SOLUTION: please help solve the equation and indicate any extraneous solutions https://ibb.co/eLKwV5 (use the link to see the problem thanks for your help and please show work)

Question 1086440: please help
solve the equation and indicate any extraneous solutions
https://ibb.co/eLKwV5
(use the link to see the problem thanks for your help and please show work)
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
2x/(x^2-1)=(4x^2+6x-6)/(x^3+x^2-x-1)-1/(x+1)
Think a common denominator will contain x+1 or x-1. Indeed, divide x^2-1 into x^3+x^2-x-1 and get (x+1)
Therefore, multiply the whole equation by (x^2-1)(x+1).
2x(x+1)=(4x^2+6x-6)-(x^2-1)
expand and collect terms
2x^2+2x=4x^2+6x-6-x^2+1, watch signs
2x^2+2x=3x^2+6x-5
0=x^2+4x-5
0=(x+5)(x-1)
x=-5, 1
If x=1, the denominator for the left side becomes 0, so extraneous.
if x=-5, -10/24=64/104-1/-4
-10/24=64/-96+24/96=-40/96, and they are equal.
x=-5

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