Elimination is a method of simplification. Trying on clothes at a store helps us eliminate options and decide what to purchase. We eliminate items on a to-do list throughout the day to simplify our schedules. In mathematics, we use the elimination method for solving systems of equations to eliminate one variable and solve for the other.

This post will explain the process for solving systems of equations by elimination.

Solving Systems of Linear Equations by Elimination

Substitution and elimination are two ways to solve systems of linear equations algebraically. In general, substitution is the best choice when one equation has a variable isolated. Thus, all things being equal, we would probably use substitution if our system of equations were:

y=1-x

x-5y=3.5

So, when is it best to use the elimination method? We might use the elimination method to solve the system of equations if rewriting the equations to isolate a variable is more complicated.

Consider the following system:

2x+5y=12.5

6x-5y=3.5

It’s clear that if we try to solve one of these equations for either x or y, we will end up with fractions. Fractions can complicate the process of substitution. So, with this system of equations, the better option would be to solve the system of equations by elimination.

What is the Elimination Method?

So, how do you solve systems of equations by elimination? Let’s start our overview of how to solve a system of equations by elimination with a list of steps. 

1. Determine which variable will be eliminated. If necessary, rewrite one or both equations to make the coefficients of that variable additive inverses.
2. Combine the equations using addition. The new equation should have only one remaining variable.
3. Solve the new equation for the remaining variable.
4. Substitute the value for that variable into one of the original equations and solve.
5. Check your work by graphing or substitution.

Check out this video for a full walkthrough of the elimination method in action.

Next, let’s put these steps to work with some examples of the elimination method.

Solving Systems of Equations by Elimination Examples

Let’s take a closer look at the system below.

2x+5y=12.5

6x-5y=3.5

First, notice that the coefficients for the y terms are 5 and -5. These two numbers are additive inverses, which means they have a sum of 0. Therefore, if we were to add these equations, the y terms will cancel each other out.

But is adding equations a legitimate means to solve a system of equations? Consider the Addition Property of Equality, which states:

If a=b, then a+c=b+c.

We’re going to use this property to solve systems of linear equations by elimination.

To start, let’s think about the second equation in two parts. We could separate each part and rewrite them as follows:

6x-5y=z\quad\text{ and }\quad z=3.5

We know these statements are true because 6x-5y=3.5. Therefore if one side of this equation equals z, then the other equals z as well.

Now, we can use the Addition Property of Equality to rewrite the first equation in our system, 2x+5y=12.5, as:

2x+5y+{\color{red}{z}}=12.5+{\color{red}{z}}

…and then substitute in our values of z to get:

2x+5y+{\color{red}{(6x-5y)}}=12.5+{\color{red}{(3.5)}}

We’ve just added our two equations! When we simplify our new equation, we’ll eliminate the y term so that we can solve for x.

2x+5y+6x-5y=12.5+3.5

 8x=16

 \dfrac{8x}{{\color{red}{8}}}=\dfrac{16}{{\color{red}{8}}}

 x=2

Next, we can substitute 2 for x in one of our original equations and solve for y.

6{\color{red}{(2)}}-5y=3.5

12-5y=3.5

-5y=-8.5

y=1.7

So, the solution to our system is (2,1.7). We can check our work by graphing, as shown below.

Solving Systems of Equations by Elimination with Multiplication

What happens if the coefficients of a variable aren’t additive inverses in a system of equations? For example, consider the system below.

4x-2y=5

7x-2y=9

If we add these equations as they are now, the y terms would become -2y+(-2y)=-4y and the x terms would become 4x+7x=11x. We won’t have eliminated any variables.

However, if we multiply one of the equations by the factor -1, we end up with coefficients that are additive inverses. Let’s try it: 

{\color{red}{-1}}(4x-2y)={\color{red}{-1}}(5)

-4x+2y=-5

Now, we can add our equations to eliminate the y variable and solve for x.

To solve other systems of equations by elimination, we’ll need to rewrite both equations to eliminate a variable. For example, consider the system below. 

3x-9y=6

2x-2y=8

We could multiply the first equation by -2 and the second equation by 3 to allow us to eliminate x when we add the equations. Let’s use this process to solve the system by elimination.

Step	Equations
1. Determine which variable will be eliminated. If necessary, rewrite one or both equations to make the coefficients of that variable additive inverses.	{\color{red}{-2}}(3x-9y)={\color{red}{-2}}(6) -6x + 18y = -12 {\color{red}{3}}(2x-2y)={\color{red}{3}}(8) 6x - 6y = 24
2. Combine the equations using addition. The new equation should have only one remaining variable.	-6x + 18y + 6x - 6y = -12 + 24 (-6x + 6x) + (18y - 6y) = 12 12y = 12
3. Solve the new equation for the remaining variable.	12y \div 12= 12 \div 12 y = 1
4. Substitute the value for that variable into one of the original equations and solve.	2x-2y=8 2x-2({\color{red}{1}})=8 2x-2=8 2x=10 x=5
5. Check your work by graphing or substitution.	3x-9y=6 3({\color{red}{5}})-9({\color{red}{1}})=6 15-9=6 6=6

More Complex Systems of Equations with Multiplication

Finally, we’ll encounter systems where we need to rewrite equations more extensively to eliminate a variable. For example, consider the system below.

3x=4y-2

5x-2y=6

Before we add these equations, we need to move the y term to the opposite side of the equals sign in the first equation and multiply the second equation by -2.

3x - 4y =4y-2 - 4y \rightarrow 3x - 4y = -2

-2(5x-2y)=-2(6)\rightarrow -10x + 4y = -12

Now, we can eliminate y terms and solve for x.

To summarize, the key to effectively rewriting the equations is selecting the correct factor to multiply one or both equations to create additive inverses. Additive inverses will cancel each other out and thus eliminate that variable when the equations are added.

Use the link here to access more solving systems of linear equations by elimination practice problems.

Solving Systems of Equations by Elimination: Keys to Remember

• We use elimination to eliminate one variable in a two-variable system.
• One or both equations may need to be manipulated to make the coefficients of the same variable additive inverses.
• Since additive inverses sum to zero, adding the equations will eliminate one variable and allow us to solve for the other.
• Once the value of one variable is known, we substitute that value into one of the original equations and solve for the other variable.
• Finally, solutions can be confirmed algebraically or by graphing.