2.4.2 Adding Equations

The Elimination method of Solving Systems of Equations

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 $d$:

$\begin{array}{ccc}\hfill \text{ if }a& \hfill =\hfill & b\hfill \\ \hfill \text{and}c& \hfill =\hfill & d\hfill \\ \hfill \text{then}a+c& \hfill =\hfill & b+d.\hfill \end{array}$

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:

$$\begin{gathered} \underset{―}{\{\begin{array}{c}3x+y=5\hfill \\ 2x-y=0\hfill \end{array}} \\ 5x=5 \end{gathered}$$

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

Let’s try another one:

$\begin{array}{c}\hfill \{\begin{array}{ccc}\hfill 4p-2s\hfill & \hfill =\hfill & \hfill -26\hfill \\ \hfill -5p+2s\hfill & \hfill =\hfill & \hfill 29\hfill \end{array}\end{array}$

We can add the equations, so $s$ will be eliminated when we add these two equations.

$$\begin{gathered} \underset{―}{\{\begin{array}{ccc}\hfill 4p-2s\hfill & \hfill =\hfill & \hfill -26\hfill \\ \hfill -5p+2s\hfill & \hfill =\hfill & \hfill 29\hfill \end{array}} \\ -p=3 \\ p=-3 \end{gathered}$$

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.

Example

Solve the system by elimination:

$\begin{array}{c}\{\begin{array}{c}\hfill 4x+8y=12\hfill \\ \hfill 4x-5y=38\hfill \end{array}\hfill \end{array}$

Step 1 - Write both equations in standard form. If any coefficients are fractions, clear them. $4x+8y=12$ $4x-5y=38$

Step 2 - Check to see if the coefficients of one variable are opposites or equivalent. The coefficients on the $x$-variables are equivalent. We can subtract the equations.

Step 3 - Subtract the equations to eliminate one variable. $$\begin{gathered} \begin{array}{c}\underset{―}{4x+8y=12 \\ 4x–5y=38}\hfill \\ 13y=–26\hfill \end{array} \end{gathered}$$

Step 4 - Solve for the remaining variable. $\begin{array}{ccc}\hfill 13y& \hfill =\hfill & –26\hfill \\ \hfill y& \hfill =\hfill & –2\hfill \end{array}$

Step 5 - Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable. $\begin{array}{ccc}\hfill 4x+8y& \hfill =\hfill & 12\hfill \\ \hfill 4x+8(–2)& \hfill =\hfill & 12\hfill \\ \hfill 4x–16& \hfill =\hfill & 12\hfill \\ \hfill 4x& \hfill =\hfill & 28\hfill \\ \hfill x& \hfill =\hfill & 7\hfill \end{array}$

Step 6 - Write the solution as an ordered pair. $(7,-2)$

Step 7 - Check that the ordered pair is a solution to both original equations.

$4x+8y=12$

$4(7)+8(-2)=12$

$28-16=12$

$12=12✔$

$4x-5y=38$

$4(7)-5(-2)=38$

$28+10=38$

$38=38✔$