Answer
Solution set: $\{4\}$
Work Step by Step
$g(a)=4$
$ \displaystyle \frac{3a-2}{a+1}+\frac{a+2}{a-1}=4\qquad$ ... $\times$LCD$=(a+1)(a-1)$
Exclude solutions which yield 0 in denominators:
$a\not\in\{-1,1\}\qquad (*)$
$(3a-2)(a-1)+(a+2)(a+1)=4(a^{2}-1)$
$3a^{2}-3a-2a+2+a^{2}+3a+2=4a^{2}-4$
$ 4a^{2}-2a+4=4a^{2}-4\qquad$ ... add $-4a^{2}-4$
$-2a=-8$
$ a=4\qquad$ ...Checking (*), this is a valid solution
Solution set: $\{4\}$
