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
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
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