1st one: we multiply throughout by x^2-1 assuming x is different from +1 or -1. and note that x^2-1=(x+1)(x-1)

So we have:

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

or: 5x-1=4

5x = 5 = > x= 1. This is indeed a solution of 2(x+1)+3(x-1)=4 above, but cannot be a solution of the original equation 2/(x-1) + 3/(x+1) = 4/(x^2-1) because at x=1 both 2/(x-1) and 4/(x^2-1) diverges to infinity. Thus, there is really no solution in the domain of definition of the terms of the equation...

2nd one:

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

same method - multiply by (x+1)(x-3) assuming x is not -1 or +3:

Yea - that's the way I got it, The x=1 solution simply blow off the 2/(x-1) term and the 4/(x^2-1) so what I said is right. They don't always tell you everything in class, sometime you're supposed to think a little (I guess...)

Also the last seems OK. I think this is the purpose of this exercise to demonstarte unacceptable solutions.