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Elimination method review (systems of linear equations)

The elimination method is a technique for solving systems of linear equations. This article reviews the technique with examples and even gives you a chance to try the method yourself.

What is the elimination method?

The elimination method is a technique for solving systems of linear equations. Let's walk through a couple of examples.

Example 1

We're asked to solve this system of equations:
2y+7x=55y7x=12\begin{aligned} 2y+7x &= -5\\\\ 5y-7x &= 12 \end{aligned}
We notice that the first equation has a 7, x term and the second equation has a minus, 7, x term. These terms will cancel if we add the equations together—that is, we'll eliminate the x terms:
2y+7x=5+ 5y7x=127y+0=7\begin{aligned} 2y+\redD{7x} &= -5 \\ +~5y\redD{-7x}&=12\\ \hline\\ 7y+0 &=7 \end{aligned}
Solving for y, we get:
7y+0=77y=7y=1\begin{aligned} 7y+0 &=7\\\\ 7y &=7\\\\ y &=\goldD{1} \end{aligned}
Plugging this value back into our first equation, we solve for the other variable:
2y+7x=521+7x=52+7x=57x=7x=1\begin{aligned} 2y+7x &= -5\\\\ 2\cdot \goldD{1}+7x &= -5\\\\ 2+7x&=-5\\\\ 7x&=-7\\\\ x&=\blueD{-1} \end{aligned}
The solution to the system is x, equals, start color #11accd, minus, 1, end color #11accd, y, equals, start color #e07d10, 1, end color #e07d10.
We can check our solution by plugging these values back into the original equations. Let's try the second equation:
5y7x=12517(1)=?125+7=12\begin{aligned} 5y-7x &= 12\\\\ 5\cdot\goldD{1}-7(\blueD{-1}) &\stackrel ?= 12\\\\ 5+7 &= 12 \end{aligned}
Yes, the solution checks out.
If you feel uncertain why this process works, check out this intro video for an in-depth walkthrough.

Example 2

We're asked to solve this system of equations:
9y+4x20=07y+16x80=0\begin{aligned} -9y+4x - 20&=0\\\\ -7y+16x-80&=0 \end{aligned}
We can multiply the first equation by minus, 4 to get an equivalent equation that has a start color #7854ab, minus, 16, x, end color #7854ab term. Our new (but equivalent!) system of equations looks like this:
36y16x+80=07y+16x80=0\begin{aligned} 36y\purpleD{-16x}+80&=0\\\\ -7y+16x-80&=0 \end{aligned}
Adding the equations to eliminate the x terms, we get:
36y16x+80=0+ 7y+16x80=029y+00=0\begin{aligned} 36y-\redD{16x} +80&=0 \\ {+}~-7y+\redD{16x}-80&=0\\ \hline\\ 29y+0 -0&=0 \end{aligned}
Solving for y, we get:
29y+00=029y=0y=0\begin{aligned} 29y+0 -0&=0 \\\\ 29y&=0 \\\\ y&=\goldD 0 \end{aligned}
Plugging this value back into our first equation, we solve for the other variable:
36y16x+80=036016x+80=016x+80=016x=80x=5\begin{aligned} 36y-16x+80&=0\\\\ 36\cdot 0-16x+80&=0\\\\ -16x+80&=0\\\\ -16x&=-80\\\\ x&=\blueD{5} \end{aligned}
The solution to the system is x, equals, start color #11accd, 5, end color #11accd, y, equals, start color #e07d10, 0, end color #e07d10.
Want to see another example of solving a complicated problem with the elimination method? Check out this video.

Practice

Problem 1
Solve the following system of equations.
3x+8y=152x8y=10\begin{aligned} 3x+8y &= 15\\\\ 2x-8y &= 10 \end{aligned}
x, equals
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3, slash, 5
  • a simplified improper fraction, like 7, slash, 4
  • a mixed number, like 1, space, 3, slash, 4
  • an exact decimal, like 0, point, 75
  • a multiple of pi, like 12, space, start text, p, i, end text or 2, slash, 3, space, start text, p, i, end text
y, equals
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3, slash, 5
  • a simplified improper fraction, like 7, slash, 4
  • a mixed number, like 1, space, 3, slash, 4
  • an exact decimal, like 0, point, 75
  • a multiple of pi, like 12, space, start text, p, i, end text or 2, slash, 3, space, start text, p, i, end text

Want more practice? Check out these exercises:

Want to join the conversation?

