# A quadratic simultaneous equation

A charming little puzzle from Brilliant:

$x^2 + xy = 20$ $y^2 + xy = 30$

Find $xy$.

I like this in part because there are *many* ways to solve it, and none of them the ‘standard’ way for dealing with simultaneous equations.

You might look at it and say “Ah, there’s an $xy$ in each equation, we can just subtract them” - but, of course, ‘just’ is a tremendously dangerous word and you’re left with $y^2 - x^2 = 10$, which doesn’t help much.

You might decide to isolate a variable: from the first equation, $y = \frac{20 - x^2}{x}$. Substituting that into the second gives $\frac{\left(20-x^2\right)^2}{x^2} + 20-x^2 = 30$, or $0 = 50x^2 - 400$ after a bit of algebra, giving $x^2 = 8$. You can substitute this back into the first to get $xy = 12$ directly… but it’s not exactly elegant.

You might notice that the sum of the equations is $(x+y)^2 = 50$. This is a Good Step. Knowing $(x+y)=\pm 5\sqrt{2}$, you can divide $y^2-x^2$ by that to find that $y-x = \pm \sqrt{2}$ and figure out $x$ and $y$ from there. Still not the neatest of approaches, although it gives the answer.

You might notice that there’s a common factor of $(x+y)$ in both equations: $x(x+y) = 20$ and $y(x+y)=30$. Now we’re getting somewhere! If you *multiply* the expressions, you get $xy(x+y)^2 = 600$ - but you know $(x+y)^2 = 50$, so $xy = \frac{600}{50} = 12$. *That* is an elegant approach.