## College Algebra 7th Edition

Interval: $(- \frac{15}{2},\ -7)\cup\ (-7,\ -\frac{13}{2})$
$\frac{1}{|x+7|}\gt 2$ We can see that $x\neq -7$ because otherwise we would have division by 0. $1\gt 2|x+7|$ $\frac{1}{2}\gt |x+7|$ $|x+7|\lt \frac{1}{2}$ $-\frac{1}{2}\lt x+7\lt \frac{1}{2}$ $-\frac{15}{2}\lt x\lt -\frac{13}{2}$ We use interval notation and skip $x=-7$: Interval: $(- \frac{15}{2},\ -7)\cup\ (-7,\ -\frac{13}{2})$