  • leaf green style avatar for user Olivia
    what if you are using three variables and you have three different equations?
    for example: (three variables,three equations)
    x-y-2z=4
    -x+2y+z=1
    -x+y-3z=11
    (46 votes)
    Default Khan Academy avatar avatar for user
    • cacteye green style avatar for user Hamza Usman
      First of all, the only way to solve a question with 3 variables is with 3 equations. Having 3 variables and only 2 equations wouldn't allow you to solve for it. To start, choose any two of the equations. Using elimination, cancel out a variable. Using the top 2 equations, add them together. That results in y-z=5. Now, look at the third equation and cancel out the same variable that you originally cancelled out. In this case, we canceled out x. Adding the first equation to the 3rd equation would get rid of his. Adding would give -5z=15. We got lucky because both the x and the y cancelled out. If they didn't both cancel out, you would just have t solve the two equations which you should know how to do. Back to the problem, -5z=15, so z=-3.Plug that into the equation y-z=5 to solve for y. y-(-3)=5, so y+3=5. That gives y=2. Plug both of those into any of the three original equations and solve for x. You get x=0. Your final solution is x=0, y=2, and z=-3 or (0,2,-3).
      (2 votes)
  • blobby green style avatar for user Yeyka Rosario
    what about unsorted equations?
    -4y-11x=36
    20=-10x-10y
    (5 votes)
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    • aqualine ultimate style avatar for user AD Baker
      Yeka,

      Use algebraic manipulation to get the x and y terms on the same side of the =. For the second equation, add 10x to both sides, add 10y to both sides, and subtract 20 from both sides. Then, proceed with the elimination method.
      (6 votes)
  • blobby green style avatar for user kesvibp0208
    kesvi patel
    are we supposed to do last divide step ?
    (4 votes)
    Default Khan Academy avatar avatar for user
    • scuttlebug yellow style avatar for user Rin
      After you add/subtract the new equations, you eliminate one of the variables and divide. After solving one of them, plug your solved variable to one of the original problems.
      This might help you understand more clearly:

      12x + 2y = 90 ... (1)
      6x + 4y = 90 ... (2)

      (2)*2 12x + 8y = 180 ... (2)'

      (1)-(2)' 12x + 2y = 90
      - 12x + 8y = 180
      -6y = -90
      You eliminated x
      and now you solve y by dividing.
      y = 15
      y = 15 plugged into (2). (Always pick the easier problem to solve your problem accurately.)
      6x + 4(15) = 90
      6x + 60 = 90
      6x = 90 - 60
      6x = 30
      x = 5

      Your final answer will be
      (x,y) = (5,15)

      Hope this helps!
      (6 votes)
  • purple pi purple style avatar for user Omelette Kai
    when plugging in... do i plug in to the original equation or the (new) equivalent equation?
    (4 votes)
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  • blobby green style avatar for user Zainab
    Do you always have to eliminate the x term?
    (3 votes)
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  • blobby green style avatar for user Vera
    Do you always have to eliminate the x term? Or could you do it with y, too?
    (0 votes)
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  • blobby green style avatar for user Izzy Pittman
    what if my equations are like this:
    4y=2x-8
    5x-4y=20
    where my equations are not lined up. how do I solve this?
    (2 votes)
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  • blobby green style avatar for user ABDULMOEZ VENOM
    how to it is subtract or addition in equation ??
    sometimes we solve equation addition
    (2 votes)
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  • blobby green style avatar for user ME!
    what if i have a square in one of the variables of one of the equation?
    (2 votes)
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  • starky sapling style avatar for user keving94547
    hew d:d they get the answer
    (2 votes)
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