#### Answer

Interval: $(- \infty,\ \frac{7}{5}]\cup [\frac{8}{5},\ \infty)$

#### Work Step by Step

$\frac{1}{|2x-3|}\leq 5$
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: $(- \infty,\ \frac{7}{5}]\cup [\frac{8}{5},\ \infty)$