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### Course: 8th grade (Eureka Math/EngageNY) > Unit 4

Lesson 4: Topic D: Systems of linear equations and their solutions- Systems of equations: trolls, tolls (1 of 2)
- Systems of equations: trolls, tolls (2 of 2)
- Testing a solution to a system of equations
- Solutions of systems of equations
- Systems of equations with graphing
- Systems of equations with graphing
- Systems of equations with graphing: 5x+3y=7 & 3x-2y=8
- Systems of equations with graphing: y=7/5x-5 & y=3/5x-1
- Systems of equations with graphing: chores
- Systems of equations with graphing
- Systems of equations with elimination: 3t+4g=6 & -6t+g=6
- Systems of equations with elimination
- Systems of equations with elimination: x+2y=6 & 4x-2y=14
- Systems of equations with elimination: -3y+4x=11 & y+2x=13
- Systems of equations with elimination: 2x-y=14 & -6x+3y=-42
- Systems of equations with elimination: 4x-2y=5 & 2x-y=2.5
- Systems of equations with elimination: x-4y=-18 & -x+3y=11
- Systems of equations with elimination
- Systems of equations with elimination: 6x-6y=-24 & -5x-5y=-60
- Systems of equations with elimination challenge
- Systems of equations with substitution: 2y=x+7 & x=y-4
- Systems of equations with substitution
- Systems of equations with substitution: y=4x-17.5 & y+2x=6.5
- Systems of equations with substitution: -3x-4y=-2 & y=2x-5
- Systems of equations with substitution: 9x+3y=15 & y-x=5
- Systems of equations with substitution
- Systems of equations with substitution: y=-5x+8 & 10x+2y=-2
- Systems of equations with substitution: y=-1/4x+100 & y=-1/4x+120
- Systems of equations with elimination: apples and oranges
- Systems of equations with elimination: TV & DVD
- Systems of equations with elimination: King's cupcakes
- Systems of equations with elimination: Sum/difference of numbers
- Systems of equations with elimination: potato chips
- Systems of equations with elimination: coffee and croissants
- Systems of equations with substitution: coins
- Systems of equations with substitution: potato chips
- Systems of equations with substitution: shelves
- Systems of equations word problems
- Age word problem: Imran
- Age word problem: Ben & William
- Age word problem: Arman & Diya
- Age word problems
- Solutions to systems of equations: consistent vs. inconsistent
- Systems of equations number of solutions: fruit prices (1 of 2)
- Systems of equations number of solutions: fruit prices (2 of 2)
- Systems of equations number of solutions: y=3x+1 & 2y+4=6x
- Solutions to systems of equations: dependent vs. independent
- Number of solutions to a system of equations graphically
- Number of solutions to a system of equations graphically
- Forming systems of equations with different numbers of solutions
- Number of solutions to a system of equations algebraically
- Comparing Celsius and Fahrenheit temperature scales
- Converting Fahrenheit to Celsius

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# Systems of equations with elimination: 6x-6y=-24 & -5x-5y=-60

Sal solves the system of equations 6x - 6y = -24 and -5x - 5y = -60 using elimination. Created by Sal Khan.

## Want to join the conversation?

- if you multiply one equation by 1/6, shouldn't you also multiply the other equation by 1/6, not 1/5 so they are equal?(12 votes)
- You are trying to keep the two sides of the equations equal. The equations themselves are not, and do not need to be equal.

Think about it with numbers

if 2 = 2

& 5 = 5 even though these are not equal you can add them

->7 = 7 and since the two sides were equal before they will still be equal after you add them.(7 votes)

- I don't understand why you have to multiply the equations by " 1/6 & 1/5 "..

