## Intermediate Algebra (12th Edition)

$\left[ -16,10 \right]$
Since for any $a\gt0$, $|x|\lt a$ implies $-a\lt x\lt a,$ then the given inequality, $|x+3|\le13 ,$ is equivalent to \begin{array}{l}\require{cancel} -13\le x+3\le13 .\end{array} Using the properties of inequality, then \begin{array}{l}\require{cancel} -13-3\le x+3-3\le13-3 \\\\ -16\le x\le10 .\end{array} Hence, the solution is the interval $\left[ -16,10 \right] .$