## Intermediate Algebra (12th Edition)

$\left( -\infty,-3 \right] \cup \left[ \dfrac{5}{3},\infty \right)$
Since for any $c\gt0$, $|x|\gt c$ implies $x\gt c$ or $x\lt-c$, the given inequality, $|3x+2|\ge7 ,$ is equivalent to \begin{array}{l}\require{cancel} 3x+2\ge7 \text{ OR } 3x+2\le-7 .\end{array} Solving each inequality results to \begin{array}{l}\require{cancel} 3x+2\ge7 \\\\ 3x\ge7-2 \\\\ 3x\ge5 \\\\ x\ge\dfrac{5}{3} \\\\\text{ OR }\\\\ 3x+2\le-7 \\\\ 3x\le-7-2 \\\\ 3x\le-9 \\\\ x\le-\dfrac{9}{3} \\\\ x\le-3 .\end{array} Hence, the solution to the given inequality is the interval $\left( -\infty,-3 \right] \cup \left[ \dfrac{5}{3},\infty \right) .$ Note that $"\gt"$ may be replaced with $"\ge"$.