Answer
The equation has one solution, -1.
Work Step by Step
$\frac{a}{a+3}=\frac{2a}{a-3}-1$
$\frac{a}{a+3}-\frac{2a}{a-3}+1=0$
$\frac{a(a-3)}{(a+3)(a-3)}-\frac{2a(a+3)}{(a+3)(a-3)}+\frac{(a+3)(a-3)}{(a+3)(a-3)}=0$
$\frac{a^2-3a-2a^2-6a+a^2-9}{(a+3)(a-3)}=0$
$(a^2+a^2-2a^2)-(3a+6a)-9=0*(a+3)(a-3)$
$-9a-9=0$
$-9a=9$
$a=-1$
Check:
$-1/(-1+3)=-1/2;(2\times-1)/(-1-3)-1=1/2-1=-1/2$
The equation has one solution, -1.
