# How would you go about solving these set of equations?

1. Jun 2, 2012

### nonequilibrium

Say you have a set of equations of the form
$\left\{ \begin{array}{rl} x+y+z &=a \\ xy + xz + yz &= b \\ xyz &= c \end{array} \right.$
(for clarity: I'm working over the regular numbers)
how would you go about solving it elegantly? (or at least rewriting it as linear equations)

I'm thinking of something analogous to how you can solve
$\left\{ \begin{array}{rl} x+y &=a \\ xy &= b \end{array} \right.$
namely by noting that (x-y)² = (x+y)² - 4xy = a² - 4b (and after taking the square root we're left with two good ol' linear equations, i.e. x+y=... and x-y=..., a form which I regard as "being solved")

2. Jun 2, 2012

### chiro

Hey mr. vodka.

Have you tried just subsituting two of the variables to get the whole thing in a third one?

So for example you could everything in terms of z by taking (1) to get x = a - y - z and taking (2) to get y = (b - xz)(x + z) and then plug in (1) and (2) and simplify to get an equation for (3) giving z = c/xy where you get an expression in terms of only z.

You could do it in more than one way (this is only one possible way), but then you would end up with some kind of equation and at the worst you can use a root-finding algorithm, and probably use a few transformations to get the thing in surd form if it exists (using your ideas in your original post).

3. Jun 3, 2012

### Studiot

Call the equations 1, 2 and 3

Multiply 2 by z

xyz+xz2+yz2=bz

Subtract 3

(x+y)z2 = bz-c

Substitute into 1

(bz-c)/z2+z=a

Can you take the cubic from there?

Last edited: Jun 3, 2012
4. Jun 3, 2012

### nonequilibrium

Thank you both for your trouble. Studiot's method was more like something I was looking for, thank you!