I have a system of equations
$$\begin{align} xy + 3zw = 0; \\ xz + 2yw = 0; \\ xw + yz = 0. \\ \end{align}$$
Plugging it into a CAS, I see that all the rational solutions to this system have exactly one of $x,y,z,w$ arbitrary and the other three are $0$ (for example, $x=1/2,y=0,z=0,w=0$). How can I prove this to be the case?
Also, can this result be generalized?
You have three equations and four unknowns $x,y,z,w$. So, consider that $w$ is a constant.
From the first equation, extract $z=-\frac{x y}{3 w}$; plug it in the second equation from which you find, beside $y=0$, $x=\pm \sqrt{6} w$; from the third, beside $x=0$, you find $y=\pm \sqrt{3} w$. So, as a total, the solutions are $$\{x= 0,y= 0,z= 0\}$$ $$\left\{x=-\sqrt{6} w,y= -\sqrt{3} w,z= -\sqrt{2} w\right\}$$ $$\left\{x= -\sqrt{6} w,y=\sqrt{3} w,z= \sqrt{2} w\right\}$$ $$\left\{x= \sqrt{6} w,y= -\sqrt{3} w,z= \sqrt{2} w\right\}$$ $$\left\{x= \sqrt{6} w,y= \sqrt{3} w,z= -\sqrt{2} w\right\}$$