Precalculus: Mathematics for Calculus, 7th Edition

$\left( -\infty, -\dfrac{7}{2}\right) \cup \left( \dfrac{7}{2},\infty \right)$
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, $|2x|\gt7 ,$ is \begin{array}{l}\require{cancel} 2x\gt7 \\\\ x\gt\dfrac{7}{2} ,\\\\\text{ OR }\\\\ 2x\lt-7 \\\\ x\lt-\dfrac{7}{2} .\end{array} In interval notation, the solution set is $\left( -\infty, -\dfrac{7}{2}\right) \cup \left( \dfrac{7}{2},\infty \right) .$