Answer
$\left( -\infty, -\dfrac{15}{8} \right) \cup \left( \dfrac{9}{8},\infty \right)$
Work Step by Step
For any $a\gt0$, $|x|\gt a$ implies $x\gt a \text{ OR } x\lt -a$. (The symbol $\gt$ may be replaced with $\ge$.)
Using the concept above, the solutions to the given inequality, $
|8x+3|\gt12
,$ is
\begin{array}{l}\require{cancel}
8x+3\gt12
\\\\
8x\gt12-3
\\\\
8x\gt9
\\\\
x\gt\dfrac{9}{8}
,\\\\\text{ OR }\\\\
8x+3\lt-12
\\\\
8x\lt-12-3
\\\\
8x\lt-15
\\\\
x\lt-\dfrac{15}{8}
.\end{array}
In interval notation, the solution set is $
\left( -\infty, -\dfrac{15}{8} \right) \cup \left( \dfrac{9}{8},\infty \right)
.$
