Solving a system of 3 nonlinear equations

[EM_Guy]

a = xyz
b = xy+xz+yz
c = x + y + z

How do you solve x, y, and z?

 

[mfb]

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]

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]

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.

 

[blue_raver22]

To solve this system of 3 nonlinear equations, we can use a variety of methods such as substitution, elimination, or graphing. One approach is to use the substitution method, where we solve one equation for one variable and then substitute it into the other equations. In this case, we can solve for z in the first equation, which gives us z = a/(xy). We can then substitute this into the second equation to get b = xy + x(a/(xy)) + y(a/(xy)). Simplifying, we get b = 2xy + a. We can now solve for x in terms of y and a, which gives us x = (b-a)/(2y). We can then substitute this value for x into the third equation to solve for y, and then solve for z using the first equation. This approach requires some algebraic manipulation but can lead to exact solutions for x, y, and z. Another approach is to use numerical methods such as Newton's method or the bisection method to approximate the solutions. These methods involve iterative calculations and can be more time-consuming, but they can handle more complex nonlinear equations. Ultimately, the method chosen will depend on the specific equations and the desired level of accuracy.