Why is $\frac{(x-1)+2}{(x-1)(x+1)} = \frac{1}{x-1}$?

Question

According to the solution to a problem, the following equation holds true:

$$\frac{(x-1)+2}{(x-1)(x+1)} = \frac{1}{x-1}$$

I can't see a way for this to work out.

Answer 1

I realised the answer while writing the question.

$(x-1)+2$ equals $x+1$ and cancels out the same fraction in the denominator.

$$\frac{(x-1)+2}{(x-1)(x+1)} = \frac{x+1}{(x-1)(x+1)} = \frac{1}{x-1}$$

Edit: As pointed out in the commentary, this doesn't hold true if $x= \pm 1$, as that would divide by zero.