3 nonlinear equations in 3 variables - with the same variable raised to different powers Hoping someone can give me some guidance on solving the following system of 3 nonlinear equations in 3 variables:
$$\left\{\begin{align*}
&x^2+y^2=100\\
&xy+yz=-102\\
&y^2+z^2=117
\end{align*}\right.$$
Thanks!

Observe that $y ≠0$ for finite $x,z$
$z=0\implies y^2=117>100\implies x^2=-17\implies x^2z^2=117(-17)≠(-102)^2$ 
$x=0\implies y^2=100\implies z^2=17 \implies  y^2z^2=1700≠(-102)^2$
So $xyz≠0$ 
$(3)-(1)\implies z^2-x^2=17\implies z+x=\frac{17}{z-x}$
$(2)\implies y(z+x)=-102\implies y=-\frac{102}{\frac{17}{z-x}}=6(x-z)$
Putting the value of $y$ in $(1),(3)$.
$x^2+36(x-z)^2=100\implies 37x^2-72zx+z^2=100-->(4)$
and $z^2+36(x-z)^2=117\implies 37z^2-72zx+x^2=117-->(5)$
On division of $(4)$ by $(5)$, $\frac{37x^2-72zx+z^2}{37z^2-72zx+x^2}=\frac{100}{117}$
$\implies x^2(37\cdot 117 -100)-72zx(117-100)+z^2(117-37\cdot 100)=0$
$\implies 4229x^2-1224zx-3583z^2=0$
If $x=a\cdot z, 4229a^2-1224a-3583=0-->(6)$ as $xyz≠0$ 
So, $y=6(x-z)=6(az-z)=6z(a-1)$
So, from (2), $az 6z(a-1)+6z(a-1)z=-102$
or $z^2=\frac{17}{1-a^2}$  where $a$ is a root of $(6)$.
