8 there so we have negative 8 minus 5 that's negative 13. so if i had negative 13 inside of parenthesis and that was squared that would also give me positive 169 so this checks out as well all right let's try another one like that so let's Suppose we had the quantity 2x minus 5 and that's squared and this equals 225 so again simple setup i'm just trying to get rid of the squared right there so i can get x by itself in the end so let's rewrite this i'm going to

take the square root of this side and i'm going to put plus or minus the square root of this side so then this will cancel with this and i'm left with two x minus five Is equal to plus or minus the square root of two hundred twenty-five the square root of two hundred twenty-five in case you don't know this is fifteen so this sets up two different equations to solve you're going to have 2x minus 5 is equal to positive 15. then or you're going to have 2x minus 5 is equal to negative 15. so

let's go ahead and add 5 to both Sides of the equation here you'll have 2x is equal to 20. divide both sides of the equation by 2 you're going to get x is equal to 10. okay then or we have plus 5 on both sides here we'll have 2x is equal to negative 10. when we scroll down a little bit we divide both sides by 2 and we're going to get x is equal to negative 5. so here's my two solutions x equals 10 Or x equals negative 5. so let's erase everything and check so

again x equals 10 or x equals negative 5. okay so we'll have inside of parentheses 2 times 10 minus 5 and this is squared this equals 225 20 is 20 20 minus 5 is 15. 15 squared is 225 [Music] so this checks out what about x equals negative five so if we multiply there 2 times negative 5 is negative 10 negative 10 minus 5 is negative 15 so you'd end up with negative 15 squared is equal to 225 and that also gives me a true statement you'll get 225 equals 225 and so this checks out

as well all right For the next one we have 3k minus 1 that quantity squared is equal to 24. so again i'm going to take 3k minus 1. this is squared equals 24. i'm going to take the square root of this side so that this will cancel with this and i'm left with this so i've got 3k minus 1 over here and then over here i've got my plus or minus The square root of 24 so equals plus or minus the square root of 24. now here we're going to have one that's a little bit

more tedious and the reason is that 24 is not a perfect square so we have to do some simplification here so let's write 3k minus 1 is equal to plus or minus if i think about the square root of 24 and just kind of do this off on the side this is equal to the square root of 4 Times the square root of 6 and the square root of 4 is 2. so really this is plus or minus 2 times the square root of 6. now again i've got to set up two different equations so

i would have 3k minus 1 is equal to 2 times the square root of 6. then or i'll have 3k Minus 1 is equal to the negative of this so negative 2 times the square root of 6. all right so to solve this equation i would add 1 to each side so i'll have 3k is equal to 2 times the square root of 6 plus 1 and then i would divide each side by 3 so i could isolate k and i'm going to have k is equal to 2 times the square root of 6 plus

1 over 3 and then over here i would add 1 to each side of the equation so i'd have 3k is equal to negative 2 times the square root of 6 plus 1 and then divide again both sides of the equation by 3 so i'll put or k is equal to negative 2 times the square root of 6 plus 1 over 3. so this is an example of some sloppy Solutions that you're going to get kind of moving forward unfortunately as you get into higher math a lot of solutions are much worse than this and

when you go back to try to check things in your original equation it becomes really really tedious so because this is the last problem we have some time we're going to go back and check these two solutions okay so Plugging this in will be kind of tedious we're going to have an open parenthesis here we have 3 and then times k so this is our value for k so our value for k is two times the square root of six plus one over three and then we have minus one so then minus one This whole

thing is squared and then it equals twenty four so before i square anything i want you to realize that you can cancel this 3 with this 3. so what would i have then inside well inside the parentheses now i have 2 times square root of 6 plus one minus one this is squared and it should equal 24. now these two would cancel And i'm really just left with two times the square root of six this is squared and it equals 24. let's keep scrolling down now you realize that this is multiplication here so really i

have 2 squared times the square root of 6 squared equals 24. 2 squared is 4. the square root of 6 squared is 6. so 4 times 6 equals 24 and it does so you get 24 equals 24. and again this gets more and more Tedious as you get higher in math so i can't tell you that you're going to have time to check everything on your test because in some situations you simply just won't have time well you want to check as many things as you can particularly if you go through your chest and you

finish everything up and you have time left please take that time to go back and check as many answers as you Can because even if you catch one or two mistakes that can save you from getting a b when you could have got an a all right so let's erase this we'll put a check mark here and i'm going to reuse this with the only exception of this being a negative out in front so the same thing is basically going to happen let me kind of scroll down a little bit this again would cancel with

this And i'd have negative 2 times square root of 6 plus 1 minus 1. this would obviously cancel so i have negative 2 times square root of 6 this is squared and it equals 24. so now negative 2 is inside of parentheses so negative 2 would be squared and then times square root of 6 which would be squared and i'm just using power to power rule here Equals 24 negative 2 squared is 4 so you get 4 times square root of 6 squared is 6. this equals 24 and it does you get 24 equals 24.

