## Precalculus: Mathematics for Calculus, 7th Edition

$\left[ 2,8 \right]$
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, $|x-5|\le3 ,$ is \begin{array}{l}\require{cancel} -3\le x-5\le3 \\\\ -3+5\le x\le3+5 \\\\ 2\le x\le8 .\end{array} In interval notation, the solution set is $\left[ 2,8 \right] .$