Answer
$\left( -4,8 \right)$
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, $
\left| \dfrac{x-2}{3} \right|\lt2
,$ is
\begin{array}{l}\require{cancel}
-2\lt\dfrac{x-2}{3}\lt2
\\\\
-6\lt x-2\lt6
\\\\
-6+2\lt x\lt6+2
\\\\
-4\lt x\lt8
.\end{array}
In interval notation, the solution set is $
\left( -4,8 \right)
.$
