# Approximate solution to a system of polynomial equations

How can I solve the following nonlinear algebraic system over the positive reals - for $x,y,z$ - perhaps with a suitable approximation for the term in parentheses?

\begin{align*} Q_1-d_1x-a_2a_3\left(\dfrac{xyz}{a_2cy+a_3z}\right) &= 0 \\ Q_2-d_2y-a_2xy &= 0 \\ Q_3-d_3z-a_3xz &= 0 \end{align*}

where $a_2,a_3,d_1,d_2,d_3,Q_1,Q_2,Q_3 \in \mathbb{R}_{> 0}$ and $c \in \mathbb{Z}_{> 0}$.

Mathematica is not able to solve this system even with the assumptions on the parameters!

Any ideas would be appreciated.

• Multiply the 1st equation by the denominator. – Rodrigo de Azevedo May 11 '17 at 16:18
• the solution of this System Looks ugly – Dr. Sonnhard Graubner May 11 '17 at 16:22
• Excellent! That means there is a solution. I'm trying with the suggestion by @RodrigodeAzevedo. – user85362 May 11 '17 at 16:23
• @RodrigodeAzevedo Yes - I just want to solve the system with the parameters. That way, I can compare the values of $x,y,z$ (which comprise the steady-state of a system of ODEs in $x,y,z$) to my numerical simulation results - both deterministic and stochastic. This is but one part of a larger research problem. – user85362 May 11 '17 at 16:30

## 2 Answers

Note that if $a_2 = 0$ or $a_3 = 0$, the system decouples and you can solve for $x, y, z$ separately. Thus you may assume that $a_2 \ne 0 \ne a_3$.

In this case, after dividing the second equation by $a_2$ and the third equation by $a_3$, you may assume that $a_2 = a_3 = 1$. Solving for $y$ and $z$ and substituting into the first equation gives $$Q_1 - d_1x -\frac{Q_2 Q_3 x}{c Q_2 (d_3+x)+ Q_3 (d_2 +x)} = 0$$ which becomes a quadratic equation after multiplying with the denominator. This can be solved with the usual formula.

• Wait a second - can you clarify why $a_2=a_3=1$ can be assumed? – user85362 May 11 '17 at 17:49
• Divide the second equation by $a_2$ and replace $Q_2$ with $Q_2/a_2$ and $d_2$ with $d_2/a_2$. It' s rescaling. Same thing for the third equation. – Hans Engler May 11 '17 at 18:34

In the same spirit as Hans Engler's answer.

Starting with the equations $$Q_1-d_1x-a_2a_3\left(\dfrac{xyz}{a_2cy+a_3z}\right) = 0\tag 1$$ $$Q_2-d_2y-a_2xy = 0 \tag 2$$ $$Q_3-d_3z-a_3xz = 0\tag 3$$ and using brute force, $y$ and $z$ can be eliminated and expressed as functions of $x$. This gives $$y=\frac{{Q_2}}{{a_2} x+{d_2}}\tag 4$$ $$z=\frac{{Q_3}}{{a_3} x+{d_3}}\tag 5$$ Multiplying $(1)$ by $(a_2cy+a_3z)$ and replacing $y$ and $z$ by their expressions, we end with $$Ax^2+Bx+C=0$$ where $$A=-{a_2}{a_3} {d_1} (c {Q_2}+{Q_3})$$ $$B={a_2} {a_3} (c {Q_1} {Q_2}+ {Q_1} {Q_3}- {Q_2} {Q_3})-d_1({a_2} c {d_3} {Q_2}-{a_3} {d_2} {Q_3})$$ $$C=-{Q_1} ({a_2} c {d_3} {Q_2}+{a_3} {d_2} {Q_3})$$