Ancient Babylonian problem: solve the system $x + y = 50$, $x^2 + y^2 + (x - y)^2 = 1400$ $x + y = 50, x^2 + y^2 + (x - y)^2 = 1400$.  [Hint: Subtract the square of the first equation from twice the second equation to get a quadratic in $x - y$.]  I have gotten it reduced to $x^2 + y^2 -3xy = 550$ but am not sure how to get it in terms of $x - y$ to solve it.

The hint is saying that you should take two times $x^2 + y^2 + (x - y)^2$ (the second equation) and subtract the square of the first, $(x + y)^2 = x^2 + 2xy + y^2$.
So the result is
$$
2(x^2 + y^2 + (x-y)^2) - (x^2 + 2xy + y^2)
$$
You should be able to simplify this into something simple in terms of just $(x-y)$.  But then, we also know that the above is equal to
$$
2(1400) - (50)^2
$$
So from here you can solve for $x-y$.  Then once you know $x - y$, since you already know $x + y$ you can find out what $x$ and $y$ are individually.
