## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$a=\left\{ -\dfrac{3}{5},5 \right\}$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $|3a-1|=|2a+4| ,$ use the definition of an absolute value equality. Then use the properties of equality to isolate the variable. $\bf{\text{Solution Details:}}$ Since $|x|=|y|$ implies $x=y \text{ or } x=-y,$ the equation above is equivalent to \begin{array}{l}\require{cancel} 3a-1=2a+4 \\\\\text{OR}\\\\ 3a-1=-(2a+4) .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} 3a-1=2a+4 \\\\ 3a-2a=4+1 \\\\ a=5 \\\\\text{OR}\\\\ 3a-1=-(2a+4) \\\\ 3a-1=-2a-4 \\\\ 3a+2a=-4+1 \\\\ 5a=-3 \\\\ a=-\dfrac{3}{5} .\end{array} Hence, $a=\left\{ -\dfrac{3}{5},5 \right\} .$