Intermediate Algebra, Chapter 2, 2.7, Section 2.7, Problem 82

Evaluate the equation $\displaystyle \left| \frac{2}{3}x - 2 \right| = \left| \frac{1}{3}x + 3 \right|$.

This equation is satisfied either if $\displaystyle \frac{2}{3}x - 2$ and $\displaystyle \frac{1}{3}x + 3$ are equal to each other or if $\displaystyle \frac{2}{3}x - 2$ and $\displaystyle \frac{1}{3}x + 3$ are negatives of each other


$
\begin{equation}
\begin{aligned}

\frac{2}{3}x - 2 =& \frac{1}{3}x + 3 && \text{or} &&& \frac{2}{3}x - 2 =& - \left( \frac{1}{3}x + 3 \right)
\\
\\
\frac{2}{3}x - \frac{1}{3}x =& 2 + 3 && \text{or} &&& \frac{2}{3}x + \frac{1}{3}x =& - 3 + 2
\\
\\
\frac{1}{3}x =& 5 && \text{or} &&& x =& -1
\\
\\
x =& 15


\end{aligned}
\end{equation}
$


The solution set is $\{ -1,15 \}$.