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.

The equation does not hold at $x = -1$. – Makoto Kato Jan 02 '13 at 02:47 — Makoto Kato, Jan 02 '13 at 02:47

Quispiam · Accepted Answer · 2013-01-05T18:17:13.997

8

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.

edited Jan 05 '13 at 18:17

answered Jan 02 '13 at 01:06

Quispiam

1

The equality does not make sense when $x = -1$. – Makoto Kato Jan 02 '13 at 01:44
1

Congratulations on the correct answer, but do mention that $x≠±1$ since the fraction would become undefined in the opposite case. – P.K. Jan 03 '13 at 13:58

1 Answers1