Solve the following system of 3 equations

Question

I need help solving this system of equations in order to prepare for my math exam. I've tried all the ways I know but I've only managed to find false solutions...

\begin{cases} y+2xz=0\\ x+4yz=0\\ x^2+2y^2-2=0 \end{cases}

Here are the correct solutions

Answer 1

The first equation gives $y^2-2xyz=0.$ The second equation gives $x^2+4xyz.$ We then derive $x^2=2y^2.$

From $x^2+2y^2-2=0$ we get $y^2=2$, hence $y= \pm \frac{1}{\sqrt{2}}.$

Can you proceed ?

Answer 2

From first and second equation: $$ 0=x+4yz=x+4(-2xz)z=x(1-8z^2) $$ Then either $x=0$ (which implies $y=0$ and contradicts the last equation) or $8z^2=1$. Then it is sufficient to replace $y=-2xz$ into the last one, to obtain $$ 0=x^2+2y^2-2=x^2+2(-2xz)^2-2=x^2(1+8z^2)-2=2(x^2-1) $$ Substitute $x=\pm 1$ and get the wanted solutions.