## Precalculus: Mathematics for Calculus, 7th Edition

$\left( -\infty, -\dfrac{15}{8} \right) \cup \left( \dfrac{9}{8},\infty \right)$
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 $\left( -\infty, -\dfrac{15}{8} \right) \cup \left( \dfrac{9}{8},\infty \right) .$