Help with simplification question: $\frac{x}{x-1} - \frac{x}{x+1}$

Question

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.

Answer 1

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: $$\frac{ab}{ad}+\frac{c}{ad}$$ $$\frac{b}{d}+\frac{c}{ad}$$

Answer 2

\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.