Question 449838
I assume the problem is really 
(x)/(x-3)+(x-1)/(x^2-9)=(2x-3)/(x+3) 
You don't want to multiply by (x+3)(x-3) which is (x^2-9)
You want to get the same denominator of (x^2-9)
so sometimes you will multiply by (x+3)/(x+3) which is 1/1
and sometimes by (x-3)/(x-3) which is also 1/1
We must discount  any solutions where x=3 or -3 
since they would lead to a zero denominator.
it ends up
(x^2+4 x-1)/((x-3) (x+3)) = (2 x-3)/(x+3)
((x-13) x+10)/(x^2-9) = 0
4/129 (x-13/2)^2 = 1

x = 1/2 (13-sqrt(129))
x = 1/2 (13+sqrt(129))