If i have a problem like " 9a - b = 15 , 3a + b = 5 " how do i solve this with elimination..(6 votes)- You don't always do elimination. You can divide both sides by the same thing or subtraction to equal out the equation to the maximum level of simplification. Then, great! You have a fully simplified equation, to find out what is greater, or finally solve for the unkwown variable.-- Hope it helped,

GEORA190(1 vote)

- How would you know what number to multiply the equations by? In the video its pretty obvious but for some reason, I can't get it on my own. For example: 4x-3y=-26 5x-4y=-35

Or: -3x+5y=29 -3x+6y=39(3 votes)- 4x-3y=-26

5x-4y=-35

You want eliminate either the x or the y. Let's say we eliminate the x first, we have to make the top (4x) and (5x) equal to each other. Multiply the top by 5 and the bottom by 4. Now both are (20x) and you can subtract to cancel them out. Remember that whatever number you multiply by is distributed to the whole equation.(8 votes)

- How do I solve this :" if a - b = 2, a - c = 1/2, then what is the answer for (b - c)^3 - 3(b - c) + 9/4"(4 votes)
- Let equation (1) be a - b = 2 and equation (2) be a - c = 1/2

equation (2) - equation (1) = (a - c) - (a - b) = (1/2) - 2 = b - c = -3/2

we now have b - c = -3/2 SO

(b - c)^3 = (-3/2)^3 = -27/8

3(b - c) = 3(-3/2) = -9/2

SO (b - c)^3 - 3(b - c) + 9/4 = -27/8 -3(-3/2) + 9/4

= -27/8 + 9/2 + 9/4 = (-27 + 36 + 18)/4 = 27/8 = (3/2)^3(2 votes)

- Can you do this WITHOUT multiplying or dividing the equations first?

I tried, but it isn't working.

Also how do you know when you have modify them? If I came across a random equation, how would I know I need to multiple/divide it first?(5 votes)- You will know you won't have to modify them if one variable will cancel out during substitution. For example, 5x+3y=10 and -5x+-8y=-14 won't have to be modified because the 5x and -5x will cancel.(1 vote)

- Is this also known as the combination method?(3 votes)
- Ask a question...can you cancel out the y instead of the x?(2 votes)
- Yes you can. That would mean that you solve for x first and then substitute the value to get y. Same method.(5 votes)

- whats the difference between substitution and elimination methods in solving systems of equations(2 votes)
- Substitution is when you isolate a variable and substitute it into another equation. Elimination is when you add or subtract an equation to another.(4 votes)

- When you're multiplying/dividing to get the equations ready to cancel out a variable, do you have to multiply/divide both equations by something? Can't you just do it to one of the equations?(3 votes)
- You can do anything to just one equation, as long as you do it to both sides of it.(1 vote)

- At around2:10, how do I know when the fractions will work? Is it simply when everything is divisible by the fraction, or is there another way to figure it out?(1 vote)
- It will always work, but first let me explain why, in this equation, it made sense to multiply both sides by fractions. By multiplying the 1st equation by 1/6 and the second equation by 1/5, the x's canceled out and it's easier to solve because ALL the variables ended up having 1 as a coefficient. I would not use fractions if it would create fractions in the equation, as fractions are a pain to work with. If the equation were, for instance,

(1/6 )6x-4y=18 (1/6)

(1/5 ) -5x-3y=20 (1/5)

x-2/3y=3

-x-3/5y=4

I could multiply the first equation by 1/6 and go from a 6x to 1x and I could multiply the second equation by 1/5 and go from -5x to -1x. But then I would create fractions and it would be difficult to solve.(3 votes)

