## Intermediate Algebra (6th Edition)

$(1,5)$
Since $|x|\lt c$ implies $-c\lt x\lt c,$ then the solution to the given inequality, $|x-3|\lt2 ,$ is equivalent to \begin{array}{l}\require{cancel} -2\lt x-3\lt2 \\\\ -2+3\lt x-3+3\lt2+3 \\\\ 1\lt x\lt5 .\end{array} The graph of the solution set is shown below. In interval notation, this is equivalent $(1,5) .$