Approximate solution to a system of polynomial equations

Question

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.

Answer 1

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.

Answer 2

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 $$A=-{a_2}{a_3} {d_1} (c {Q_2}+{Q_3})$$ $$B={a_2} {a_3} (c {Q_1} {Q_2}+ {Q_1} {Q_3}- {Q_2} {Q_3})-d_1({a_2} c {d_3} {Q_2}-{a_3} {d_2} {Q_3})$$ $$C=-{Q_1} ({a_2} c {d_3} {Q_2}+{a_3} {d_2} {Q_3})$$