Answer
The solution is $(-\infty,\frac{3}{2}]$
Work Step by Step
$\frac{a-5}{a+2}\lt-1$
First solve the corresponding equation
$\frac{a-5}{a+2}=-1$
$a-5=-a-2$
$2a=3$
$a=\frac{3}{2}$
For $(-\infty,-2)$, choose $-1$: $\frac{-1-5}{-1+2}=-6\lt-1$
For $(-2,\frac{3}{2})$, choose $0$: $\frac{0-5}{0+2}=\frac{-5}{2}\lt-1$
For $(\frac{3}{2},\infty)$, choose $4:$, $\frac{4-5}{4+2}=-\frac{1}{6}\gt-1$
The solution is $(-\infty,-2) \cup (-2,\frac{3}{2}]$ which is also $(-\infty,\frac{3}{2}]$
