# Solving systems of linear equations by elimination

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This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr. Beginning with the graphical solution of systems, this chapter includes an interpretation of independent, inconsistent, and dependent systems and examples to illustrate the applications for these systems. The substitution method and the addition method of solving a system by elimination are explained, noting when to use each method. The five-step method is again used to illustrate the solutions of value and rate problems (coin and mixture problems), using drawings that correspond to the actual situation.Objectives of this module: know the properties used in the addition method, be able to use the addition method to solve a system of linear equations, know what to expect when using the addition method with a system that consists of parallel or coincident lines.

## Overview

• The Properties Used in the Addition Method
• Addition and Parallel or Coincident Lines

## The properties used in the addition method

Another method of solving a system of two linear equations in two variables is called the method of elimination by addition . It is similar to the method of elimination by substitution in that the process eliminates one equation and one variable. The method of elimination by addition makes use of the following two properties.

1. If $A$ , $B$ , and $C$ are algebraic expressions such that

$\begin{array}{ccc}\frac{\begin{array}{rrr}\hfill A& \hfill =& \hfill B\\ \hfill C& \hfill =& \hfill D\end{array}}{\begin{array}{rrr}\hfill A+C& \hfill =& \hfill B+D\end{array}}& & \begin{array}{l}\text{and}\\ \text{then}\\ \\ \end{array}\end{array}$
2. $ax+\left(-ax\right)=0$

Property 1 states that if we add the left sides of two equations together and the right sides of the same two equations together, the resulting sums will be equal. We call this adding equations . Property 2 states that the sum of two opposites is zero.

To solve a system of two linear equations in two variables by addition,

1. Write, if necessary, both equations in general form, $ax+by=c.$
2. If necessary, multiply one or both equations by factors that will produce opposite coefficients for one of the variables.
3. Add the equations to eliminate one equation and one variable.
4. Solve the equation obtained in step 3.
5. Do one of the following:
(a)  Substitute the value obtained in step 4 into either of the original equations and solve to obtain the value of the other variable,
or
(b)  Repeat steps 1-5 for the other variable.
6. Check the solutions in both equations.
7. Write the solution as an ordered pair.

The addition method works well when the coefficient of one of the variables is 1 or a number other than 1.

## Sample set a

Solve  $\left\{\begin{array}{rrr}\hfill x-y=2& \hfill & \hfill \left(1\right)\\ \hfill 3x+y=14& \hfill & \hfill \left(2\right)\end{array}$

Step 1:  Both equations appear in the proper form.

Step 2:  The coefficients of $y$ are already opposites, 1 and $-1,$ so there is no need for a multiplication.

$\frac{\begin{array}{c}x-y=2\\ 3x+y=14\end{array}}{4x+0=16}$

Step 4:  Solve the equation $4x=16.$

$4x=16$

$x=4$

The problem is not solved yet; we still need the value of $y$ .

Step 5:  Substitute $x=4$ into either of the original equations. We will use equation 1.

$\begin{array}{rrrrr}\hfill 4-y& \hfill =& \hfill 2& \hfill & \hfill \text{Solve\hspace{0.17em}for\hspace{0.17em}}y.\\ \hfill -y& \hfill =& \hfill -2& \hfill & \hfill \\ \hfill y& \hfill =& \hfill 2& \hfill & \hfill \end{array}$

We now have $x=4,\text{\hspace{0.17em}}\text{\hspace{0.17em}}y=2.$

Step 6:  Substitute $x=4$ and $y=2$ into both the original equations for a check.

$\begin{array}{rrrrrrrrrrrr}\hfill \left(1\right)& \hfill & \hfill x-y& \hfill =& \hfill 2& \hfill & \hfill \left(2\right)& \hfill & \hfill 3x+y& \hfill =& \hfill 14& \hfill \\ \hfill & \hfill & \hfill 4-2& \hfill =& \hfill 2& \hfill \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}& \hfill & \hfill & \hfill 3\left(4\right)+2& \hfill =& \hfill 14& \hfill \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\\ \hfill & \hfill & \hfill 2& \hfill =& \hfill 2& \hfill \text{Yes,\hspace{0.17em}this\hspace{0.17em}is\hspace{0.17em}correct}\text{.}& \hfill & \hfill & \hfill 12+2& \hfill =& \hfill 14& \hfill \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\\ \hfill & \hfill & \hfill & \hfill & \hfill & \hfill & \hfill & \hfill & \hfill 14& \hfill =& \hfill 14& \hfill \text{Yes,\hspace{0.17em}this\hspace{0.17em}is\hspace{0.17em}correct}\text{.}\end{array}$

Step 7:  The solution is $\left(4,2\right).$

The two lines of this system intersect at $\left(4,2\right).$