so this guy checks out as a valid solution as well hello and welcome to algebra 1 lesson 61 in this video we're going to learn about solving quadratic equations by completing the square so so far in this course we've learned how to solve a quadratic equation By factoring which we're not going to be able to do in every scenario and then also in the last lesson we learned how to solve a quadratic equation by using something known as the square root property now the thing about the square root property you can use it on every

quadratic equation but you have to have the quadratic equation in a certain format to do that so the purpose of this lesson is to learn how to complete the Square and that's going to put it in a format where you can use the square root property so before we kind of jump in and start learning about this i want to make sure that everybody is on the same page with the square root property so in the last lesson i started you off with something generic and i said if k was greater than 0 Meaning k

was positive and we had something like x squared was equal to k then what we said was that x could be equal to the principal square root of k or x could be equal to the negative square root of k and it's important that you understand why we have these two solutions to use an example let's say we saw something like x squared is equal to 64. Well if i take the square root of this side right here let me just rewrite this that's going to undo the squaring so i know that i will isolate

x and have it by itself but if i just take the principal square root of 64 that's not going to give me the whole picture the principal square root of 64 by definition is 8. so x would equal 8 here if i plugged in An 8 and i squared it i would get 64. but just as it shows you here i have the principal square roots and the negative square root so what i'd want here is a symbol for the negative square root of 64 as well so we put a plus and then a minus

underneath and this notation just tells me that i want the positive or principal square root of 64 along with the negative square root of 64. and so x would be 8 or X would also be the negative square root of 64 which is negative 8. and if i plugged in a negative 8 for x negative 8 squared would also give me 64. so this is just a way to account for both solutions and if i use this on a more complex example it's the same thing let's say i had something like the quantity x minus

5 this is squared and it's equal to 4. now If i take the square root of this side right here i would cancel this with the squared right so that's going to go away and i'm left with x minus 5. but over here i've got to make sure that i account for the positive and the negative square root of 4. and when i solve this i'm going to explain y so let's say this is equal to we know the square root of 4 is 2. so positive or negative two right You'd read that as plus

or minus two so this is two different scenarios it's x minus five is equal to two or x minus five is equal to negative two so let's pause for a minute and see why that would make sense just without even doing anything else let me erase this we're going to go back to the original equation if you find a value for x that when plugged in We subtract away 5 and get two well that means that this value here would be two two squared would give me four same thing over here if i find a

value for x that when plugged in we subtract away five we get negative two well this in here would be negative 2 negative 2 squared will also be 4. so we're accounting for those two possibilities so let me just erase this real quick and we're going to solve this and i'm Going to show you that so if i add 5 if i add 5 to both sides of the equation i get x is equal to 7. if i add 5 to both sides of the equation here i would get x is equal to 3. plug

in a 7 there 7 minus 5 is 2 2 squared is 4. plug in a 3 there 3 minus 5 is negative 2 the negative 2 and just to show you this The negative 2 would be inside of parentheses negative 2 squared would also be 4. so that's why it's so important to make sure we put that plus or minus out in front as we've shown here we could write this as x is equal to plus or minus the square root of k so now that we've reviewed the square root property i want to just

take a minute and talk about a perfect square trinomial so before we Get to that remember a perfect square in general is just a number with a rational square root so something like 4 or let's say 100 or 169 if i take the square root of that number i'm going to get a rational number back so something like 4 if i take the square root i get 2. something like 100 if i take the square root i get 10. something like 169 if i take the square root i get 13. But something that's not a

perfect square let's say i take the square root of 5 square root of 5. well we know that this is an irrational number and if we were to try to write this out in decimal form the decimal does not terminate and it does not repeat the same pattern forever and ever and ever so we end up having to approximate these so that's the difference between having Something that's a perfect square and not a perfect square when we get more advanced into this we talk about something known as a perfect square trinomial and i gave you

this back when we talked about special products back when we talked about special factoring but you might not have understood what it was yet because we hadn't really developed everything now you should have a complete Understanding of what a perfect square trinomial is a perfect square trinomial is just a trinomial that can be factored as the square of a binomial so as an example x squared plus 2xy plus y squared can be factored as the square of a binomial x squared minus 2xy plus y squared can be factored as the square of a binomial something

like x squared plus 10x Plus 25 can be factored into x plus 5 that quantity squared now why would something like this be important for us in this lesson well remember i just showed you that if you have something in this format let's say i had x plus 5 that quantity squared and it was equal to some number let's just say it's equal to 3. well i know i can solve this using the Square root property so what we're going to be doing in this lesson is we're going to take trinomials that are not perfect

square trinomials and we're going to make them into perfect square trinomials through a process known as completing the square okay so you're going to see we're going to do that and then we're going to get the left side into this format the right Side as just a number or a constant and then we're going to use the square root property to end up getting our solution all right so here's the official procedure for completing the square and i would advise you to write this down and use it as we work through these problems and then

in your textbook i would just refer to this sheet over and over and over again until you are just flowing through these problems and you've memorized this process So you want to write the quadratic equation where the variable terms okay the variable terms ax squared and bx are on one side and the constant is on the other so all the terms with variables on one side all the constants on the other and you're just going to use the addition property of equality to move things around now the next thing you want to do and This

may or may not be necessary depends on what kind of problem you're looking at you want to make the coefficient on the squared term one by dividing both sides by the original coefficient in the beginning of your textbook the problems they're going to give you will be easy and you won't have to do this right you'll just have an x squared but later on you'll get something like 6x squared or 10x squared or 9x squared Where you're going to have to deal with that but you want the coefficient on the squared term to be 1.

