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Algebra 1

Course: Algebra 1 > Unit 6

Lesson 3: Solving systems of equations with elimination
Math
Algebra 1
Systems of equations
Solving systems of equations with elimination
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Elimination method review (systems of linear equations)

CCSS.Math:
HSA.REI.C.6
Google Classroom
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=−55y−7x=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+ 5y−7x=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=−52⋅1+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:
5y−7x=125⋅1−7(−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+4x−20=0−7y+16x−80=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:
36y−16x+80=0−7y+16x−80=0\begin{aligned} 36y\purpleD{-16x}+80&=0\\\\ -7y+16x-80&=0 \end{aligned}
Adding the equations to eliminate the x terms, we get:
36y−16x+80=0+ −7y+16x−80=029y+0−0=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+0−0=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:
36y−16x+80=036⋅0−16x+80=0−16x+80=0−16x=−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
  • Current
Solve the following system of equations.
3x+8y=152x−8y=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:

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