SOLUTION: How would I solve this equation? x - 1 = 2 ___ ___ _____ x+1 x-1 x^2-1 I used the LCD of x+1, x-1 and i know the answer cannot be x=1 or x=-1. I came up w

Algebra ->  Coordinate Systems and Linear Equations  -> Lessons -> SOLUTION: How would I solve this equation? x - 1 = 2 ___ ___ _____ x+1 x-1 x^2-1 I used the LCD of x+1, x-1 and i know the answer cannot be x=1 or x=-1. I came up w      Log On


   

Question 372647: How would I solve this equation?
x - 1 = 2
___ ___ _____
x+1 x-1 x^2-1

I used the LCD of x+1, x-1 and i know the answer cannot be x=1 or x=-1. I came up with the answer of x= -1 but since it cannot be that, i put the empty set symbol of 0. Please help me, if this is incorrect? thank you.
Answer by Jk22(389) About Me  (Show Source):
You can put this solution on YOUR website!
the LCD of x-1 and x+1 is (x+1)(x-1)=(x^2-1)

Domain of definition x is not 1 or -1.

+x%2F%28x%2B1%29-1%2F%28x-1%29=2%2F%28x%5E2-1%29 times (x^2-1)=(x+1)(x-1)
x(x-1) - (x+1) = 2

x^2-x-x-1 = 2

x^2 -2x -3 = 0

(x-3)(x+1)=0

the solutions are : x=3 or x=-1, x=-1 is no solution since it would be a division by zero)

Solution : x=3

Verification : 3/4 - 1/2 = 1/4

and RHS : 2/(9-1)=2/8=1/4, which is the same.