Answer
The solution is $(-\infty,-\frac{3}{2}) \cup [\frac{-13}{9},\infty)$
Work Step by Step
$\frac{a+2}{3+2a} \leq 5$
First solve the corresponding equation
$\frac{a+2}{3+2a} = 5$
$a+2=15+10a$
$9a=-13$
$a=\frac{-13}{9}$
For $(-\infty,-\frac{3}{2})$, choose $-3$: $\frac{-3+2}{3+2(-3)}=\frac{1}{3} \lt 5$
For $(-\frac{3}{2},\frac{-13}{9})$, choose $-1.45$: $\frac{-1.45+2}{3+2(-1.45)}=\frac{11}{2}\gt5$
For $(\frac{-13}{9},\infty)$, choose $4:$, $\frac{4+2}{3+2(4)}=\frac{6}{11}\lt5$
The solution is $(-\infty,-\frac{3}{2}) \cup [\frac{-13}{9},\infty)$
