## College Algebra 7th Edition

Interval: $(- \infty,\ \frac{7}{5}]\cup [\frac{8}{5},\ \infty)$
$\frac{1}{|2x-3|}\leq 5$ We see that $x\ne\frac{3}{2}$ because otherwise we would get division by 0. $1\leq 5|2x-3|$ $\frac{1}{5}\leq|2x-3|$ We split the absolute value into the positive and negative: $\frac{1}{5}\leq 2x-3$ $\frac{16}{5}\leq 2x$ $\frac{8}{5}\leq x$ $2x-3 \leq-\frac{1}{5}$ $2x\leq\frac{14}{5}$ $x\leq\frac{7}{5}$ Interval: $(- \infty,\ \frac{7}{5}]\cup [\frac{8}{5},\ \infty)$