Solving the system $x^2+2y^2-y-2z=0$, $x^2-8y^2+10z-1=0$, $x^2-7yz=0$ How to solve this system using elimination:
$$\begin{align}
x^2+2y^2-y-2z&=0 \\
x^2-8y^2+10z-1&=0 \\
x^2-7yz&=0
\end{align}$$
I feel like it should be easy but I have tried many different ways and cannot seem to solve for any of the variables.
If you take equation 3 and subtract equation 2 you end up getting
$8y^2-7yz-10z+1=0$ and using the quadratic I could solve for y with respect to z but it didn't get me any closer to solving the system.
 A: I actually think eliminating $x^2$ is the easiest first step, since $x$ only appears as $x^2$ and is not multiplied by any other variable.  So from the third equation, $x^2 = 7yz$, hence the first two equations become
$$\begin{align}
2y^2 - y + 7yz - 2z &= 0, \\
-8y^2 + 7yz + 10z - 1 &= 0. \end{align}$$  Then the next step is to observe that these equations are linear in $z$ but quadratic in $y$, so we collect like terms in $z$ and get
$$\begin{align}
(7y-2)z &= y - 2y^2, \\
(7y+10)z &= 8y^2 + 1. \end{align}$$
Therefore, $$(7y+10)(y - 2y^2) = (7y-2)(8y^2+1).$$  Now we have a cubic in $y$ that we can solve:
$$70y^3 - 3y^2 - 3y - 2 = 0.$$  Unfortunately, this cubic is not "nice," so the roots are very complicated to write out.  The unique real root is
$$y = \frac{1}{210} \left(3+\sqrt[3]{135162-945 \sqrt{20165}}+3 \sqrt[3]{5006+35
   \sqrt{20165}}\right) \approx 0.368933448565128\ldots,$$ which can then be substituted back into the earlier equations to get
$$z \approx 0.166015453844704, \\ x \approx \pm 0.654782847449611.$$  Substitution into the system verifies this result.  There are other solutions but they are complex-valued.
A visualization of the three equations in $\mathbb R^3$ is shown below.  The solution set is depicted as purple spheres (they're a bit hard to see but you can follow the surfaces and locate where all three mutually intersect).