all right so here comes the next step and this is the one that everybody gets wrong when they first start out you're going to complete the square by adding one half of the coefficient of the bx term squared to both sides okay let me let me just reread that complete the square by adding one half Of the coefficient of the bx term squared to both sides so i want you to write in your notes cut it in half then square it this is how my eighth grade teacher taught it to me and it's stuck in

my head for so so many years she would say cut it in half and square it okay so every time that i Would go to a problem and i struggled with this i remember what do i do cut it in half and square it caught it in half and squared just keep repeating that to yourself and then completing the square will just be you know many years from now 10 years from now you're you know going to help your kid on their homework and how do i complete the square cut in half and square okay

very very good technique to remember this you may say what is the bx term This is just your term where the variable has an exponent of one sometimes in your book if you're reading the procedure there will say the first degree okay it just means that the exponent on the variable is a one then once you've gone through these previous steps the final thing you're going to do is you're going to factor and solve the equation using the square root property all right so let's start Out with our first example and i gave you one

that was really really easy in fact i gave you one that you could solve using factor so if you wanted to pause the video and factor this in fact we can just do it really quickly we could have an x here and an x here give me two integers whose sum is negative 8 and whose product is negative 33. well i know i could do negative 11 and positive 3. so Negative 11 and positive 3. we know at this point that the solutions would be x is equal to 11 or x is equal to negative

3. right we can eyeball that and see it at this point hopefully so let's not use this though let's use our completing the square method and let's start out at step number one so for step number one we want to write the quadratic equation where the variable terms ax squared and bx are on One side and the constant is on the other so in other words these are your variable terms i want them by themselves so what i've got to do is some basic algebra and i'm going to add 33 to both sides of the

equation so that's going to give me x squared minus 8x is equal to 33. now The next thing is to make the coefficient on the square term one but on the squared term here i already have a coefficient of a one so i can just skip that and in the case of algebra one most of your problems will be pretty easy where you don't have to deal with that pretty much towards the end of your section on this you'll get a few of them but mostly this is going to Occur in algebra 2. now the

next thing i want to do is complete the square when i complete the square it's the most kind of tedious thing that you're going to do because you have a lot of things that are going on so let's let's kind of go slowly here so we complete the square by adding one half of the coefficient of the bx term squared to both sides okay this is where you hear me say cut it in half and square it so this is Your coefficient for the bx term okay you're basically your x raised to the first power

so if i cut it in half meaning i multiply negative eight by one half and i square that i'm going to get what this is going to be equal to negative 8 times 1 half is negative 4 so i'd have negative 4 squared so this is equal to 16. So essentially i want to add 16 to each side of the equation so i'd have x squared minus 8x plus 16 is equal to 33 plus 16. now what i've done here is i've just completed the square if you'll notice now this is a perfect square this

is a perfect Square trinomial so i know that this is going to factor into a binomial squared and it would be what it would be x minus i have that minus there and the square root of 16 would be 4. so x minus 4 of that quantity squared is equal to over here 33 plus 16 is 49. so now i have something where i can use my square root property and in the final step it says to factor which we just did and then to solve using our square root Property so i'm going to take

the square root of this side and then over here i'm going to put plus or minus the square root of this side and so what i'm going to have is x minus 4 because this would cancel with this is equal to plus or minus the square root of 49 is seven so this gives me two different scenarios Two different scenarios it gives me x minus four is equal to seven or x minus four is equal to negative 7. so then i solve this add 4 to each side of the equation i get x equals 11

or add 4 to each side of the equation i get x is equal to negative 3. now you'll recall at the very beginning of this problem i solved it using factoring and i got the same thing And you might say why in the world would you use completing the square that took way longer than factoring well a lot of cases you're not going to be able to use factoring and additionally we're not going to be using completing the square for more than a week in our algebra 1 class in the next lesson we're going to

learn about something that was a quadratic formula and once you learn that you're not going to use factoring and you're not going to Use completing the square pretty much you're going to use that you just plug numbers into a generic formula and it spits out an answer for you but before you can get to that you need to understand this here so in the interest of time i'm not going to check this normally i check everything but we have a lot more problems that we want to cover i don't make the video too long so

