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## 8th grade (Eureka Math/EngageNY)

### 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: -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.

## Want to join the conversation?

- solve the system of equation by the subtraction methos

15x+5y=28

-3x=y+6(7 votes)- Subtract the last equation by y on both sides, then multiply by 5 to obtain:

15x+5y=28

-15x-5y=30

Then add the first equation to the second from which it follows that:

15x-15x+5y-5y=0x+0y=0=58.

The system of equation has no solutions.(6 votes)

- At3:02, Sal says you could have solved this with substitution. How could you do that?(3 votes)
- EpicOne,

The system of equations was

-3y+4x=11

y+2x=13

If you subtracted 2x from both sides of the second equation to get

y=13-2x

Now you can substitute (13-2x) for the y in the first equation.

-3(13-2x)+4x=11

Then you can solve for x.

After finding x,

Then substitute that answer for x in either equation and solve for y.

The substitute both answers into the other equation to double check.

If the equation is in balance, you have the correct answers.

I hope that helps.(2 votes)

- Are there worksheets on this topic?(4 votes)
- yeah there is an exercise it's called systems of equations with simple elimination, however you don't need to multiply it is already set to be able to be eliminated without that ext a step.

Another thing you could do is you could ask your Math teacher to give you some examples or you could use a Math book.(3 votes)

- Why is it that when you only want to multiply a certain "term" in an equation, you have to multiply that entire side of that equation, rather than let's say for example, the term was a fraction and you needed to multiple it to make it NOT a fraction, you don't multiply that entire side of that equation?

For example, like the equation Khan gave us:

y+2x=13

Why is it when we need to multiply the "y" it to make it cancel out with the "y" in the other equation, -3y+4x=11, we multiply the entire left side of the equation?

That now gives us:

3y+6x=39

Now, if the equation was, 6/3x+2x=13, and we wanted to take away that fraction, we would do this:

3*6/3x+2x=13*3

...right? We would NOT multiply the entire left side of the equation, only multiply that fraction (so we can "get rid" of it) and the other side of the equation because we have to do what we did on one side, to the other. Why is that? Why do we have the multiply the ENTIRE side of the equation in the one Khan gave us and not in one with a fraction?

I hope my question wasn't too lengthy!(2 votes)- Actually in your second example you would also need to multiply the entire left side, otherwise the equation doesn't hold anymore.

To see this, I'll use numbers instead of x or y.

As an example, take the simple equation 5/2 + 1/2 = 3 (can you see that this is correct?), and say you want to get rid of that first fraction. If you multiply only the first fraction, what do you get?

2*5/2 + 1/2 = 2*3

=> 5 + 1/2 = 6

Hang on, that's definitely not correct.

Let's try again, this time multiplying the whole left side, and of course the right side as well:

2*5/2 + 2*1/2 = 2*3

=> 5 + 1 = 6.

Yep, that looks good.

It doesn't matter whether you use x and y, or just numbers (because x and y are just placeholders for numbers we don't know yet). Every time you multiply or divide a term of an equation, you have to do the same to the other terms, and to the other side of the equation.

In your example (6/3)x+2x=13, we would do the same, multiplying both sides by 3:

3*(6/3)x + 3*2x = 3*13

=> 6x + 6x = 39

=> 12x = 39 = x = 39/12 = 13/4.

Multiplying only the first term gives a completely different, and incorrect, result of x=39/8 (try it!)(3 votes)

- x+2y= 10 & 3x-y= 9 (elimination)

how would you solve this?(2 votes) - couldn't x be 8 and y be -3?(1 vote)
- Let's try using your x and y values for the second equation:
`x = 8`

y = -3

y + 2x = 13

-3 + 2 * 8 = 13

-3 + 16 = 13

13 = 13

Great! It works! Now let's try using them for the first equation:`x = 8`

y = -3

-3y + 4x = 11

(-3 * -3) + (4 * 8) = 11

9 + 32 = 11

41 = 11

Uh-oh... Something went wrong there! As you can see, just because it works for one equation, doesn't mean it will work for both! That's the tricky part of these questions. In fact, you could find lots of different values that satisfy each equation alone, but there will only be one set of values that will work for both :)(3 votes)

- why did you multiply by three and why did you chose to turn it into three y(1 vote)
- −3𝑦 + 4𝑥 = 11

𝑦 + 2𝑥 = 13

The reason Sal multiplies the second equation by 3 is that he wants the 𝑦-term in both equations to have the same coefficient (albeit different signs):

−3𝑦 + 4𝑥 = 11

3𝑦 + 6𝑥 = 39

The first equation tells us that adding 11 is the same thing as adding (−3𝑦 + 4𝑥).

This means that we can add 11 to the right-hand side of the second equation and (−3𝑦 + 4𝑥) to the left-hand side without changing the equality.

3𝑦 + 6𝑥 + (−3𝑦 + 4𝑥) = 39 + 11**Since the two 𝑦-terms have the same coefficient they will cancel each other out**, and we are left with:

10𝑥 = 50 ⇔ 𝑥 = 5(2 votes)

- Without graphing complete parts a and b

X= -y

Y+x = 4

Is it identical lines, parallel or intersecting lines at a single point and ihow many solutions does this system have(1 vote)- Rewrite the two equations like this

x=-y --> y = -x

y+x=4 --> y = -x +4**Looking at these equations**you can see that the slope of both lines are the same (-1) and that the y-intersections of the two lines are different (0 for the first line and 4 for the second line). Try to visualize this!

When you have the same slope (ex. -1) on two lines that intersect with the y-axis on different places (ex. 0 and 4), these lines will never intersect with eachother - they are parallel.

Since these two lines never intersect with eachother, there is no point on their lines that they have in common and therefore there is no solution to this system.

You could also see that the system does not have a solution by**solving the system**, you will get an inconsistent result (0=4 or 0=-4) - the lines are parallel.(2 votes)

- The sum of three consecutive integers is 54. Find the integers using the elimination method(1 vote)
- Why multiply the equation by 3?(1 vote)
- Sal multiplied the entire bottom equation by 3 to cancel out the "y" variables. The top equation had a -3y while the bottom equation had a 3y (after multiplying by 3).

3y+(-3y) = 0. So when you add the equations together, the y's cancel out leaving you with just the "x" variables to make the equation solvable.(1 vote)

## 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.