Answer
The equation has two solutions, -5 and 2.
Work Step by Step
$\frac{3}{s-1}+1=\frac{12}{s^2-1}$
$\frac{3}{s-1}+1=\frac{12}{(s-1)(s+1)}$
$(s-1)(s+1)(\frac{3}{s-1}+1)=(s-1)(s+1)\frac{12}{(s-1)(s+1)}$
$3(s+1)+(s-1)(s+1)=12$
$3s+3+s^2-1-12=0$
$s^2+3s-10=0$
$s^2+5s-2s-10=0$
$s(s+5)-2(s+5)=0$
$(s-2)(s+5)=0$
$s-2=0$ or $s+5=0$
$s=2$ or $s=-5$
Check:
$3/(2-1)+1=3/1+1=4; 12/(2^2-1)=12/3=4$
$3/(-5-1)+1=-1/2+1=1/2; 12[(-5)^2-1]=12/24=1/2$
The equation has two solutions, -5 and 2.