pause the video go back up to the original Equation and plug in an 11 for x and plug in a negative 3 for x and you'll see that both of these would be a valid solution all right let's take a look at the next one so we have x squared minus 10x minus 39 equals 0. so again the first thing i do is i move the variable terms on one side and i move the constant to the other so in this case i would add 39 to each side of the equation And i would have

x squared minus 10x is equal to 39 and then the next thing is to make sure the coefficient on the square term is one in this case we already have that then we're going to complete the square by adding one half of the coefficient of the bx term squared to both sides so again what i want to do cut this in half and square it cut it in half and square it so negative 10 Caught it in half that's like multiplying by half and then square so this is equal to that would be negative 10

over 2 is negative 5 negative 5 squared is 25. so i'm going to add 25 to each side of the equation so i'm going to have x squared minus 10x plus 25 Is equal to 39 plus 25 and 39 plus 25 is 64. and over here i would factor this this is now a perfect square trinomial so it can be factored into a binomial squared and this would be x i have minus there square root of 25 is 5. so x minus 5 that quantity squared so now we can use our square root property and

let me just rewrite this we have x minus 5 this quantity squared is equal to 64. And i want to take the square root of this side and then i want to do plus or minus the square root of this side and so what i'm going to get is this cancels with this i'll have x minus 5 is equal to plus or minus the square root of 64 is 8. so i would have two different scenarios x minus 5 is equal to 8 or x minus 5 is equal to negative 8. so we'd solve these

add 5 to each side Of the equation here and i would get x is equal to 13 add 5 to each side of the equation here and i will get x is equal to negative 3. so again in the interest of time i'm not going to check these so you can pause the video come back to this original equation x squared minus 10x minus 39 equals 0. and then plug in for x you had x equals 13 that's one solution and then another solution again is x Equals negative 3. in each case you're going to

see that the left and the right sides would be equal and so those are two valid solutions all right let's take a look at another one so we have negative 3x squared minus 3x minus 21 is equal to negative 4x squared minus 5. so knowing that i want all the variable terms on one side and all the constants on the other if i add 4x squared to each side of the equation [Music] This is going to go away and on this side i'm going to just have x squared then minus 3x and what i want

to do here i'm going to move this over here so let me add 21 to each side of the equation that's gone so this will be equal to negative 5 plus 21 is 16. so now i have x squared minus 3x equals 16. now the coefficient on the x squared is a 1. so i need to do anything there So now i want to move into my complete the square step so i take the coefficient for the x to the first power here so that's negative three i cut it in half and i square it

cut it in half so i multiply it by one half and then i square so negative three times one half is just negative three halves And then if i square that i'm gonna get nine fourths okay nine fourths so i'm going to be adding that to both sides of the equation so i'm going to add 9 4 here and i'm going to add 9 4 here let me scroll down so if i add that over here i'll have x squared minus 3x plus 9 4. then over here if i get a common denominator going i'll

multiply this by 4 over 4. 4 times 16 is of course 64. so i'd have 64 plus 9 over 4 and 64 plus 9 is 73 so this would be 73 4. now at this point this is a perfect square trinomial i can factor this and i know a lot of you aren't used to seeing fractions when you go to factor Stuff it's not any more difficult just follow your pattern so this is going to be x you have a minus here so this is minus what's going to go in that position well just take

the square root of this the square root of 9 is 3 and the square root of 4 is 2. so this is going to be 3 halves and go ahead and foil that out if you don't believe me you're going to end up with x squared minus 3x plus 9 4. Okay so this equals 73 4. and now we use our square root property so i take the square root of this side and then plus or minus the square root of this side and what i'm going to end up with is x minus 3 halves

is equal to plus or minus now i can split this up into the square root of 73 over the square root of 4 which is 2. so square root of 73 over 2. so this is two Different scenarios i would have x minus 3 halves is equal to the square root of 73 over 2 and then or x minus 3 halves is equal to the negative square root of 73 over 2. all right so the next thing i want to do is add 3 halves to both sides of the equation over here so i will

have x is equal to the square root of 73 plus 3 Over 2. can't do anything else really here that's kind of the best i can do and then or over here i'm going to add 3 halves to both sides of the equation [Music] and so i'm going to have x is equal to the negative square root of 73 plus 3 over 2. so really i can condense this and say x Is equal to plus or minus the square root of 73 plus 3 over 2. all right let's take a look at another one so

i have negative 2k squared minus 6k plus 3 is equal to negative 4k squared so what i want to do let me just add 4k squared to each side of the equation to start and let me subtract 3 away from each side of the equation So negative 2k squared plus 4k squared is 2k squared then minus 6k this would cancel so that's gone and this is equal to this would cancel just negative 3. so now we have a scenario that we haven't come across before notice that the coefficient on the squared term is a 2.

