Intermediate Algebra (12th Edition)

$x=\left\{ -\dfrac{13}{4},\dfrac{11}{4} \right\}$
Using the properties of equality, the given equation, $\left|\dfrac{2}{3}x+\dfrac{1}{6}\right|+\dfrac{1}{2}=\dfrac{5}{2} ,$ is equivalent to \begin{array}{l}\require{cancel} \left|\dfrac{2}{3}x+\dfrac{1}{6}\right|=\dfrac{5}{2}-\dfrac{1}{2} \\\\ \left|\dfrac{2}{3}x+\dfrac{1}{6}\right|=\dfrac{4}{2} \\\\ \left|\dfrac{2}{3}x+\dfrac{1}{6}\right|=2 .\end{array} Since for any $a\gt0$, $|x|=a$ implies $x=a$ OR $x=-a$, then the equation above is equivalent to \begin{array}{l}\require{cancel} \dfrac{2}{3}x+\dfrac{1}{6}=2 \text{ OR } \dfrac{2}{3}x+\dfrac{1}{6}=-2 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} \dfrac{2}{3}x+\dfrac{1}{6}=2 \\\\ 6\left(\dfrac{2}{3}x+\dfrac{1}{6}\right)=(2)6 \\\\ 4x+1=12 \\\\ 4x=12-1 \\\\ 4x=11 \\\\ x=\dfrac{11}{4} \\\\\text{ OR }\\\\ 6\left(\dfrac{2}{3}x+\dfrac{1}{6}\right)=(-2)6 \\\\ 4x+1=-12 \\\\ 4x=-12-1 \\\\ 4x=-13 \\\\ x=-\dfrac{13}{4} .\end{array} Hence, the solutions are $x=\left\{ -\dfrac{13}{4},\dfrac{11}{4} \right\} .$