Answer
$\left( -2,\dfrac{2}{3} \right)$
Work Step by Step
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)
.$
