Solving simultaneous equations in terms of variables If $x+y = m$ and $x-y=n$ then $(x^2-y^2) -2x$ is equal to
in terms of $m$ and $n$ only!
How do you solve? 

Notice that
$$(x^2 - y^2) - 2x = (x + y)(x - y) - [(x + y) + (x - y)]$$
But you know what $x + y$ and $x - y$ are ($m$ and $n$ respectively). I believe it is very, very simple to continue on from here.
This is the simplest method to solve this particular problem. Of course, you could also choose to express $x$ and $y$ individually in terms of $m$ and $n$ and then substitute them in. If we do a bit of manipulation, we get $x = \frac{m + n}{2}$ and $y = \frac{m - n}{2}$. Substituting into $(x^2 - y^2) - 2x = (x + y)(x - y) - 2x$:
$$\left(\frac{m + n + m - n}{2}\right)\left(\frac{m + n - m + n}{2}\right) - 2\left(\frac{m + n}{2}\right)\\
=\left(\frac{2m}{2}\right)\left(\frac{2n}{2}\right) - (m + n)\\
= mn - m - n$$
The key idea is to split $x^2 - y^2$ into $(x + y)(x - y)$ instead of actually going all the way to evaluate two squares.