so we don't want that so the way we're going to get rid of it is we're Going to divide both sides of the equation by 2. so i divide this side of the equation by 2 and i divide this side of the equation by 2 and what that's going to give me is you can just split this up and have 2 k squared over 2 minus 6 k over 2 is equal to negative 3 over 2. and then we can see that This cancels with this and i just have k squared so that's what i

want i want k squared so the coefficient out here is a 1 minus 6 over 2 is 3 so minus 3 k is equal to negative 3 halves so now that i've done that i want to get into completing the square so i want to look at the coefficient for the variable raised to the first power so that's negative 3 here And i want to cut it in half i want to cut it in half so that means multiply by half and then i want to square it okay so negative three times one half is

negative three halves and if i square that i get 9 4 okay 9 4. so i'm going to add 9 4 to each side of this equation so i'll have k squared minus 3 k plus 9 4 is equal to negative three halves Plus nine fourths okay so over here on the right i would get a common denominator going multiply this by two over two and i'm going to have negative 6 plus 9 which is 3 over 4. this is 3 4. over here i'm going to go ahead and factor this this is a perfect

square Trinomial now so i'm going to factor this into i have k i have minus and then what's the square root of nine fourths well we know it's three halves so i get k minus three halves that quantity is squared and this equals 3 4. so now i can use my square root property take the square root of this side and then plus or minus the square root of this side And i'll end up with k minus 3 halves right because this would cancel with this and this is equal to plus or minus the square

root of three fourths now the square root of four is two so really i could rewrite this as plus or minus the square root of 3 over 2. so let's scroll down And we'll look at these two scenarios so we'll have k minus 3 halves is equal to square root of 3 over 2 or we'll also see k minus 3 halves is equal to the negative square root of 3 over 2. so in each scenario to isolate k i'm going to be adding 3 halves to each side of the equation And what's going to happen

is this is going to cancel and this is going to cancel and what i'm going to be left with is k is equal to the square root of 3 plus 3 over that common denominator of 2. or we could say k is equal to the negative square root of 3 Plus 3 over the common denominator of 2 where you could say k is equal to plus or minus the square root of 3 plus 3 all over 2. hello and welcome to algebra 1 lesson 62. in this video we're going to learn about solving quadratic equations

using the quadratic formula so when we first started learning about quadratic equations we only dealt with quadratic equations that we could solve via Factoring and we kept it that way for a while and then in the last lesson we learned how to solve a quadratic equation whether we could factor it or not using a process known as completing the square now completing the square is one of those things where it's it's very very tedious and there's so many steps involved it's very easy to make a mistake so in this lesson we're going to look at

an easier process to solve any Quadratic equation that you're going to come across so this is known as solving a quadratic equation using the quadratic formula but before we begin using the formula let's take a look at where it actually comes from so we see ax squared plus bx plus c equals zero this is standard form and of course this is a generic example all i'm looking at here is i have my Squared variable first my variable to the first power second and then my constant third so i could have something like 6x squared plus

2x minus 7 equals 0. this is standard form and one of the things we're going to have to do is match things up so in other words this right here the coefficient for x squared that's a this 2 right here the coefficient for x to the first power That's b this negative 7 is going to be c or the constant term and it's very important that you include the sign with this because remember i could write this as a plus and then a negative set okay so that c there is negative 7. and the variable

does not have to be x again that's just a generic example you can see something like 3m squared plus 4m minus 2 equals 0. now you have 3 as the coefficient for m squared so 3 is a 4 is the coefficient for m to the first power so 4 is b and then negative 2 would be c that's your constant now let me erase this real quick and i want to show you how to derive the quadratic formula and after that we're going to look at some examples using the quadratic formula and you do a

few Examples of that and basically you're good to go right you're just plugging things in as long as you can do some basic arithmetic you're good to go so for this guy right here ax squared plus bx plus c equals 0 what is the solution and i realize we don't have numbers we have an a a b and a c that represent numbers but how would i get a solution well we know already that we can use completing the square so let's go ahead And do that let's start out by doing what let's just copy

this ax squared plus bx plus c equals 0. i know i want all the variable terms on one side these are the variable terms and i want my constant on the other so i got to move that over so how to do that again i just subtract c away from each side of the equation so i would have ax squared plus bx Is equal to negative c this is going to cancel so once i've done that the next thing that i've told you is that you always want the coefficient out here to be a 1.

now when you use a generic example like this you have a that represents your coefficient a could be 1 A could really be any number that you choose it just can't be 0 because if i had a 0 here let's say i put 0 there 0 times x squared is 0 you no longer have a quadratic equation so actually that's one restriction that you have with quadratic equation your a or your coefficient for the squared variable cannot be zero right now a can be anything other than zero so to deal with that we're going to

get rid of a from right there and i do That with some basic algebra i divide both sides of the equation by a and what happens is it's going to cancel here and now the coefficient for x squared is definitely 1. so i have x squared plus b over a times x is equal to negative c over a so the next step is the one that everybody struggles with we're going to complete the square let me just copy this real quick i'm going to move it Down i have some room and to complete the square

