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Course: Algebra (all content) > Unit 5

Lesson 3: Equivalent systems of equations and the elimination method

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:

\begin{aligned} 2 y + 7 x & = - 5 \\ 5 y - 7 x & = 12 \end{aligned}

We notice that the first equation has a

7 x

term and the second equation has a

- 7 x

term. These terms will cancel if we add the equations together—that is, we'll eliminate the

x

terms:

\begin{aligned} 2 y + 7 x & = - 5 \\ + 5 y - 7 x & = 12 \\ 7 y + 0 & = 7 \end{aligned}

Solving for

y

, we get:

\begin{aligned} 7 y + 0 & = 7 \\ 7 y & = 7 \\ y & = 1 \end{aligned}

Plugging this value back into our first equation, we solve for the other variable:

\begin{aligned} 2 y + 7 x & = - 5 \\ 2 \cdot 1 + 7 x & = - 5 \\ 2 + 7 x & = - 5 \\ 7 x & = - 7 \\ x & = - 1 \end{aligned}

The solution to the system is

x = - 1

y = 1

We can check our solution by plugging these values back into the original equations. Let's try the second equation:

\begin{aligned} 5 y - 7 x & = 12 \\ 5 \cdot 1 - 7 (- 1) & \overset{?}{=} 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:

\begin{aligned} - 9 y + 4 x - 20 & = 0 \\ - 7 y + 16 x - 80 & = 0 \end{aligned}

We can multiply the first equation by

- 4

to get an equivalent equation that has a

- 16 x

term. Our new (but equivalent!) system of equations looks like this:

\begin{aligned} 36 y - 16 x + 80 & = 0 \\ - 7 y + 16 x - 80 & = 0 \end{aligned}

Adding the equations to eliminate the

x

terms, we get:

\begin{aligned} 36 y - 16 x + 80 & = 0 \\ + - 7 y + 16 x - 80 & = 0 \\ 29 y + 0 - 0 & = 0 \end{aligned}

Solving for

y

, we get:

\begin{aligned} 29 y + 0 - 0 & = 0 \\ 29 y & = 0 \\ y & = 0 \end{aligned}

Plugging this value back into our first equation, we solve for the other variable:

\begin{aligned} 36 y - 16 x + 80 & = 0 \\ 36 \cdot 0 - 16 x + 80 & = 0 \\ - 16 x + 80 & = 0 \\ - 16 x & = - 80 \\ x & = 5 \end{aligned}

The solution to the system is

x = 5

y = 0

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.

\begin{aligned} 3 x + 8 y & = 15 \\ 2 x - 8 y & = 10 \end{aligned}

x =

y =

Want more practice? Check out these exercises:

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