## Precalculus: Mathematics for Calculus, 7th Edition

$\left( -\infty, -1 \right] \cup \left[ \dfrac{7}{3},\infty \right)$
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, $|3x-2|\ge5 ,$ is \begin{array}{l}\require{cancel} 3x-2\ge5 \\\\ 3x\ge5+2 \\\\ 3x\ge7 \\\\ x\ge\dfrac{7}{3} ,\\\\\text{ OR }\\\\ 3x-2\le-5 \\\\ 3x\le-5+2 \\\\ 3x\le-3 \\\\ x\le-\dfrac{3}{3} \\\\ x\le-1 .\end{array} In interval notation, the solution set is $\left( -\infty, -1 \right] \cup \left[ \dfrac{7}{3},\infty \right) .$