1/(X+3)+4/(x+5)= 2/(x^2+8x+15)
a. the solution set is
b. the solution set is all real numbers
c. the solution set is empty
Did you notice that x^2 + 8x + 15 = (x+3)(x+5) which happen to be the denominators on the left side
So let's multiply everybody by (x+3)(x+5)
(x+5) + 4(x+3) = 2
Now how easy is that from there ?