I need help on this algebra problem.?

Question

Explain how to do this problem.

(18/(n²-9))+1 = (n/(n+3))

Answer 1

im redoing it...

(18/(n²-9))+1 = (n/(n+3))

thenchange 1 to (n²-9)/(n²-9) and change (n/(n+3)) to (n²-3n)/(n²-9)

(18/(n²-9))+ (n²-9)/(n²-9) = (n²-3n)/(n²-9)

now simplify...cuz of common denominater..

(18 + n² - 9 + n² +3n) / (n² - 9) = 0

then simplify some more..

(3n +9) / (n² - 9) = 0

then

3(n +3) / (n +3) (n - 3) = 0

3 / (n - 3) = 0

ok ummm thats what i get and i have no idea if thats right still cuz you say the answer is -3...sorry if im no help

Answer 2

(18/(n²-9))+1 = (n/(n+3))

this becomes

(18+n²-9)/(n²-9) = n/(n+3)

the multiply both sides by n²-9 and n+3)

so it becomes

(18+n²-9)/(n-3) = n(n-3)

multiply both sides by n-3

18+n²-9= n(n²-6n+9)

18+n²-9=n^3-6n^2+9n

Subtracting 18+n²-9 from both sides

0=n^3-7n^2+9n-9

You do the factoring. Can't think of the factors.