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# Systems of equations with substitution: y=-1/4x+100 & y=-1/4x+120

Learn to solve the system of equations y = -1/4x + 100 and y = -1/4x + 120 using substitution.  Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

Y=3x+7 and...........Y=3x+7
• X can be any real number and y will be whatever that real number is times three and plus seven.
• Wht about a problem like this
-5x+y=-3
3x-8y=24
• Let -5x+y=-3 be the first equation, and 3x-8y=24 be the second. Let's manipulate the first equation by adding 5x to both side; resulting in y=-3+5x, or 5x-3. Now let's plug in 5x-3 for wherever y has been in the second formula. 3x-8(5x-3)=24. This is now pretty straightforward to solve for x. We distribute the 8 into (5x-3) and get 3x-40x-24=24. -48=27x. x=-48/27, x=-16/9.

Now let's plug in the x for either one of the equations to get y. 3(-16/9)+8y=24. -16/3+8y=24. 8y=24+16/3. 24y=72+16. 24y=88. y=88/24, y=11/3.

Hope it helped!
• How do you know what you're supposed to multiply the equations by?
• give an example and i will probably be able to help
• solve for y in terms of x
5x +380y= -2280
• Regina,
Solve for y in terms of x just means "isolate y on one side of the equation".

You do the same thing to both sides of the equation to keep it equal while eliminating everything but the y on the left.

If you first subtract 5x from both sides, the left will have 5x-5x which becomes 0 and disappears leaving you with only 380y on the left.

If you then multiply each side by the reciprocal of 380 (which is 1/380, the 380/380 on the left becomes 1 and disappears leaving you with only y on the left.

Make sure as you do this, you do the same thing to the right side.

When the y becomes isolated on the left, you will have solved for y in terms of x.

I hope that helps make it click for you.
• What is x-y=12
• You can't figure this out with the provided information. It could have multiple answers, which include: (12,0), (100, 88), and infinitely many others.
• I am still a little bit confused about why the substitution method actually works. Can someone please explain it to me more?
• How do you solve with x/3-y/2=1 and y=2x-2
• Gabby,
x/3-y/2=1
y=2x-2
From the second equation you know that y and 2x-2 are the same so you can substitute (2x-2) for the y in the first equation
x/3-y/2=1 Substiture (2x-2) for the y
x/3 - (2x-2)/2 = 1
Now distribute the -1/2
x/3 - 2x/2 + 2/2 = 1 The 2/2s both become 1
x/3 - 1x +1 = 1 Subtract 1 from both sides
x/3 - 1x = 0
Now, you might just see that x=0 is the solution for x,
Or to combine the x terms, you must find a common denominator.
x/3 - (3/3)x = 0 now 1/3 - 3/3 = -2/3 so
(-2/3)x = 0 Now multiply both sides by the reciprocal of -2/3 which is -3/2
x=0

Now find y by substituting 0 for x in either equation.
y= 2(0) -2
y= 0-2
y= -2

Now double check by substituting 0 for x and -2 for y in the other equation
0/3 - (-2/2) = 1
0-(-1) = 1
1=1
This is true so the answers double check.
x=0 y=-2 or (0,-2)

I hope that helps make it click for you.
• So, what if the X was solved already for us and the Y would be the thing we would have to solve. Would this apply the same way it does for solving X? Solving Y?
• Yes, solving Y is the same thing as solving X. Lets say that you have x already solved lets say x=3. Now you have to solve for Y. It is the exact same thing. Lets plug what we know into an equation. Lets say that 3+4y=15. Do the same thing you would to find X. If you do your same steps, you will get that Y=4 even doing the same thing you use for X.
• How do i solve this problem?
6x - 6y = -12
-4x + y = 11
• Solve one of the equations for one of the variables, then substitute into the second equation and solve for the second variable, then complete the problem by solving for the first variable.
y = 11 + 4x
6x - 6(11 + 4x) = -12
6x - 66 - 24x = -12
-18x = 54
x = -3
y = 11 + 4x-3
y = 11 -12
y = -1
6(-3) - 6(-1) = -12
-18 + 6 = -12
-12 = -12
-4(-3) - 1 = 11
12 - 1 = 11
11 = 11