## Precalculus: Mathematics for Calculus, 7th Edition

$\left( -2,\dfrac{2}{3} \right)$
For any $a\gt0$, $|x|\lt a$ implies $-a\lt x\lt a$. (The symbol $\lt$ may be replaced with $\le$.) Using the concept above, the solutions to the given inequality, $|3x+2|\lt4 ,$ is \begin{array}{l}\require{cancel} -4\lt3x+2\lt4 \\\\ -4-2\lt3x\lt4-2 \\\\ -6\lt3x\lt2 \\\\ -\dfrac{6}{3}\lt x\lt\dfrac{2}{3} \\\\ -2\lt x\lt\dfrac{2}{3} .\end{array} In interval notation, the solution set is $\left( -2,\dfrac{2}{3} \right) .$