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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 substitution: 9x+3y=15 & y-x=5

Learn to solve the system of equations 9x + 3y = 15 and y - x = 5 using substitution. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

- i am told to solve each system using substitution .

how would i do this problem ?

m=2n

m+4n=30(3 votes)- brian,

Your first equation, m=2n, is already solved for m in terms of n, so plug that in your second equation like this:

m=2n, so

m+4n=30 is the same thing as (2n)+4n=30

Therefore 6n=30, so n= 5. And if n=5 and m=2n, m= 2(5) =10.

Rechecking:

m+4n=30. If m=10 and n=5

(10)+4(5)=30

10+20=30

30=30. The answer m=10 and n=5 checks good.

This method is called solving by substitution, because you first solve for one variable in terms of the other using 1 equation, and then you substitute that solution in the other equation to solve for the second variable explicitly. Then you go back and solve for the first variable explicitly.

Does that help?(12 votes)

- I have watched almost all the videos for substitution and I still do not understand. For example, for the system of equations x=2y+7, and x=y+4, I got x=-6 and y=-3, but I'm not sure I'm right.(2 votes)
- Your y is correct, but not your x.

Plug the y in to find x

x = y + 4 = -3 + 4 = 1

Then plug them both in the other equation to check

x = 2y + 7 = 2(-3) + 7 = -6 + 7 = 1(5 votes)

- no one's asked a question in 5+1 years.

Why is that?(3 votes)- i dont even know becuase you said this 4 years ago(1 vote)

- how do i solve

y=x-6

y=2x-8

via substitution(2 votes)- Since y = x-6 and y = 2x-8, set x-6 equal to 2x-8.

x-6 = 2x-8

-6 = x-8 Subtract x from both sides.

x = 2 Add 8 to both sides.

Plug in x for 2 into the first equation y = x-6: y = 2-6 = -4

Verify that the second equation y = 2x-8 holds true: y = 2(2)-8 = -4, -4 = -4!(2 votes)

- How do you solve this equation -3x-8y= 20 -5x + y=19(2 votes)
- How do I solve -8x + y= -3

3x + 2y =13

I can't seem to find something similar.(2 votes)- This problem can be solved using substitution. It takes a few more steps than simpler ones, but here's how you do it:

The first equations seems simpler, since the y has a coefficient (number before the variable) of 1. If we add -8x to both sides of the equation, we get y = 8x - 3. Now we can substitute this in for y in the second equation to get: 3x + 2(8x-3) = 13, and then we just solve for x. If you did the steps right, you should get x= 1. Then, by putting in 1 for x in any of the two equations, you should get y=5.

Let's check our work. -8(1)+(5) = -8+5 = -3, and 3(1)+2(5) = 3 + 10 = 13. This is right, so our solution set is x=1, y=5, or (1,5).(2 votes)

- Having trouble solving this

-4x+y=6

-5x-y=21(2 votes) - Is 0/12 possible? What about 12/0? What's the difference?

(I am referencing2:30in the video.)

And if two equations have the same y-intercept (b), does that make the system of equations' point to be that y-intercept point?(2 votes) - i have this question that is bugging me SO BAD can you help?

x=2y-2

3y=x+6

Thank You(2 votes)- x=6

y=4

if you rearrange the first equation to be -2y=-x-2, then you add the two equations together you will see y=4. Then place that value in either of the original equations and solve for x. hth.(1 vote)

- solve using substitution method 6x +3y=12, 9x + 6y=15(2 votes)
- 6x+3y=12 ====> 3y=12-6x ===> y=4-2x

Substitute the value of y in the second equation

9x+6y=15 (as all the coefficients are multiples of 3, we can divide the whole equation (both the sides) by 3

3x+2y=5 ===> 3x+2(4-2x) = 5

3x+8-4x=5 ====> -x = -3 ====> x=3

Substitute the value of x in the first equation

y=4-2x =====> 4-2(3) = -2

Sol : x=3 ; y=-2(1 vote)

