Answer
$x=\left\{ -\dfrac{5}{3},\dfrac{1}{3} \right\}$
Work Step by Step
Using the properties of equality, the given equation, $
\left|\dfrac{1}{2}x+\dfrac{1}{3}\right|+\dfrac{1}{4}=\dfrac{3}{4}
,$ is equivalent to
\begin{array}{l}\require{cancel}
\left|\dfrac{1}{2}x+\dfrac{1}{3}\right|=\dfrac{3}{4}-\dfrac{1}{4}
\\\\
\left|\dfrac{1}{2}x+\dfrac{1}{3}\right|=\dfrac{2}{4}
\\\\
\left|\dfrac{1}{2}x+\dfrac{1}{3}\right|=\dfrac{1}{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{1}{2}x+\dfrac{1}{3}=\dfrac{1}{2} \text{ OR } \dfrac{1}{2}x+\dfrac{1}{3}=-\dfrac{1}{2}
.\end{array}
Solving each equation results to
\begin{array}{l}\require{cancel}
\dfrac{1}{2}x+\dfrac{1}{3}=\dfrac{1}{2}
\\\\
6\left( \dfrac{1}{2}x+\dfrac{1}{3} \right)=\left(\dfrac{1}{2}\right)6
\\\\
3x+2=3
\\\\
3x=3-2
\\\\
3x=1
\\\\
x=\dfrac{1}{3}
\\\\\text{ OR }\\\\
\dfrac{1}{2}x+\dfrac{1}{3}=-\dfrac{1}{2}
\\\\
6\left( \dfrac{1}{2}x+\dfrac{1}{3} \right)=\left(-\dfrac{1}{2}\right)6
\\\\
3x+2=-3
\\\\
3x=-3-2
\\\\
3x=-5
\\\\
x=-\dfrac{5}{3}
.\end{array}
Hence, the solutions are $
x=\left\{ -\dfrac{5}{3},\dfrac{1}{3} \right\}
.$