We want all denominators to be (w^2-9)
[(w-3)+(x+3)]/(w^2-9)=(w^2-3)/(w^2-9) multiply both sides by (w^2-9)
w-3+w+3=w^2-3 combine like terms on left side
2w=w^2-3 subtract 2w from both sides
w^2-2w-3=0 factor
w^2+w-3w-3=0
w(w+1)-3(w+1)=0
(w+1)(w-3)=0
w=-1 and 3
HOWEVER 3 is an extraneous solution as it would make a couple of our original fractions undefined because of division by zero, so there really is only one soluton:
w=-1