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

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.

-
The equation does not hold at $x = -1$. –  Makoto Kato Jan 2 '13 at 2:47

$(x-1)+2$ 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.
The equality does not make sense when $x = -1$. –  Makoto Kato Jan 2 '13 at 1:44
Congratulations on the correct answer, but do mention that $x≠±1$ since the fraction would become undefined in the opposite case. –  Parth Kohli Jan 3 '13 at 13:58