Solving a system of 3 nonlinear equations


by EM_Guy
Tags: equations, nonlinear, solving
EM_Guy
EM_Guy is offline
#1
Jan23-13, 01:55 PM
P: 3
a = xyz
b = xy+xz+yz
c = x + y + z

How do you solve x, y, and z?
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mfb
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#2
Jan23-13, 02:14 PM
Mentor
P: 10,840
x=c-y-z
b=(c-y-z)y + (c-y-z)z + yz = c(y+z)-y^2-zy-z^2
Solve this quadratic equation for y (or z), use both in a=xyz and hope that it has a nice solution?
EM_Guy
EM_Guy is offline
#3
Jan23-13, 03:07 PM
P: 3
It is not a quadratic equation. And it is not a "nice" solution.

I have determined that z^3-cz^2+bz-a = 0. So, if we can find the roots of the cubic function, then we have z as a function of a, b, and c. Then, it should be straightforward to find x and y in terms of a, b, and c.

But I forget how to find the roots of a cubic function.

mfb
mfb is offline
#4
Jan23-13, 04:46 PM
Mentor
P: 10,840

Solving a system of 3 nonlinear equations


b= c(y+z)-y^2-zy-z^2 is a quadratic equation in y (or z).

Solutions of a cubic equation
z^3-cz^2+bz-a = 0 looks very nice in my opinion.


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