Solving a System of Equations By Elimination

We have solved systems of linear equations by graphing and by substitution. Graphing works well when the variable coefficients are small and the solution has integer values. Substitution works well when we can easily solve one equation for one of the variables and not have too many fractions in the resulting expression.

The third method of solving systems of linear equations is called the Elimination Method. When we solved a system by substitution, we started with two equations and two variables and reduced it to one equation with one variable. This is what we’ll do with the elimination method, too, but we’ll have a different way to get there.

Solving a System of Equations By Elimination

The Elimination Method is based on the Addition Property of Equality. The Addition Property of Equality says that when you add the same quantity to both sides of an equation, you still have equality. We will extend the Addition Property of Equality to say that when you add equal quantities to both sides of an equation, the results are equal.

For any expressions a, b, c, and d,

To solve a system of equations by elimination, we start with both equations in standard form. Then we decide which variable will be easiest to eliminate. How do we decide? We want to have the coefficients of one variable be opposites, so that we can add the equations together and eliminate that variable.

Notice how that works when we add these two equations together:

The y’s add to zero and we have one equation with one variable.

Let’s try another one:

This time we don’t see a variable that can be immediately eliminated if we add the equations.

But if we multiply the first equation by −2, we will make the coefficients of x opposites. We must multiply every term on both sides of the equation by −2.

Now we see that the coefficients of the x terms are opposites, so x will be eliminated when we add these two equations.

Add the equations yourself—the result should be −3y = −6. And that looks easy to solve, doesn’t it? Here is what it would look like.

We’ll do one more:

It doesn’t appear that we can get the coefficients of one variable to be opposites by multiplying one of the equations by a constant, unless we use fractions. So instead, we’ll have to multiply both equations by a constant.

We can make the coefficients of x be opposites if we multiply the first equation by 3 and the second by −4, so we get 12x and −12x.

This gives us these two new equations:

When we add these equations,

the x’s are eliminated and we just have −29y = 58.

Once we get an equation with just one variable, we solve it. Then we substitute that value into one of the original equations to solve for the remaining variable. And, as always, we check our answer to make sure it is a solution to both of the original equations.

Now we’ll see how to use elimination to solve the same system of equations we solved by graphing and by substitution.

Example: How to Solve a System of Equations by Elimination

Solve the system by elimination. \(\begin{array}{c}2x+y=7\hfill \\ x-2y=6\hfill \end{array}\)

Solution

The steps are listed below for easy reference.

First we’ll do an example where we can eliminate one variable right away.

Example

Solve the system by elimination. \(\begin{array}{c}x+y=10\hfill \\ x-y=12\hfill \end{array}\)

Solution

Both equations are in standard form.
The coefficients of y are already opposites.
Add the two equations to eliminate y. The resulting equation has only 1 variable, x.
Solve for x, the remaining variable. Substitute x = 11 into one of the original equations.
Solve for the other variable, y.
Write the solution as an ordered pair.	The ordered pair is (11, −1).
Check that the ordered pair is a solution to both original equations. \(\begin{array}{cccc}\begin{array}{ccc}\hfill x+y& =\hfill & 10\hfill \\ \hfill 11+\left(-1\right)& \stackrel{?}{=}\hfill & 10\hfill \\ \hfill 10& =\hfill & 10\phantom{\rule{0.2em}{0ex}}✓\hfill \end{array}& & & \begin{array}{ccc}\hfill x-y& =\hfill & 12\hfill \\ \hfill 11-\left(-1\right)& \stackrel{?}{=}\hfill & 12\hfill \\ \hfill 12& =\hfill & 12\phantom{\rule{0.2em}{0ex}}✓\hfill \end{array}\end{array}\)
	The solution is (11, −1).

In the example below, we will be able to make the coefficients of one variable opposites by multiplying one equation by a constant.

Example

Solve the system by elimination. \(\begin{array}{c}3x-2y=-2\hfill \\ 5x-6y=10\hfill \end{array}\)

Solution

Both equations are in standard form.
None of the coefficients are opposites.
We can make the coefficients of y opposites by multiplying the first equation by −3.
Simplify.
Add the two equations to eliminate y.
Solve for the remaining variable, x. Substitute x = −4 into one of the original equations.
Solve for y.
Write the solution as an ordered pair.	The ordered pair is (−4, −5).
Check that the ordered pair is a solution to both original equations. \(\begin{array}{cccc}\begin{array}{ccc}\hfill 3x-2y& =\hfill & -2\hfill \\ \hfill 3\left(-4\right)-2\left(-5\right)& \stackrel{?}{=}\hfill & -2\hfill \\ \hfill -12+10& \stackrel{?}{=}\hfill & -2\hfill \\ \hfill -2y& =\hfill & -2\phantom{\rule{0.2em}{0ex}}✓\hfill \end{array}& & & \begin{array}{ccc}\hfill 5x-6y& =\hfill & 10\hfill \\ \hfill 3\left(-4\right)-6\left(-5\right)& \stackrel{?}{=}\hfill & 10\hfill \\ \hfill -20+30& \stackrel{?}{=}\hfill & 10\hfill \\ \hfill 10& =\hfill & 10\phantom{\rule{0.2em}{0ex}}✓\hfill \end{array}\end{array}\)
	The solution is (−4, −5).