i take one half of the coefficient of the first degree term and i square it okay and then i'm going to add that to both sides so this is the coefficient for the first degree term this is x to the first power and this is your coefficient and again when we work with kind of generic examples it's hard to see that right but this is your variable here x is the Variable b over a is the coefficient so i want to cut it in half so i'll multiply it by half and then i want to

square it so this would be b over 2a b over 2a and if i square that b squared is b squared and then 2 squared is 4 a squared is a squared so this is the value that i'm going to add to each side of the equation so let me just kind of slide this down And so i'm going to add b squared over 4a squared to both sides of the equation now on the right side of the equation i want to get a common denominator going so how can i do that well here i

have 4 a squared here i just have a so if i multiplied by 4 a top and bottom here i would get let me kind of do this down here b squared minus 4a times c is 4ac all over this would now be a common denominator of 4a squared okay so the right side that's good to go so now the left side is a perfect square trinomial so we're going to factor this guy we know the first position would be x and then there's a plus there and then in this position the square root of

b squared is b the square root of 4a squared is 2a so now what i want to do is use my Square root property let me take the square root of this side and i'm going to go plus or minus the square root of this side and of course this is going to cancel with this and i'm going to have x plus b over 2a is equal to plus or minus the square root of b squared minus 4ac over 4a squared all right let's scroll down and one of the things we can do is we

can break this up right i know if i have the square root of let's say 1 4 this is equal to the square root of 1 over the square root of 4 which is equal to 1 over 2 or one-half so i can do the same thing here so i can write this as plus or minus that can stay just with the numerator the square root of b Squared minus 4ac over the square root of 4a squared and over here on the left this isn't going to change we just have x plus b over 2a

now why did i say this can just stay in the numerator well just assume for a minute that everything was positive right i have the positive version well nothing would change let me just erase that real quick you would Just have this over this if i had the negative version well only one of them can be negative if i had a negative over a negative well then that's going to give me a positive so when you have this plus or minus it's really just out in front saying i have a positive version of this and

a negative version of this so you really want 1 plus or minus okay so i just stick that in front of the numerator Now in the denominator the square root of 4a squared would be 2a so how can i get x by itself well i can just subtract b over 2a away from each side of the equation and let me scroll down a little bit what we're going to have here this will cancel is x is equal to i'll have negative b over 2a then i'll put plus i'll have plus or minus The square root

of b squared minus 4ac over 2 a now i have a common denominator here i have 2a here and 2a here so really i can write this in a more compact form and in the format you're going to see in your textbook as x is equal to negative b plus or minus the square root of b squared minus 4ac over the common denominator of 2a And you might say well yeah you solved it for x but what is this this looks like a bunch of nonsense all i need to do to get the solution for

a quadratic equation is plug in so you remember when we saw the quadratic equation in standard form that a b and c represented real numbers right it's just a restriction where a cannot be 0. but If i plug in for a b and c i'm going to get a solution for my quadratic equation so let me copy this real quick and we're going to go down and i'm going to label this as the quadratic formula and let me just give you a quick example before we start kind of going into some more detail here if

i saw something easy like x squared minus 5x Plus 6 equals 0 this is something we can solve with factoring so two integers whose sum is going to be negative five and whose product is six would be negative two and negative three so we can look at this at this point and hopefully know that the solutions would be x equals 2 or x equals 3. all right i'd set each one of these equal to 0 and i'd solve if i plugged into the quadratic formula All i would need to do is say okay a generic

quadratic equation in standard form is ax squared plus bx plus c equals zero so in this case in this case a is equal to what it's equal to one b is equal to what it's equal to negative 5 and c is equal to what it's equal to 6. right i'm just lining things up With that generic example so now we just plug into the formula so i would have x is equal to negative b the negative of negative 5 is positive 5. so i'd have negative negative 5. again that's positive 5. then you have plus

or minus so i'm going to do the positive scenario first so plus the square root of you have b squared negative 5 squared is 25 minus 4 times a times c 4 times 1 is 4 times 6 is 24. then this is all over 2 times a a in this case is 1 so all over 2. so 25 minus 24 is 1. so you'd have the square root of 1 which is 1. so 5 plus 1 is 6 so you'd have 6 over 2 which equals 3 which is one of your solutions right there Now

if we do the negative scenario the negative scenario you'd have again negative b so negative of negative 5 is 5. then minus the square root of we know this is going to be the same b squared you have negative 5 squared that's 25 minus 4 times a a is 1 times c c is 6. so 4 times 1 times 6 is 24. so this ends up being the square root of 1 again so this is 1. over 2 times a again a is 1 so over 2. 5 minus 1 is 4. 4 over 2 is

