# Solution for a polynomial system of three equations

1. Nov 24, 2015

### Bruno Tolentino

I desire to know tho solution (solution for a, b and c in terms of B', C' and D') of this system of equation:

B' = - 2 a - c
C' = 2 a c + b²
D' = - b² c

I don't know none method for solve this kind of system, therefore, I came to ask here...

2. Nov 24, 2015

### Staff: Mentor

Hi Bruno,

What is the context of this question? Is this for schoolwork?

3. Nov 25, 2015

### Mark Harder

Bruno,
Since time is short, I ran the problem without any prior tricks through Mathematica. If you know MMA, I naively entered the equations you provided into it's Solve[ ] function. The symbolic answer is a very, very ugly set. This kind of brute force, followed by some simplification, gives 6 solution sets for the 3 variables. When I substituted B=1, C=1, D=1 into these, there were 2 sets of real solutions. The other 4 were complex solutions. In some cases, one of the variables = 0.

If you are trying to find the answer to some coursework, then I suggest you find some trick that radically simplifies the equations. I'd start by solving one of these for one variable in terms of all others and substitute into the remaining eqns. Play around a bit. There's probably some symmetry that makes the problem radically simple. Good luck.

4. Nov 25, 2015

### HallsofIvy

Why are you under the impression that you have to learn a different "method" for any new kind of system? The basic ideas of "systems of equations" work nicely here. Your equations are B' = - 2 a - c, C' = 2 a c + b², and D' = - b² c. "Eliminate" one unknown at a time, just as you would for systems of linear equations until you have a single equation in one unknown. From the last equation, b²= -D'/c. Substituting into the second equation, C'= -2ac+ D'c. So 2ac= D'c- C' and a= (D'c- C')/2c. Then the first equation becomes B'= (C'- D'c)/c- c so B'+ c= (C'- D'c)/c, (B'+ c)c= c²+ B'c= C'- D'c, c²+ (B'+ D')c- C'= 0. That is a quadratic equation in c. Use the quadratic formula to find two solutions to that, the "back substitute" into a= (D'c- C')2c and b²= -D'/c to find corresponding solutions for and b.

5. Nov 26, 2015

### micromass

I would advise you to study elimination theory and Gröbner bases.