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Section 16.1 Make it Go Away

We have seen that we can solve systems of linear equations using substitution, but that one of the challenges of this method was the computational problem of manipulating fractions. The method of elimination is another approach to solving systems of equations that mostly sidesteps the issue (at least at this level), which often makes it a more efficient approach.

The core concept of this method is a slightly different use of axioms of equality that were introduced in Section 1.1 mixed in with a substitution. We can add the same value to both sides of an equation without losing the equality. The only twist is that we use two different representations in the addition step. Here is how it looks:

\begin{equation*} \text{If $a = b$ and $c = d$, then $a + c = b + d$.} \end{equation*}

The challenge is to find useful equations so that adding the equations together gives a helpful result. Here are two examples to compare:

First, notice that the addition in columns is simply an organizational tool. By putting like terms together, it simplifies the process of combining them and avoids errors. Second, notice that the final equation on the right only has one variable even though the two initial equations had two variables. The goal of the method of elimination is to eliminate one of the variables. This allows you to solve for the remaining variable.

Activity 16.1.1. Solving a System by Elimination (Part 1).

Here is the worked solution to the system of equations on the right (see above):

Try it!

Solve the system of equations below using the method of elimination.

\begin{equation*} \left\{ \begin{array}{rcrcr} 3x \amp + \amp y \amp = \amp 3 \\ -3x \amp + \amp 3y \amp = \amp -7 \\ \end{array} \right. \end{equation*}
Solution.

Activity 16.1.2. Solving a System by Elimination (Part 2).

Not all systems of equations are in a form that immediately lends itself to the method of elimination. Let's take another look at the example on the left from earlier:

\begin{equation*} \left\{ \begin{array}{rcrcr} 2x \amp + \amp 3y \amp = \amp 3 \\ 3x \amp + \amp 2y \amp = \amp 5 \\ \end{array} \right. \end{equation*}

If we add these equations together directly, nothing will cancel out. But we can multiply either equation by any non-zero number to change its form. And if this is done correctly, it will lead to cancellation.

From here, we can plug in \(y = -\frac{1}{5}\) into either of the original equations, but that will lead to fraction manipulations. If we wanted to avoid those, we can just use the same idea to eliminate the terms instead.

This leads to the solution \((x,y) = ( \frac{9}{5}, -\frac{1}{5} )\text{.}\)

Try it!

Solve the system of equations below using the method of elimination.

\begin{equation*} \left\{ \begin{array}{rcrcr} 2x \amp + \amp 3y \amp = \amp -7 \\ -3x \amp + \amp y \amp = \amp 3 \\ \end{array} \right. \end{equation*}
Solution.

There's no real "trick" to figuring out the numbers that you need to get the terms to cancel, but there is a general heuristic. Once you've identified the variable you want to eliminate, find the least common multiple of the coefficients, and multiply the equations by the appropriate value to make match, and then make sure that the signs are opposite each other. While some instructors suggest that you can "subtract" the equations, this ends up causing a lot of computational errors, and so it's usually better to use addition.