How would you go about solving these set of equations?

[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")

 

[chiro]

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")

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).

 

[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?

 

[nonequilibrium]

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

 

[CellCoree]

First, we can rewrite the original set of equations as:

x + y + z = a
xy + (x + y)z = b
xyz = c

Next, we can use substitution to eliminate one variable at a time. For example, we can solve for z in the second equation by substituting x + y for z:

xy + (x + y)(x + y) = b
x² + 2xy + y² = b

Now, we can use the same method as in the given example to solve for x and y:

(x - y)² = (x + y)² - 4xy = a² - 4b
x - y = ±√(a² - 4b)

Substituting this value into the first equation, we can solve for x and y:

x + y = a ±√(a² - 4b)

Finally, we can use these values to solve for z in the first equation:

z = a - x - y = a - (a ±√(a² - 4b)) = ±√(a² - 4b)

Therefore, the solutions for x, y, and z are:

x = (a ±√(a² - 4b))/2
y = (a ±√(a² - 4b))/2
z = ±√(a² - 4b)

This method elegantly solves the set of equations and reduces it to linear equations, similar to the given example. It also works for any set of equations in this form, as long as the variables are not raised to powers higher than 1.