Solving the equation system $\{x+z=-6,y+zx=-1,yz=30\}$? I am interested in solving the following system of non-linear equations.
I am looking for $$x,y,z \in \mathbb{C}$$  with
$$x+z = -6$$
$$y+zx=-1$$
$$yz=30$$
In fact I am actually interested in just establishing a solution exists even without finding it. Also the constants on the right side are somewhat arbitrary so I would like an argument that can apply for any numbers there.  I also realize I could do a direct substitution which could reduce this to solving a single cubic equation but that is not ideal because it would then rest on me being able to solve an arbitrary cubic if I let the constants be arbitrary.
Is there some sort of concept, analogous to determinants for linear systems, that would help here?
 A: Let us consider the following system of nonlinear equations:
$$x+z=\alpha,\ y+zx=\beta,\ yz=\gamma.$$
If $\gamma=0$, then either $z=0$ or $y=0$. In the first case, we then have that $x=\alpha$ and $y=\beta$. In the second case, we have the expressions $x+z=\alpha$ and $zx=\beta$ which means that $x$ and $z$ are the two roots of the quadratic polynomial $p(s)=s^2-\alpha s+\beta$.
Now consider the case that $\gamma\ne0$. In this case, we have that $z\ne0$ as well. If we multiply by $z$ the two first expressions we get that
$$z(x+z)=zx+z^2=z\alpha,\quad z(y+zx)=zy+z^2=\beta z.$$
Combining those expressions together with $yz=\gamma$ yields the following cubic polynomial equation $q(z)=0$ where
$$q(z)=-z^3+\alpha z^2-\beta z+\gamma.$$
This equation has three solutions in $\mathbb{C}$ which we denote by $z_1,z_2,z_3$. So, we have three solutions $(x,y,z)=(\alpha-z_i,\gamma/z_i,z_i)$, $i=1,2,3$.
To summarize:

*

*If $\gamma=0$, the solutions are $(\alpha,\beta,0)$, $(s_1,0,s_2)$, and $(s_2,0,s_1)$ where $s_1$ and $s_2$ are the two roots of the polynomial $p(s)$.

*If $\gamma\ne0$, the solutions are $(\alpha-z_i,\gamma/z_i,z_i)$, $i=1,2,3$, where $z_i$ are the roots of $q(z)$.

A: Let $a, b , c \in {\Bbb C}$ with $c \not= 0$. The system
\begin{align}
x+z &= a \\
y + zx &= b \\
yz &= -c
\end{align}
is equivalent to $x = -z - a$, $y = -c/z$ and
$zy + z^2x - bz = 0$ and this latter equation can be rewritten as $z^2(-z-a)-bz -c = 0$, that is $z^3 + az^2 + bz + c = 0$.
In other words, a suitable choice of the right hand side of your equations yields a generic cubic equation, so your problem is essentially equivalent to solving a cubic equation. If you just want to prove the existence of a solution, then of course every polynomial equation in $\Bbb C$ has a solution.
