<s>[INST] 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.
 [/INST] 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.
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