2 so that's 2 and that's how we get our other solution right there so you can see that plugging into the quadratic formula is pretty easy i don't need to factor anything i don't need to complete any squares i don't need to do anything other than line up my equation in standard form Figure out what a is b is and c is and then plug in and get an answer okay it's very very simple all right so let's go through some technical details so b squared minus 4ac this is the part that's under the square

root symbol is called the discriminant write that down very very important and it comes up a lot especially in algebra 2 and college algebra if b squared minus 4ac equals 0 There is one solution one solution okay if b squared minus 4ac is greater than 0 like we just had in the last example there are two solutions okay two solutions and then if b squared minus four ac is less than zero there is no real solution no real solution and in some algebra 1 courses you're going to deal with negative square roots In other algebra

courses like this one i'm not going to take it on i'm going to wait for algebra 2 so that you can completely understand what's going on i've taught in algebra 1 before and i feel like students just don't have a good enough understanding of the basic processes to get into that so in the next course i'm going to do which is algebra 2 we are going to talk about what to do When b squared minus 4ac is less than 0. there is a solution for it but involves the complex number system which we just haven't

covered yet all right so let's take a look at some practice problems so we have 4x squared minus 5x equals 6. the first thing i'm going to do is i'm going to write the generic form for the quadratic equation in standard form you should have this on a card and it should be something that you memorize so it's Ax squared plus bx plus c is equal to zero now this guy right here does not match this guy okay what we need to do is move our constant term over to the left and again we do

that with basic algebra we just subtract 6 away from each side of the equation and i'm going to have and let me just write it over here 4x squared Minus 5x i subtracted 6 away so minus 6 this would cancel so equals 0. now the next thing that i want to do is line everything up and i want to figure out what is a what is b and what is c okay i need those to plug into the formula so a is the coefficient for the x squared so that's 4. b is the coefficient for

the x to the First power so that's negative 5. you need the negative there don't just put 5. and then c is negative 6. it's the constant you need the negative don't just put 6. so once we have this information we plug it into the quadratic formula so again the quadratic formula is x is equal to negative b plus or minus The square root of b squared minus 4ac this is all over 2a so what are we going to have here we'll have x is equal to negative b so i put negative and then for

b i put negative 5 in there plus or minus the square root of we have b squared so let's put negative 5 and let's square that minus 4 times a that's 4 Times c that's negative 6 and this is all over 2 times a a is 4 so 2 times 4. all right so this is the first scenario so then x is equal to the negative of negative 5 is 5 plus the square root of negative 5 squared is 25 minus 4 times 4 is 16 16 times negative 6 is negative 96. so you'd have

minus negative 96 or plus 96. so if i add these two Together i'm going to get 121. [Music] and this is over 2 times 4 is 8. now what we just talked about was that if the result of this in here is greater than 0 i'm going to have two solutions if it equals zero i have one solution why is that the case because the square root of zero would be zero this would be gone and i just have in this case if that did happen i'd have Five over eight okay that's not the case

here i have the square root of 121 which is 11. so this is 11. 5 plus 11 is 16. 16 over 8 is 2. so i get x equals 2 as one of my solutions then for the other solution this would all be the same so negative of negative 5 is 5. i'd use minus now so minus the square root of we know this turns out to be 121 so Let's just put minus 11 over 8. so 5 minus 11 is negative 6 so this would be negative 6 over 8 which would reduce to negative

3 4. so we get x equals negative 3 4. okay so pretty easy overall let me erase all this with these they're pretty tedious to check so i'm not going to check all of them but i am going to check one of them So let's go ahead and check this one so we know how to check things at this point we take our answer which in this case is 2 4x and we plug it in so plug a 2 in there and there we'll make sure the left and the right side are equal so 4

times 2 squared would be 4 minus 5 times two should be equal to six four times four is sixteen Five times two is ten sixteen minus ten is six so you get six equals six so this solution does check out then the other one is negative three-fourths so let's erase this and so we have four times you'd have negative three fourths and this is squared and then minus 5 times negative 3 4 and this equals 6. So if i square this guy i'd have 9 16. so times 9 16 and of course this would cancel

with this and give me a 4. so i'd have 9 4 there you have basically negative 5 times negative 3 4 negative 5 times negative 3 is positive 15 over 4 equals 6. 9 plus 15 is 24 you'd have 24 over 4 which is 6 again you get 6 equals 6 so this checks out as well So x here is equal to 2 or x is equal to negative 3 4. okay let's go ahead and take a look at the next example so we have 5r squared is equal to negative 10r minus 5. so the

main thing here is that you don't want to get confused because the variable is not an x it doesn't matter if you have an x an r q or z whatever it is you just follow the generic formula so if i have ax squared plus bx plus c equals 0 i could rewrite This as a r squared plus b r plus c equals 0. it doesn't matter what the variable is we're just saying that we need to figure out what a is b is and c is so we can plug it into that formula okay

so what i want to do is move these two terms over here and i'm going to do that by adding 10r To each side of the equation and i'm going to add 5 to each side of the equation so i would have 5 r squared plus 10 r plus 5 is equal to 0 and then i can see that my a would be 5 this is a my b would be 10 this is b and my c would be 5. this is c all right so once i know that i can plug into the quadratic

formula let me scroll down and for that Quadratic formula it's usually x is equal to but let's just say r is equal to negative b plus or minus the square root of b squared minus 4ac this is all over 2a all right so i'm going to plug in for b so i'm going to plug in a 10 here and i'm going to plug in a 10 here so 10 squared let's go ahead and put 100 and then minus 4ac so a is 5 c is 5. So minus 4 times 5 times 5. and then this

is over 2a a is 5 so 2 times 5 is 10. now the one thing you're going to notice here is that 4 times 5 is 20 20 times 5 is 100 100 minus 100 is zero square root of zero is zero so do i need plus or minus zero no because if i add zero or i subtract zero i'm not doing anything right it's not going to change anything so you can erase this And i told you that a minute ago if b squared minus 4ac if the discriminant as it's called is equal to

