Finding equilibrium points of system of equations I'm asked to solve system of ODE for finding equilibrium points of this
$$\frac{dx}{dt} = a_1x(1-x) + a_2x\frac{z}{1+z}+a_3xy$$
$$\frac{dy}{dt} = b_1y(1-y) + b_2y\frac{z}{1+z}+b_3xy$$
$$\frac{dz}{dt} = c_1z(1-z) + c_2z\frac{x}{1+x}+c_3z\frac{y}{1+y}$$
I checked some related links to solve this problem and i know that i need to extract variables from this system to matrix form: link:https://www.math24.net/linear-autonomous-systems-equilibrium-points/
but problem how to extract them
My question is how correctly do change of variable for further extraction to matrix form
 A: Hint
An equilibrium point is a triple $\left(x , y , z\right)$ satisfying the non-linear system
\begin{equation}
\renewcommand{\arraystretch}{1.5}  \left\{\begin{array}{rcl}{a}_{1} x \left(1-x\right)+{a}_{2} x \displaystyle  \frac{z}{1+z}+{a}_{3} x y&=&0\\
{b}_{1} y \left(1-y\right)+{b}_{2} y \displaystyle  \frac{z}{1+z}+{b}_{3} x y&=&0\\
{c}_{1} z \left(1-z\right)+{c}_{2} z \displaystyle  \frac{x}{1+x}+{c}_{3} z \frac{y}{1+y}&=&0
\end{array}\right.
\end{equation}
The most interesting case is when $x \neq  0 , y \neq  0 , z \neq  0$. Then the system reduces to
\begin{equation}
\renewcommand{\arraystretch}{1.5}  \left\{\begin{array}{rcl}{a}_{1} \left(1-x\right)+{a}_{2} \displaystyle  \frac{z}{1+z}+{a}_{3} y&=&0\\
{b}_{1} \left(1-y\right)+{b}_{2} \displaystyle  \frac{z}{1+z}+{b}_{3} x&=&0\\
{c}_{1} \left(1-z\right)+{c}_{2} \displaystyle  \frac{x}{1+x}+{c}_{3} \frac{y}{1+y}&=&0
\end{array}\right.
\end{equation}
From the two first equations, one deduces in general a relation of the form
$x = {\alpha} y+{\beta}$. Substituting again in these two first
equations leads to
\begin{equation}
x = \frac{P \left(z\right)}{1+z} \qquad  y = \frac{Q \left(z\right)}{1+z}
\end{equation}
where $P$ and $Q$ are polynomials of degree $1$ which coefficients
depend on ${a}_{1} , {a}_{2}, a_3 , {b}_{1} , {b}_{2}, b_3$.
Injecting these values of $x$ and $y$ in the third equation leads to
\begin{equation}
{c}_{1} \left(1-z\right)+{c}_{2} \displaystyle  \frac{P \left(z\right)}{1+z+P \left(z\right)}+{c}_{3} \frac{Q \left(z\right)}{1+z+Q \left(z\right)} = 0
\end{equation}
This gives a cubic equation in $z$ with potentially three complex
roots, producing three equilibrium points.
With the help of a computer algebra system all the equilibrium points of this system can be computed.
