## College Algebra 7th Edition

Interval: $[\frac{9}{2},\ 5)\cup (5,\ \frac{11}{2}]$
$0\lt|x-5| \leq\frac{1}{2}$ $\frac{1}{2}\leq x-5\leq\frac{1}{2}$ $\frac{1}{2}+5\leq x-5+5\leq\frac{1}{2}+5$ $\frac{9}{2}\leq x\leq\frac{11}{2}$ $4.5\leq x\leq5.5$ However, we exclude $x=5$ because then we would have $0\lt0$, which is false. Interval: $[\frac{9}{2},\ 5)\cup (5,\ \frac{11}{2}]$