#### Answer

$x=\left\{ -\dfrac{32}{3},8 \right\}$

#### Work Step by Step

Since for any $a\gt0$, $|x|=a$ implies $x=a$ OR $x=-a$, then the given equation, $
\left|1+\dfrac{3}{4}x\right|=7
,$ is equivalent to
\begin{array}{l}\require{cancel}
1+\dfrac{3}{4}x=7 \text{ OR } 1+\dfrac{3}{4}x=-7
.\end{array}
Solving each equation results to
\begin{array}{l}\require{cancel}
1+\dfrac{3}{4}x=7
\\\\
4\left(1+\dfrac{3}{4}x\right)=(7)4
\\\\
4+3x=28
\\\\
3x=28-4
\\\\
3x=24
\\\\
x=\dfrac{24}{3}
\\\\
x=8
\\\\\text{ OR }\\\\
1+\dfrac{3}{4}x=-7
\\\\
4\left(1+\dfrac{3}{4}x\right)=(-7)4
\\\\
4+3x=-28
\\\\
3x=-28-4
\\\\
3x=-32
\\\\
x=-\dfrac{32}{3}
.\end{array}
Hence, the solutions are $
x=\left\{ -\dfrac{32}{3},8 \right\}
.$