Solving a cubic system of equations

Question

I have the following equation system:

$A_1 x + B_1 y + C_1 z + D_1 xy + E_1 xz + F_1 yz + G_1 xyz = M_1$ $A_2 x + B_2 y + C_2 z + D_2 xy + E_2 xz + F_2 yz + G_2 xyz = M_2$ $A_3 x + B_3 y + C_3 z + D_3 xy + E_3 xz + F_3 yz + G_3 xyz = M_3$

$A_1$, $B_1$, ..., $M_1$, $A_2$, $B_2$, ..., $M_2$, $A_3$, $B_3$, ..., $M_3$ are known.

Trying to get $x$ based on $y$ and $z$ from the first equation, then substituting it in the second equation, then getting $y$ based on $z$ and substituting it in the third equasion seems a nightmare. How to solve this equation system?

Answer 1

A possible workaround can be the following:

1. Let's multiply the first equation with with $-G_2$, the second equation with $G_1$, and add them.
2. Now multiply the second equation with $-G_3$, the third one with $G_2$, and add them.
3. Finally multiply the third equation with $-G_1$, and the first one with $G_3$ and add them.

Theoretically we have now tree new equations, but we escaped from the $xyz$ part.

In the same way let's escape from xy, xz, xz.

We have now a simple equation system with 3 equations and 3 unknowns:

$AA_1x + BB_1y + CC_1z = MM_1$

$AA_2x + BB_2y + CC_2z = MM_2$

$AA_3x + BB_3y + CC_3z = MM_3$

that's easy to solve.

If you think this wouldn't always work please let me know, or if you have a better solution it is welcomed!

Answer 2

Maybe a bit ambiguous, but you can also try to set up the equations for $M_1+M_2$, $M_1+M_3$, $M_2+M_3$ and $M_1+M_2+M_3$, which is easy, in total you then have 7 equations, with 7 unknowns ($x,y,z,xy,xz,yz,xyz$). But i dont think this is even allowed.