0 you're going to have only one solution so here r is equal to negative 10 over 10 which is negative 1. and that's a simple one to check so let's go ahead and do that so again r is equal to negative 1. so if i plug that in here and here i'd have 5 times negative 1 squared is Equal to negative 10 times negative 1 minus 5. negative 1 squared is 1 5 times 1 is 5. this equals negative 10 times negative 1 is and then minus 5 10 minus 5 is 5 so you get

5 equals 5. so this solution checks out r is equal to negative 1. all right let's take a look at another one so we have 18x squared minus 10x minus 2 is equal to 9x squared minus 4x minus 10. really all i have to do is subtract 9x Squared away from each side of the equation that'll cancel add 4x to each side of the equation that'll cancel and add 10 to each side of the equation that'll cancel over there so i'd have 18x squared minus 9x squared is 9x squared negative 10x plus 4x is negative

6x negative 2 plus 10 is positive eight and this equals zero And so this is in standard form right we have ax squared plus bx plus c equals zero we have a is nine b is negative six and c is 8. so if i plug into my quadratic formula what am i going to get i get that x is equal to negative b so in this case b is negative 6. so the negative of negative 6 is plus 6 Plus or minus the square root of b squared if i square negative 6 i get 36

minus 4 times a a is 9 4 times 9 is 36 times c c is 8. so 36 times 8 is 288. now you already can see you're going to have a problem but let's continue this is over 2 times a a is 9. so 2 times 9 is 18. now why did i say that you're going to Have a problem if 36 minus 288 is a negative value and it's negative 252 can i take the square root of a negative number well i can but not yet the solution is not a real number in

this particular case we'd stop and say there's no real solution okay so this is going to come up pretty often Until you start dealing with complex numbers and then you actually will be able to get a solution but we haven't gotten to that yet so for right now we just stop and say there's no real solution all right let's take a look at note so we have negative 5x squared minus 6x minus 19 equals negative 2 minus 8x squared so i'm going to add 8x squared to each side of the equation that'll cancel i'm going

to add 2 to Each side of the equation that'll cancel so i'll have 0 on the right on the left 8x squared minus 5x squared is 3x squared minus 6x and then negative 19 plus 2 is negative 17. and at this point you should be kind of getting this this is ax squared plus bx plus c equals 0. a is the coefficient for the squared Variable b is the coefficient for the variable to the first power and then c is the constant okay make sure you include the negatives otherwise you won't get the right answer

all right so the quadratic formula remember it's x is equal to the negative of b b is negative six so the negative of negative six is plus six plus or minus the square root of B squared if i square b i get 36 minus 4 times a a is 3 4 times 3 is 12 times c 12 times negative 17 is negative 204 so minus the negative 204 is plus 204 so i would get 36 plus 204 which is 240. so this is plus or minus the square root of 240 and this is over 2

times a a is 3 so 2 times 3 would be 6. all right so now what is the square root Of 240 well 240 is not a perfect square so here we get something that's going to be kind of messy so for 240 i'll factor that to 24 times 10 and let me factor this into 6 times 4 i know this is 5 times 2 this is 2 times 2 this is 3 times 2. so i have 1 2 3 4 factors of 2 so i'm going to have 16 times 15 16 times 50. so

i know 16 is a perfect square square root of 16 would be 4. so basically i could write this as 4 times the square root of 15. now i just have two scenarios i will have x is equal to 6 plus 4 times square root of 15 over 6 then or x equals 6 minus okay let me scroll down 4 times the square root of 15 Over 6. and we can further simplify these what we can do is we can factor a 2 from the numerator so i can put 2 times 3 plus 2 times

square root of 15 over 2 times 3. so this would cancel with this and i'd be left with 3 plus 2 times square root of 15 over 3. it'll be the same thing over here if i canceled a common factor of 2 this would Be a 3 this would be a 2 and this would be a 3 down here so i can have x equals 3 plus 2 times square root of 15 over 3 or x equals 3 minus 2 times square root of 15 over 3. those would be your two solutions and of course

you could write that in a more compact format by just saying x equals 3 plus or minus 2 times square root of 15 Over 3. now i could check this answer but in the interest of time i'm not because something like that gets really really tedious if you want some practice and i think it is good to practice with things like that pause the video take each one of these and plug it in for x in the original equation and see that you get the left and the right side As the same value