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. 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.

5 comments on “A quadratic simultaneous equation”

• 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…

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