# 5.3 Solve systems of equations by elimination  (Page 2/6)

 Page 2 / 6

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.

## How to solve a system of equations by elimination

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

## Solution

Solve the system by elimination. $\left\{\begin{array}{c}3x+y=5\hfill \\ 2x-3y=7\hfill \end{array}$

$\left(2,-1\right)$

Solve the system by elimination. $\left\{\begin{array}{c}4x+y=-5\hfill \\ -2x-2y=-2\hfill \end{array}$

$\left(-2,3\right)$

The steps are listed below for easy reference.

## How to solve a system of equations by elimination.

1. Write both equations in standard form. If any coefficients are fractions, clear them.
2. Make the coefficients of one variable opposites.
• Decide which variable you will eliminate.
• Multiply one or both equations so that the coefficients of that variable are opposites.
3. Add the equations resulting from Step 2 to eliminate one variable.
4. Solve for the remaining variable.
5. Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable.
6. Write the solution as an ordered pair.
7. Check that the ordered pair is a solution to both original equations.

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

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

## 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).

Solve the system by elimination. $\left\{\begin{array}{c}2x+y=5\hfill \\ x-y=4\hfill \end{array}$

$\left(3,-1\right)$

Solve the system by elimination. $\left\{\begin{array}{c}x+y=3\hfill \\ -2x-y=-1\hfill \end{array}$

$\left(-2,5\right)$

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

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

## 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).

Solve the system by elimination. $\left\{\begin{array}{c}4x-3y=1\hfill \\ 5x-9y=-4\hfill \end{array}$

$\left(1,1\right)$

Solve the system by elimination. $\left\{\begin{array}{c}3x+2y=2\hfill \\ 6x+5y=8\hfill \end{array}$

$\left(-2,4\right)$

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.

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

## 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?

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