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.