Question

How do you solve `2/(x+3)-4/(x^2+2x-3)=1/(1-x)`?

Answer 1

The given equation is invalid.
There are no possible solutions.

Explanation:

Given
`color(white)("XXX")2/(x+3)-4/(x^2+2x-3)=1/(1-x)`

Noting that
`color(white)("XXX")x^2+2x-3=(x+3)(x-1)`
the given equation can be re-written as
`color(white)("XXX")(2(x-1))/(x^2+2x-3)-4/(x^2+2x-3)= ((-1)(x+3))/(x^2+2x+3)`

Provided `x!=-3` and `color(red)(x!=1)`
then `x^2+2x-3 != 0`

and we have
`color(white)("XXX")2(x-1)-4=-(x+3)`

`color(white)("XXX")2x-6=-x-3`

`color(white)("XXX")3x = 3`

`color(white)("XXX")x=1` ...but this is one of the "forbidden values".

Both the terms `4/(x^2+2x-3)` and `1/(1-x)` are undefined when `x=1`