Simple Algebra.. solve the value of x?

2/(x - 1) + 3/(x + 1) = 4/(x^2 - 1)

(x - 1)(x + 1)[2/(x - 1) + 3/(x + 1)] = (x - 1)(x + 1){4/[(x + 1)(x - 1)]

2(x + 1) + 3(x - 1) = 4

2(x) + 2(1) + 3(x) - 3(1) = 4

2x + 2 + 3x - 3 = 4

2x + 3x = 4 - 2 + 3

5x = 5

x = 5/5

x = 1
(answer a, x^2 - 1 cannot be 0)

2 ( x + 1 ) + 3 ( x - 1 ) = 4

5 x - 1 = 4

5 x = 5

x = 1

OPTION c

Hi !
2/(x-1) + 3/(x+1) = 4/(x^2 -1)
(2*(x+1) + 3*(x-1)) / ((x+1)*(x-1)) = 4/(x^2 -1)
2x + 2 + 3x - 3 = 4
2x + 2 + 3x - 3 - 4 = 0
5x - 5 = 0
5x = 5
x = 1

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Good Luck !

Answer's letter A. 0 cannot be in the denominator or it wiil be an undefined answer.

If you simplify the equation it will look like this:

5(x-1)/[(x=1)(x-1)] = 0

Substitute 1 or -1 the values of x and you're going to have a null set solution.

The answer is a.
You sub. the answers in:
1 is impossible since 1-1=0, and anything divided by 0 is meaningless, so it doesn't satisfy the right-hand side of the equation.
-1 is also impossible since -1+1=0, reasons for not accepting this answer is the same as above.
By method of elimination, only A is the answer.

answer = a

if x is either 1 or -1, you will have a term whose denominator is zero, which makes the term (and the whole expression as well) undefined :)

The answer is A
You cannot divide by 0, so..
x-1 =/= 0, x cannot be 1. x+1 cannot be 0, so x=/= -1. This leaves A as the answer.

2/(x-1) + 3/(x+1) = 4/(x^2 -1)
2(x+1)+3(x-1)=4
5x-1=4
5x=5
x=1 (a) x can't be one

d