Ask Uncle Colin: A Three-Variable Simultaneous Equation

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,

I have a system of equations I can't solve!

$x + y + z = 100$;
$.08x +.1y +.2z = 12$;
$y - z = 25$

I keep tripping up on the decimals and negative signs!

-- We're Extremely Stressed Solving Equations Linear

Hello, WESSEL! My best advice for not tripping over the decimals is to get rid of them -- multiply the whole of the second equation by 100 to get

$8x + 10y + 20z = 1200$ (B')

I wouldn't start there for my solution, though: I'd begin by finding $y$ from the third equation:

$y - z = 25$, so $y = 25 + z$.

Substituting this into the first equation, $x + y + z = 100$, gives $x + (25 + z) + z = 100$, so $x + 2z = 75$ (A')

Doing the same for (B') gives $8x + 10(25+z) + 20z = 1200$, or
$8x + 30z = 950$ (B'')

Rearranging (A') gives $x = 75 - 2z$, which you can substitute into (B'') to get:

$8(75 - 2z) + 30z = 950$, or $14z=350$, so $z = 25$.

Now working backwards, $x = 75 - 50 = 25$, and $y = 25+25=50$.

The solution is $x=25, y=50, z=25$. Hope that helps!

-- Uncle Colin

* Edited 2017-07-12 to add categories.

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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