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.

## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008.
He lives with an espresso pot and nothing to prove.

## Sean

Hi Uncle Colin, I was wondering how you got to… y-x= + or – root 2

I’m assuming you’re implying for us to divide y²-x² = 10 by x+y = ±5 √2

in the following statement…

Knowing (x+y)=±52‾√, you can divide y2−x2 by that to find that y−x=±2‾√

## Sean

Hi again. I somehow arrived at y – x = 10/±5√2 after figuring that y² + x² = 10

x²-y² = (x+y) (x-y)

& x + y = ±5√2

so I put y²-x² = (±5√2)(x-y)

10 = (±5√2)(x-y)

x-y = 10/(±5√2)

..any help on how you arrived at: x-y = ±√2

would be appreciated!

## Colin

You’re practically there! You have $x-y = \frac{10}{\pm 5\sqrt{2}}$; divide top and bottom by 5 to get $x-y = \pm \frac{2}{\sqrt{2}}$. Now, $\frac{2}{\sqrt{2}} = \sqrt{2}$, so you have $\pm \sqrt{2}$.

## Sean/Vahid

y−x = ±√2

yes, i made it to this point with your kind help. May I ask, how you would solve for XY from this point on, as you infer in your VLOG as being one possible route to the solution of XY=12.

Eagerly awaiting your response Uncle Colin, although no rush at all.

May you be well.

## Colin

We have:

* $x+y = \pm 5\sqrt{2}$

* $x-y = \pm \sqrt{2}$

So, $x = y \pm \sqrt{2}$ from the second.

Substituting that into the first gives $2y \pm \sqrt{2} = \pm 5\sqrt{2}$

So $2y = \pm 4\sqrt{2}$, and $y = \pm 2\sqrt{2}$.

Then, since $x = y \pm \sqrt{2}$, we get $x = 3\sqrt{2}$; that points to $xy = \left(2\sqrt{2}\right)\left(3\sqrt{2}\right) = 6\times 2 = 12$

The one thing I’ve glossed over here is that all of the $\pm$s “point the same way” – if $x+y$ is positive, so is $x-y$. I imagine I could make that clearer…