#### Answer

$x=\left\{ -\dfrac{15}{13},-\dfrac{13}{29} \right\}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given equation, $
\left| \dfrac{8x-1}{3x+2} \right|=7
,$ use the definition of absolute value equality.
$\bf{\text{Solution Details:}}$
Since for any $c\gt0$, $|x|=c$ implies $x=c \text{ or } x=-c,$ the equation above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{8x-1}{3x+2}=7
\\\\\text{OR}\\\\
\dfrac{8x-1}{3x+2}=-7
.\end{array}
Solving each equation results to
\begin{array}{l}\require{cancel}
\dfrac{8x-1}{3x+2}=7
\\\\
(3x+2)\left( \dfrac{8x-1}{3x+2}\right)=7(3x+2)
\\\\
1(8x-1)=7(3x+2)
\\\\
8x-1=21x+14
\\\\
8x-21x=14+1
\\\\
-13x=15
\\\\
x=\dfrac{15}{-13}
\\\\
x=-\dfrac{15}{13}
\\\\\text{OR}\\\\
\dfrac{8x-1}{3x+2}=-7
\\\\
(3x+2)\left( \dfrac{8x-1}{3x+2}\right)=-7(3x+2)
\\\\
1(8x-1)=-7(3x+2)
\\\\
8x-1=-21x-14
\\\\
8x+21x=-14+1
\\\\
29x=-13
\\\\
x=-\dfrac{13}{29}
.\end{array}
Hence, the solutions are $
x=\left\{ -\dfrac{15}{13},-\dfrac{13}{29} \right\}
.$