SOLUTION: Please explain how to find real solutions and check for extraneous solutions: 1/(w+3) + 1/(w-3)= (w^2-3)/(w^2-9)

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Please explain how to find real solutions and check for extraneous solutions: 1/(w+3) + 1/(w-3)= (w^2-3)/(w^2-9)      Log On


   

Question 1139129: Please explain how to find real solutions and check for extraneous solutions:
1/(w+3) + 1/(w-3)= (w^2-3)/(w^2-9)
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
1%2F%28w%2B3%29+%2B+1%2F%28w-3%29=+%28w%5E2-3%29%2F%28w%5E2-9%29+
%28w-3+%2Bw%2B3%29%2F%28%28w%2B3%29%28w-3%29%29=+%28w%5E2-3%29%2F%28w%5E2-9%29+
2w%2F%28w%5E2-9%29=+%28w%5E2-3%29%2F%28w%5E2-9%29+
since denominator cannot be equal to zero, exclude solution that makes
w%5E2-9=0
w%5E2=9
=>w=3 and w=-3=>extraneous solutions
real solution will be w that makes numerators equal
2w=+w%5E2-3
w%5E2-2w-3=0
w%5E2%2Bw-3w-3=0
%28w%5E2%2Bw%29-%283w%2B3%29=0
w%28w%2B1%29-3%28w%2B1%29=0
%28w-3%29%28w%2B1%29=0
solutions:
w=3 =>this solution is already excluded
or
w=-1=> this is your real solution