Now we’ll do an example where we need to multiply both equations by constants in order to make the coefficients of one variable opposites.

Example

Solve the system by elimination. \(\begin{array}{c}4x-3y=9\hfill \\ 7x+2y=-6\hfill \end{array}\)

Solution

In this example, we cannot multiply just one equation by any constant to get opposite coefficients. So we will strategically multiply both equations by a constant to get the opposites.

Both equations are in standard form. To get opposite coefficients of y, we will multiply the first equation by 2 and the second equation by 3.
Simplify.
Add the two equations to eliminate y.
Solve for x. Substitute x = 0 into one of the original equations.
Solve for y.
Write the solution as an ordered pair.	The ordered pair is (0, −3).
Check that the ordered pair is a solution to both original equations. \(\begin{array}{cccc}\begin{array}{ccc}\hfill 4x-3y& =\hfill & 9\hfill \\ \hfill 4\left(0\right)-3\left(-3\right)& \stackrel{?}{=}\hfill & 9\hfill \\ \hfill 9& =\hfill & 9\phantom{\rule{0.2em}{0ex}}✓\hfill \end{array}& & & \begin{array}{ccc}\hfill 7x+2y& =\hfill & -6\hfill \\ \hfill 7\left(0\right)+2\left(-3\right)& \stackrel{?}{=}\hfill & -6\hfill \\ \hfill -6& =\hfill & -6\phantom{\rule{0.2em}{0ex}}✓\hfill \end{array}\end{array}\)
	The solution is (0, −3).

What other constants could we have chosen to eliminate one of the variables? Would the solution be the same?

When the system of equations contains fractions, we will first clear the fractions by multiplying each equation by its LCD.

Example

Solve the system by elimination. \(\begin{array}{c}x+\frac{1}{2}y=6\hfill \\ \frac{3}{2}x+\frac{2}{3}y=\frac{17}{2}\hfill \end{array}\)

Solution

In this example, both equations have fractions. Our first step will be to multiply each equation by its LCD to clear the fractions.

To clear the fractions, multiply each equation by its LCD.
Simplify.
Now we are ready to eliminate one of the variables. Notice that both equations are in standard form.
We can eliminate y multiplying the top equation by −4.
Simplify and add. Substitute x = 3 into one of the original equations.
Solve for y.
Write the solution as an ordered pair.	The ordered pair is (3, 6).
Check that the ordered pair is a solution to both original equations. \(\begin{array}{cccc}\begin{array}{ccc}\hfill x+\frac{1}{2}y& =\hfill & 6\hfill \\ \hfill 3+\frac{1}{2}\left(6\right)& \stackrel{?}{=}\hfill & 6\hfill \\ \hfill 3+6& \stackrel{?}{=}\hfill & 6\hfill \\ \hfill 6& =\hfill & 6\phantom{\rule{0.2em}{0ex}}✓\hfill \\ \\ \\ \\ \\ \end{array}& & & \begin{array}{ccc}\hfill \frac{3}{2}x+\frac{2}{3}y& =\hfill & \frac{17}{2}\hfill \\ \hfill \frac{3}{2}\left(3\right)+\frac{2}{3}\left(6\right)& \stackrel{?}{=}\hfill & \frac{17}{2}\hfill \\ \hfill \frac{9}{2}+4& \stackrel{?}{=}\hfill & \frac{17}{2}\hfill \\ \hfill \frac{9}{2}+\frac{8}{2}& \stackrel{?}{=}\hfill & \frac{17}{2}\hfill \\ \hfill \frac{17}{2}& =\hfill & \frac{17}{2}\phantom{\rule{0.2em}{0ex}}✓\hfill \end{array}\end{array}\)
	The solution is (3, 6).

In the Solving Systems of Equations by Graphing section we saw that not all systems of linear equations have a single ordered pair as a solution. When the two equations were really the same line, there were infinitely many solutions. We called that a consistent system. When the two equations described parallel lines, there was no solution. We called that an inconsistent system.