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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: x+2y=6 & 4x-2y=14

Learn to use elimination to solve the system of equations x + 2y = 6 and 4x - 2y = 14. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

- how would i solve -14x+9y=46

14x-9y=102(5 votes)- No solution means the lines are parallel... They will never cross.(7 votes)

- Hi I'm new to this topic so I am kind of confused in the first part. How do you know whether you add or subtract the two equations together.

(sorry if its a stupid question :)(4 votes)- Here's an example. We have two equations, x+2y=6 and 4x-2y=14. Your question is how do we know whether you add or subtract the two equations together. Well, to know when to subtract or add the two equations, you want to know which variable to cancel out. Now what does that mean?

x+2y=6

4x-2y=14

As you can see in the second equation, it has the coefficient '2y' as the second term, right? Right now, its a negative term. In the first equation, '2y' is positive. Perfect. What is 2y -2y? It equals 0, or in other words, cancels out. Here's what it looks like.

1x+2y=6

+ 4x -2y = + 14 <-------- We add the equation to the first equation, but remember to subtract 2y.

This gives us 5x = 20, right? Right now, I'm going to assume you know how to solve the rest.

So lets recap, we add or subtract a variable to cancel out the other variable in elimination. If 4x-2y=14 was actually 4x+2y=14, and do what we did, it wouldn't cancel out the first equation.

To get it to cancel out 2y, we multiply the equation by (-1) by both sides.

-1(4x + 2y) = (14)-1

Turns into:

-4x -2y = -14

Do our magic.

x + 2y= 6

- 4x- 2y = -14 <--- Just solve and your done. We just need to cancel out one of the variables, x or y.

Ask any question if you still don't get it :)(4 votes)

- how do i solve these using elimination method?4x-5y=14;-4x-y=-6(3 votes)
- You add both equations together to get -6y=20, divide by -6 on both sides to get y=-10/3(3 votes)

- How do you know when to add or subtract?(3 votes)
- You can do either, its up to you, your preference. As long as you eliminate one of the variables.(1 vote)

- 2x-y=8

3x+2y=20 (5,2)(3 votes) - how to solve 4x+7y=-25 -12x-7y=19(2 votes)
- 7y and -7y are already opposites, so you can just add them together. You'll get -8x= -6. Solve for x, and then substitute and solve for y.(1 vote)

- Can you use division to simplify an equation into elimination method structure? Example:

(6x-3y=15) / -3

2x+3y=-8(2 votes)- Yes, the top equation just becomes -x=2x+y=-5(1 vote)

- how would you solve using the addition method 3x+ 2y+18 and 2x-3y=-14(2 votes)
- What if in one of the equations y is divided by 3? what should I do then? What if the equations include fractions?(2 votes)
- Multiply by three. then it becomes 3y/3 which is the same thing as 3/3 times y , which is the same thing as 1 times y , which is the same ting as just Y. you have now eliminated the fraction :)(1 vote)

- How do you solve 3x+5y=20

4x-y=19?(2 votes)

## Video transcript

Use elimination to
solve for x and y. And they gave us two equations
here-- x plus 2y is equal to 6 and 4x minus 2y is equal to 14. So to solve by
elimination, what we do is we're going to add
these two equations together so that one of the two variables
essentially gets eliminated, gets canceled out. And what we could do right here,
we see we have a plus 2y here, and we see we have a
negative 2y right over here. So clearly if we added
these two together, the y's would cancel
out, and so that's exactly what we're going to do. We're going to add the
left side of this equation to the left side
of this equation, and the right-hand
side of that equation to the right-hand side
of the bottom equation. And just to make it clear
that this should make sense is we're just using both of
these constraints, whenever you learn about any
type of equation. So if I have x plus
2y is equal to 6, you learned early on
in algebra that you can manipulate this
equation in any way as long as whatever you
do to the left-hand side of the equation, you
do the right-hand side. If this is equal to
that, the only way that the equality
will still hold is whatever you do to this--
whatever you add to this or multiply it by-- you also
do to the right-hand side. So when we're adding
these two equations, that's exactly what we're doing. We could say, hey, let's add 14
to both sides of this equation. So you could add
14 on this side, you could add 14 on that side. That wouldn't be anything new,
but the second equation right here tells us that 4x minus
2y is the same thing as 14. So instead of adding 14
on the left-hand side, I could add 4x minus 2y. When we're adding
these two equations, we're really just
adding the same thing. You could view it as we're
starting with this equation, and then we're adding the
same thing to both sides. On the right-hand side it looks
like we're adding 14 to the 6. On the left-hand side it looks
like we're adding 4x minus 2y to whatever is on
the left-hand side. But the second
constraint tells us that 14 and 4x minus
2y are the same thing. So we're adding the same
thing to both sides. So with that said,
let's just do it. So the left-hand side, if we add
it up, we have x plus 4x is 5x, and then the 2y cancels
out with the negative 2y. And then on the right-hand
side we have 6 plus 14. 6 plus 14 Is 20. So we're left with one
equation with one unknown. 5x is equal to 20. We can divide both
sides by 5, and we are left with x is equal to 4. Now we can go back
and substitute in x equals 4 into either of
these equations to solve for y. So let's use this top one. So we have 4 plus
2y is equal to 6. We can subtract 4
from both sides. So then we get 2y is equal to 2. Divide both sides by 2. We get y is equal to 1. So the solution,
the x's and y's that satisfy both of these
equations, are x is equal to 4, and y is equal to 1. So this is the solution
for this system, or this coordinate
would be the point of intersection of
these two lines. And we can verify it. Let's verify that when
we put x is equal to 4, and y is equal to 1, in this
first equation it satisfies it. So we have 4 plus 2 times 1. That's 4 plus 2. That does, indeed, equal 6. And then the second
equation right over here, you have 4 times
4 minus 2 times 1. This is equal to 16 minus 2,
which does indeed equal 14. So it definitely does satisfy
both of these equations. So we're done. X is equal to 4,
and y is equal to 1.