## Video transcript

We're asked to solve and graph
this system of equations here. And just as a bit of a review,
solving a system of equations really just means figuring out
the x and y value that will satisfy both of these
equations. And one way to do it is to use
one of the equations to solve for either the x or the y, and
then substitute for that value in the other one. That makes sure that you're
making use of both constraints. So let's start with this bottom
equation right here. So we have y minus
x is equal to 5. It's pretty straightforward
to solve for y here. We just have to add x to both
sides of this equation. So add x. And so the left-hand side,
these x's cancel out, the negative x and the positive x,
and we're left with y is equal to 5 plus x. Now, the whole point of me doing
that, is now any time we see a y in the other equation,
we can replace it with a 5 plus x. So the other equation was-- let
me do it in this orange color-- 9x plus 3y
is equal to 15. This second equation told us, if
we just rearranged it, that y is equal to 5 plus x, so we
can replace y in the second equation with 5 plus x. That makes sure we're making
use of both constraints. So let's do that. We're going to replace
y with 5 plus x. So this 9x plus 3y equals 15
becomes 9x plus 3 times y. The second equation says
y is 5 plus x. So we're going to put 5 plus
x there instead of a y. 3 times 5 plus x
is equal to 15. And now we can just
solve for x. We get 9x plus 3 times 5
is 15 plus 3 times x is 3x is equal to 15. So we can add the 9x and the 3x,
so we get 12x, plus 15, is equal to 15. Now we can subtract 15 from both
sides, just so you get only x terms on the
left-hand side. These guys cancel each other
out, and you're left with 12x is equal to 0. Now you divide both sides by 12,
and you get x is equal to 0/12, or x is equal to 0. So let me scroll down
a little bit. So x is equal to 0. Now if x equals 0, what is y? Well, we could substitute
into either one of these equations up here. If we substitute x equals 0 in
this first equation, you get 9 times 0 plus 3y is
equal to 15. Or that's just a 0, so you
get 3y is equal to 15. Divide both sides by 3, you get
y is equal to 15/3 or 5. y is equal to 5. And we can test that that also
satisfies this equation. y, 5, minus 0 is also
equal to 5. So the value x is equal to--
I'll do this in green. x is equal to 0, y is equal
to 5 satisfies both of these equations. So we've done the first part. Let's do the second part, where
we're asked to graph it. The second equation is pretty
straightforward to graph. We actually ended up putting
it in mx plus b form right there. And actually, let
me rewrite it. Let me just switch the x and the
5, so it really is in that mx plus b form. So y is equal to x plus 5. So its y-intercept is 5--
1, 2, 3, 4, 5-- and its slope is 1, right? There's a 1 implicitly being
multiplied, or the x is being multiplied by 1. So it looks like Let me see
how well I can draw it. The line will look like that. It has a slope of 1. You move back 1,
you go down 1. You move forward
1, you go up 1. That's a pretty good job. So that right here
is this equation. Now let's graph that
top equation. And we just have to put it
in mx plus b form, or slope-intercept form. And I'll do that in green. So we have 9x plus 3y
is equal to 15. One simplification we can do
right from the get-go is every number here is divisible by
3, so let's just divide everything by 3 to make
things simpler. So we get 3x plus
y is equal to 5. Now we can subtract 3x
from both sides. We are left with y is equal
to negative 3x plus 5. So that's what this first
equation gets turned into, if you put it in slope-intercept
form. y is equal to negative 3x plus 5. So if you were to graph it,
the y-intercept is 5. 0, 5. And then its slope
is negative 3. So you move 1 in the
x-direction, you move down 3 in the y-direction. Move 2, you would move down 6. 2, 4, 6. Move 2, you go 2, 4, 6. So this line is going to look
something like this. It's going to look something
like that right there. As you can see, the solution to
this system is the point of intersection of these
two lines. It's the combination of x and y
that satisfy both of these. Remember, this pink line, or
this red line, is all of the x's and y's that satisfy
this equation: y minus x is equal to 5. This green line is all of the
x's and y's, or all the combinations of them, that
satisfy this first equation. Now, the one x and y combination
that satisfies both is their point
of intersection. And we figured it out
algebraically using substitution. That happens at x is equal to 0,
y is equal to 5. x is equal to 0-- this is the x-axis-- y
is equal to 5 right there.