Brutal simultaneous equations

I recently became aware of the IYGB papers, available from Madas Maths. Like the Solomon papers, they're intended to stretch you a bit -- they're ranked by difficulty from standard to extremely hard. My student, being my student, demanded we go through one of the extremely hard ones.

There were several stretchy questions, but one that was hard enough for me to tweet it was this:

I thought it might be good to look at several ways of approaching it.

The Ninja Way

"What's ugly?" is the first question the Ninja always asks: here, it's the brackets in the second equation. How about we try a substitution?

Let $X = (x+3)$ and $Y = (y-1)$. Then the first equation becomes $15(Y+1) - 8(X-3) = 39$, or $15Y = 8X$ Now we're getting somewhere!

The second equation is $X^2 + Y^2 = 289$; the Ninja, of course, spotted the result from here because he knows his Pythagorean triples and his square numbers. However, we lesser mortals might substitute in:

$X^2 + \left(\frac {8}{15} X\right)^2 = 289$

$X^2 + \frac{64}{255} X^2 = 289$

$\frac{ 225+64}{225} X^2 = 289$

$\frac{289}{225}X^2 = 289$

$\frac{1}{225}X^2 = 1$

$X^2 = 225$ so $X = \pm 15$ and $ Y = \pm 8$. Converting those back into the original variables, you get $x = -3 \pm 15$ and $y = 1 \pm 8$, or $(12, 9)$ and $(-18, -7)$.

The standard way

The standard way, I have to warn you, is a Great Big Mess. It's worth doing, though, just to see what tricks we can employ to keep things nice.

The standard way is to get $x$ or $y$ on its own and substitute.

Since $15y - 8x = 39$, we can make $y = \frac{8x + 39}{15}$, and $y^2 = \frac{64x^2 + 39\times 16 x + 39^2}{225}$. Because $40 \times 16 = 640$, thirty-nine sixteens are 624; thirty-nine squared is 1,521, as everyone knows.

So, what happens to $(x+3)^2 + (y-1)^2 = 289$? We get $x^2 + 6x + 9 + \frac{64x^2 + 624x + 1521}{225} - \frac{2(8x + 39)}{15} + 1 = 289$. How lovely! Let's group like terms:

$\left( 1 + \frac{64}{225}\right)x^2 + \left(6 + \frac{624}{225} - \frac{16}{15}\right)x + 9 + \frac{1521}{225} - \frac{78}{15} + 1 - 289 = 0$. It gets nicer with every step, doesn't it? I suppose we'd better simplify the fractions:

$\left(\frac{225 + 64}{225}\right) x^2 + \left(\frac{ 1350 + 624 - 240 }{225}\right)x + \frac{1521 - 1170 - 279 \times 225} {225}= 0$.

Now. $279 \times 225$? Let's apply some Ninja tricks there. Because $225 = 25 \times 9$, he would suggest multiplying by 900 and dividing by 4. $279 \times 9 = 2790 - 279 = 2511$, and $251,100 / 4 = 62,775$. So where were we?

$ \frac {289}{225}x^2 + \frac{ 1,734 }{225} x - \frac{62,424}{225} = 0$ -- and we can get rid of the 225s.

$289x^2 + 1,724x - 62,424$. It'd be awfully nice if there was another common factor to all of those, wouldn't there? 289 is $17^2$, of course, but what about the others? Let's factorise!

1,734 is even, so we can halve it: it's $2 \times 867$; 867 is a multiple of 3 (its digits add up to 21), so $1734 = 2 \times 3 \times 289$: what do you know?! It's got a factor of 289. That could come in handy, but only if 62,424 works out too.

Again, it's even (a multiple of 8, if you know the rules), so $62,424 = 8 \times 7,803$. However, 7,803 is a multiple of 9 (its digits add up to 18), so $62,424 = 8 \times 9 \times 867$, which looks familiar! The 'numbers' term is $2^3 \times 3^3 \times 17^2$.

So, divide out the 289 and you have $x^2 + 6x - 216 = 0$, and that factorises at $(x+18)(x-12)=0$ and it's simple from there.

A more cunning way

That's a lot of heavy-duty arithmetic, especially getting and dealing with the 62,424. Is there a simpler way? I don't know about simpler, but I do know about more cunning: rather than work out the numbers, work with them.

Let's start from $x^2 + 6x + 9 + \frac{64 x^2 + 39 \times 16 x + 39^2}{225} - \frac{2 (8x + 39)}{15} + 1 - 289 = 0$.

The $x^2$ terms are $x^2$ and $\frac{64}{225}x^2$, which add to $\frac{289}{225}x^2$, as before -- nothing much to see here.

The $x$ terms are $6$, $\frac{39 \times 16}{225}$ and $-\frac{16}{15}$. Putting them over a common denominator gives $\frac{6 \times 225 + 39 \times 16 - 15\times 16}{225}$. All three terms on top have a factor of 6 that comes out: $\frac{6(225 + 13\times 8 - 5 \times 8)}{225}$. The last two terms combine into $8\times8 = 64$, and 225 + 64 = 289, so the $x$ terms simplify to $\frac{6 \times 289}{225}$.

The unit terms are $9$, $\frac{39^2}{225}$, $-\frac{2\times 39}{15}$, $1$ and $-289$. Starting with the fractions, they combine into $\frac{39 \times 39 - \frac 30 \times 39}{225} = \frac{9 \times 39}{225}$. That does cancel further, but everybody else has got 225s on the bottom, and it seems to be the trend. I'll leave the -289 alone, but think about the 10: that's $\frac{10 \times 225}{225}$, so I have $\frac{10 \times 225 + 9 \times 39}{225} + 289$ as my unit term. There's a factor of 9 on top, clearly: $\frac{9 (250 + 39)}{225} - 289$ and, hey presto! There's a 289 on the top.

Putting it all together, we've got $\frac{289}{225} x^2 + \frac{6 \times 289}{225}x + \frac{9 \times 289}{225} - 289 = 0$. Multiplying by $\frac{225}{289}$ gives:

$ x^2 + 6x + (9 - 225) = 0$. The unit term is -216, as we had earlier, and the rest comes out as before.


Phew!

That's an awful question -- and not something you're going to see in your C1 paper, ever, I promise. However, the tricks you use for making it drop out nicely are ones that can save you a huge amount of time and work. (The final way of doing that took me a small fraction of the time of the 'standard' way, and the sums were much easier.)

Have you got any other interesting ways to solve this?

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.

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