How to solve this nonlinear equation? This is the system of equations:
$$xy + z = -30\\
yz + x = 30\\
zx + y = -18$$ 
I have also done some work: if we add the first two equations we'll get:
$$xy + z + yz + x = 0\\
\implies y (x+z) + z + x = 0\\
\implies (x+z)(y+1) = 0$$
Now, a product equals zero if at least one of the factors equals zero.
Thus we have:
$$y+1=0 \implies y=-1$$
and
$$x+z = 0 \implies x = -z \ \textrm{ or }\ z = -x$$
Now if we substitute this into the equations we get:
$$xy + z = -30 \implies x(-1)+(-x) = -30 \implies x=15$$
We have:
$$x = 15 \ \textrm{ and }\ y = 1$$
But then the third equation makes no sense.
$$zx + y = -18$$
How can I solve this and is there a strategy that does not use numerical algorithms? 
Thanks!

As you observe, from the first two equations, $y=-1$ or $z=-x$.  If $y=-1$, then we have two equations to solve for $x$ and $z$, namely, $x-z=30$ and $xz=-17$.   Therefore, $x$ and $-z$ are the roots of the quadratic polynomial $t^2-30t+17$, whereby it follows that $$(x,y,z)=(15-4\sqrt{13},-1,-15-4\sqrt{13})$$
and
$$(x,y,z)=(15+4\sqrt{13},-1,-15+4\sqrt{13})$$
are solutions.  
If $z=-x$, then we have two equations to solve for $x$ and $y$, i.e., $x(y-1)=-30$ and $x^2-y=18$.  Thus,
$$x^2+\frac{30}{x}=x^2-(y-1)=(x^2-y)+1=19\,.$$
That is, $$(x+5)(x-2)(x-3)=x^3-19x+30=0\,.$$
Thus,
$$(x,y,z)=(-5,7,5)\,,$$
$$(x,y,z)=(2,-14,-2)\,,$$
and $$(x,y,z)=(3,-9,-3)$$
are the solutions in this case.
