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.
 A: 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.
A: This is a symmetric system, so it can be reformulated in terms of $x+y=50$ and $xy=?$. Just observe that $x^2+y^2=(x+y)^2-2xy$.
When we know the sum and product of two numbers, it's easy to find them.
A: As you have now figured out the answer on your own, I'll post the way I was hinting at in a comment.
I would say your book is pointing you towards the hard way. It's easier just to solve for $y$ in the first equation and substitute that into in the second, giving you a quadratic in $x$, trivially solved by inspection or the quadratic equation.
$$
x + y = 50 \\
y = 50 - x \\
--- \\
x^2 + y^2 + (x - y)^2 = 1400 \\
x^2 + (50-x)^2 + (x - (50 - x))^2 = 1400 \\
x^2 + x^2 - 100x + 50^2 + 4x^2 - 200x + 50^2 = 1400 \\
6x^2 - 300x + 3600 = 0 \\
x^2 - 50x + 600 = 0 \\
(x - 20)(x - 30) = 0 \\
x = 20~or~30\\
y = 50 - x\\
y = 30~or~20
$$
So x and y are 20 and 30, in either combination.
