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

$x=\left\{ -\dfrac{11}{2},\dfrac{13}{2} \right\}$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $\left| \dfrac{2x-1}{3} \right|=4 ,$ use the definition of an absolute value equality. Then use the properties of equality to isolate the variable. $\bf{\text{Solution Details:}}$ Since for any $c\gt0$, $|x|=c$ implies $x=c \text{ or } x=-c,$ the given equation is equivalent to \begin{array}{l}\require{cancel} \dfrac{2x-1}{3}=4 \\\\\text{OR}\\\\ \dfrac{2x-1}{3}=-4 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} \dfrac{2x-1}{3}=4 \\\\ 3\cdot\dfrac{2x-1}{3}=3\cdot4 \\\\ 2x-1=12 \\\\ 2x=12+1 \\\\ 2x=13 \\\\ x=\dfrac{13}{2} \\\\\text{OR}\\\\ \dfrac{2x-1}{3}=-4 \\\\ 3\cdot\dfrac{2x-1}{3}=3\cdot(-4) \\\\ 2x-1=-12 \\\\ 2x=-12+1 \\\\ 2x=-11 \\\\ x=-\dfrac{11}{2} .\end{array} Hence, $x=\left\{ -\dfrac{11}{2},\dfrac{13}{2} \right\} .$