Question

How do you solve `1/(x-1)+3/(x+1)=2`?

Answer 1

`x_1=0` and `x_2=2`

Explanation:

`1/(x-1)+3/(x+1)=2`

`[1*(x+1)+3*(x-1)]/[(x-1)(x+1)]=2`

`(4x-2)/(x^2-1)=2`

`4x-2=2*(x^2-1)`

`x^2-1=2x-1`

`x^2-2x=0`

`x*(x-2)=0`

Hence `x_1=0` and `x_2=2`

Answer 2

`x=0`, `x=2`

Explanation:

`1/(x-1)+3/(x+1)=2`

`rArr ((x+1)+3(x-1))/((x-1)(x+1))=2`

`rArr (x+1+3x-3)/((x-1)(x+1))=2`

`rArr (4x-2)/((x-1)(x+1))=2`

`rArr 4x-2=2(x+1)(x-1)`

`rArr 4x-2=2x+2(x-1)`

`rArr 4x-2=2x^2-2x+2x-2`

`rArr 4x-2=2x^2-2`

`rArr 2x^2-4x=0`

`rArr 2x(x-2)=0`

`rArr x-2=0 -> x=2`

`rArr 2x=0 -> x=0`