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## 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: -3y+4x=11 & y+2x=13

Sal solves the system of equations -3y + 4x = 11 and y + 2x = 13 using elimination. Created by Sal Khan and Monterey Institute for Technology and Education.

## Video transcript

Solve for x and y. And we have two equations
with two unknowns. Negative 3y plus
4x is equal to 11, and y plus 2x is equal to 13. What we could do is try to
solve this using elimination. Maybe we can add these
two equations together to get some variables
to cancel out. But if we just add
y and negative 3y, those won't cancel out. And if we just add a 2x and
4x, those won't cancel out. But maybe we can scale
one of these equations, so that we will get
some cancellation. If this y could
somehow become a 3y, then 3y and negative
3y would cancel out. And so the easiest way
to turn this y into a 3y is to just multiply
this entire equation, so we don't actually change
the information in it, by 3. So let's multiply
both sides by 3. If we multiply the left-hand
side by 3, we get 3y plus 6x. We have to multiply
every term by 3. 3y plus 6x is equal
to 3 times 13 is 39. So this equation
and this equation are the exact same equation. I've just multiplied both sides
by 3 to go from here to here. So I multiplied by 3. And of course, we have this
original equation up here. That negative 3y plus
4x is equal to 11. And now, if we add the left-hand
side and the right-hand sides of this equation, something
interesting happens. The negative 3y and the
positive 3y cancel out. That was the whole point behind
multiplying this second guy over here by 3. So let's add them. So if we add, these
guys cancel out. We have 4x plus 6x is 10x. And that is equal to 11
plus 39, which is 50. And then we can divide
both sides by 10. We get x is equal to
50 divided by 10 is 5. And then we could go
substitute back to solve for y. If x is 5, we could use the
second equation right over here. We get y plus 2 times 5. I'll do it in that same color. 2 times 5 is equal to 13. Or y plus 10 is equal to 13. We can subtract 10
from both sides. And we will get y. These cancel out.
y is equal to 3. So our solution that
we get for x and y is that x is equal to 5. And y is equal to 3. And we can verify that it
works in both equations. In this top equation-- let
me do it in a new color-- negative 3 times 3 plus 4
times 5 is-- this is negative 9 plus 20, which is,
indeed, equal to 11. So both of these satisfy
the first equation. And then if we take the second
equation, 3 plus 2 times 5, that's 3 plus 10, that
does, indeed, equal 13. So it definitely satisfies both. And I want to be clear. You could have solved
this with substitution. Or you could have changed which
equation you scale so that we get different
cancellations when we add or different eliminations. So another way that we
could have done this, we could've kept the
first equation the same. Negative 3y plus
4x is equal to 11. But maybe we want this 2x
to cancel with this 4x. And the only way
we could do that is if this 2x could
become a negative 4x. And the way to get from
a 2x to a negative 4x is to multiply it by negative 2. So let's multiply this entire
equation by negative 2. So it would become
negative 2y minus 4x is equal to negative 26. And if we then add
the two equations, the x's will cancel out. You have negative
3y minus 2y, which is negative 5y is
equal to 11 minus 26. That's negative 15. Now, you can divide both
sides by negative 5. And you get-- the negatives
cancel out-- y is equal to 3. So in this situation,
we eliminated the x's first to solve for y. And then you could go back,
substitute, and solve for x. You'll get x is equal to 5. The first time we did it,
we eliminated the y's. We scaled this
equation right here so that the y's would cancel out. And so we solved for the x
first and then substituted back. Either way works. You get x is equal to 5. y is equal to 3.