Ask Uncle Colin: Simultaneous Equations

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 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 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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