## Precalculus: Mathematics for Calculus, 7th Edition

$\left( -4,8 \right)$
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, $\left| \dfrac{x-2}{3} \right|\lt2 ,$ 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 $\left( -4,8 \right) .$