*Ask Uncle Colin* is a chance to ask your burning, possibly embarrassing, maths questions -- and to show off your skills at coming up with clever acronyms. Send your questions to colin@flyingcoloursmaths.co.uk and Uncle Colin will do what he can.

Dear Uncle Colin,

Could you please tell me how to solve simultaneous equations? I have a rough idea, but I get confused about it.

-- Stuck In Mathematical Examinations/Qualifications

Hello, SIMEQ!

Here’s how I attack linear simultaneous equations, such as:

$5x + 6y = -34$ (A)

$7x + 2y = 10$ (B)

First thing: rearrange get either the $x$s or the $y$s on their own on one side of each equation. Let’s pick $y$s here — it really doesn’t matter, we'll get the same answer either way.

$6y = -34 - 5x$ (A')

$2y = 10 - 7x$ (B')

Now get the same NUMBER of $y$s in each equation. If I multiply (A’) by $2$ and (B’) by $6$, this will automatically work:

$12y = -68 - 10x$ (A'')

$12y = 60 - 42x$ (B'')

If $12y$ equals one thing in (A’’) and $12y$ equals another thing in (B’’), the one thing and the other thing have to be the same.

$-68 - 10x = 60 - 42x$ (C)

(Take a moment to convince yourself that’s true).

Now rearrange to find $x$:

$ 32x = 128$ (C’)

$x = 4$

And substitute back into anything that has an $x$ and a $y$ in to find $y$:

$7x + 2y = 10$ (B)

$28 + 2y = 10$

$2y = -18$

$y = -9$

Lastly, check with the other initial equation to make sure it works:

$5x + 6y = -34$ (A)

$20 - 54 = -34$, which is true!

Hope that helps!

-- Uncle Colin

## 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.