Answer
$x=\left\{ -\dfrac{13}{4},\dfrac{11}{4} \right\}$
Work Step by Step
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\}
.$
