Lesson 14: Solving Systems by Elimination (Part 1)

Solving Systems by Elimination (Part 1)

Student Summary

Another way to solve systems of equations algebraically is by elimination. Just like in substitution, the idea is to eliminate one variable so that we can solve for the other. This is done by adding or subtracting equations in the system. Let’s look at an example.

$\begin {cases} \begin{aligned}5x+7y=64\\ 0.5x - 7y = \text-9 \end{aligned} \end {cases}$

Notice that one equation has $7y$ and the other has $\text-7y$ .

If we add the second equation to the first, the $7y$ and $\text-7y$ add up to 0, which eliminates the $y$ -variable, allowing us to solve for $x$ .

$\begin{aligned} 5x+7y&=64\\ 0.5 x - 7y &= \text-9 \quad+\\ \overline {5.5 x + 0} &\overline {\hspace{1mm}= 55}\\ 5.5x &= 55 \\x &=10 \end{aligned}$

Now that we know $x = 10$ , we can substitute 10 for $x$ in either of the equations and find $y$ :

$\begin{aligned} 5x+7y&=64\\ 5(10)+7y &= 64\\ 50 + 7y &= 64\\ 7y &=14 \\y &=2 \end{aligned}$

$\begin{aligned} 0.5 x - 7y &= \text-9\\ 0.5(10) - 7y &= \text-9\\ 5 - 7y &= \text-9\\ \text-7y &= \text-14\\ y &= 2 \end{aligned}$

In this system, the coefficient of $y$ in the first equation happens to be the opposite of the coefficient of $y$ in the second equation. The sum of the terms with $y$ -variables is 0.

What if the equations don't have opposite coefficients for the same variable, like in the following system?

$\begin {cases}\begin{aligned}8r + 4s &= 12 \\ 8r + \hspace{2.3mm}s &= \hspace{1mm}\text-3 \end{aligned} \end {cases}$

Notice that both equations have $8r$ , and if we subtract the second equation from the first, the variable $r$ will be eliminated because $8r-8r$ is 0.

$\begin{aligned} 8r + 4s &= 12\\ 8r + \hspace{2.3mm}s &= \hspace{1mm}\text-3 \quad-\\ \overline {\hspace{2mm}0 + 3s }& \overline{ \hspace{1mm}=15} \\ 3s &= 15 \\s &=5 \end{aligned}$