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 firstname.lastname@example.org 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.