Help with simplification question: $\frac{x}{x-1} - \frac{x}{x+1}$ Write $\frac{x}{x-1} - \frac{x}{x+1}$ as a single fraction in its simplest form.
I got $\frac{x}{x-1}$ but according to the mark scheme this is incorrect. Could someone analyze my working out and see where I went wrong? 
Working out:
Combined both fractions, then expanded brackets
$$\frac{x^2}{(x - 1)(x + 1)} = \frac{x^2}{x^ 2 - 1}$$
Further simplified
$$\frac{x(x)}{x(x) - 1}$$
Removed $x$ from top and bottom
$$\frac{x}{x-1}$$
Any help would be much appreciated.
 A: You messed up a little bit of algebra at the beginning, and then you made a mistake at the end by cancelling on the top and bottom. Here is how this should be done:
First get a common denominator:
$$\frac{x(x+1)}{(x-1)(x+1)}-\frac{x(x-1)}{(x-1)(x+1)}$$
Then combine:
$$\frac{x(x+1)-x(x-1)}{(x-1)(x+1)}$$
Multiply the factors:
$$\frac{x^2+x-x^2+x}{x^2-1}$$
$$\frac{2x}{x^2-1}$$
This should be the correct answer. 
FOR FUTURE REFERENCE:
If, in a fraction, you have something like this:
$$\frac{ab+c}{ad}$$
You CANNOT cancel the $a$ on the top and bottom. You can only cancel if $a$ can be factored out of all of the numerator; for example, you could cancel if you had
$$\frac{a(b+c)}{ad}$$
or if you had
$$\frac{ab+c}{ad}$$
and split it up into two fractions:
$$\frac{ab}{ad}+\frac{c}{ad}$$
$$\frac{b}{d}+\frac{c}{ad}$$
A: \begin{align}
\frac{x}{x-1}-\frac{x}{x+1} &= \frac{x(x+1)-x(x-1)}{(x-1)(x+1)} \\
&=\frac{x^2+x-(x^2-x)}{x^2-1} \\
&=\frac{x^2+x-x^2+x}{x^2-1} \\
&=\frac{2x}{x^2-1}
\end{align}
Remark: Without the answer scheme, we should know that soemthing went wrong if $$a-b=a$$ if $b \neq 0$ in general.