## Video transcript

We never know when we might have
to do a little bit more party planning. So it doesn't hurt to
have some practice. And that's what this
exercise is doing for us, is generating
problems so that we can try solving systems of
equations with elimination. And so in this first
problem, it says solve for x and y
using elimination. And then this is
what they have-- 6x minus 6y is equal
to negative 24. Negative 5x minus 5y is
equal to negative 60. So let me get my scratch
pad out to solve this. Let me rewrite it. So they gave us 6x minus
6y is equal to negative 24. And negative 5x minus 5y
is equal to negative 60. So what we have to
think about, and we saw this in several
of the other videos, is when we want to
eliminate a variable, we want to manipulate
these two equations. And if we were to add
the corresponding sides, that variable might disappear. So if we just added a
6x to a negative 5x, that's not going
to cancel it out. If this was a negative
6x, that would work out. Or if this was a positive
5x, that would work out. But this isn't exactly right. So if I want to
eliminate the x, I have to manipulate these
equations so that these two characters might cancel out. And one thing that
pops into my brain is it looks like
all of this stuff up here is divisible by
6, and all of this stuff down here is divisible by 5. And if we were to divide
all this stuff up here by 6, we'd be left with
an x over here. And if we were to divide
all this bottom stuff by 5, we'd be left with a
negative x right over here. And then they just
might cancel out. So let's try that out. Let's take this first equation. And we're going to
multiply both sides by 1/6. Or another way you
could think about it is we're dividing both sides by 6. And as long as we do the
same thing to both sides, the equation holds. The equality holds. So if you multiply everything
by 1/6, 6x times 1/6 is just going to be x. 6y times 1/6 is just y. So it's negative y. Negative 24 times
1/6 is negative 4. Or you could just view it
as negative 24 divided by 6 is negative 4. So this equation, the blue one,
we've simplified as x minus y is equal to negative 4. Let's do something similar
with the second one. Here we could say we're going to
multiply everything times 1/5. Or you could say that we're
dividing everything by 5. If we do that, negative
5x divided by 5 is just negative x. Negative 5y divided
by 5 is negative y. And then negative 60
divided by 5 is negative 12. And now, this looks
pretty interesting. If we add the two left-hand
sides-- and remember, we can keep the equality,
because we're essentially adding the same
thing to both sides. You can imagine we're starting
with the blue equation. And on the left-hand side,
we're adding negative x minus y. And on the right-hand side,
we're adding negative 12. But the second equation tells
us that those two things are equal. So we're doing
the same principle that we saw when we first
started looking to algebra, that you can maintain your
equality as long as you add the same thing. On the left-hand side,
we're going to add this. And on the right-hand side,
we're going to add this. But this second
equation tells us that those two things are equal. So we can maintain our equality. So let's do that. What do we get on
the left-hand side? Well, you have a positive
x and a negative x. They cancel out. That was the whole point behind
manipulating them in this way. And then you have negative y
minus y, which is negative 2y. And then on the right-hand
side, you have negative 4 minus 12, which is negative 16. And these are going to
be equal to each other. Once again, we're adding the
same thing to both sides. To solve for y, we can divide
both sides by negative 2. And we are left with y
is equal to positive 8. But we are not done yet. We want to go and substitute
back into one of the equations. And we can substitute
back into this one and to this one, or
this one and this one. The solutions need to satisfy
all of these essentially. This blue one is another way of
expressing this blue equation. This green equation
is another way of expressing this
green equation. So I'll go for whichever one
seems to be the simplest. And this one seems to be
pretty simple right over here. So let's take x
minus y-- we just solved that y would be positive
8-- is equal to negative 4. And now to solve for x, we just
have to add 8 to both sides. And we are left with,
on the left-hand side, negative 8 plus 8 cancels out. You're just left with an x. And negative 4 plus 8
is equal to positive 4. So you get x is equal
to 4, y is equal to 8. And you can verify
that it would work with either one of
these equations. 6 times 4 is 24 minus 6 times
8-- so it's 24 minus 48-- is, indeed, negative 24. Negative 5 times 4 is negative
20, minus negative 40, if y is equal to 8, does,
indeed, get you negative 60. So it works out
for both of these. And we can try it out by
inputting our answers. So x is 4, y is 8. So let's do that. So let me type this in. x is going to be equal to 4. y is going to be equal to 8. And let's check our answer. It is correct. Very good.