Discovering "hidden solution" in system of equations

Question

For the system of equations

$x(1-4x-y)=0$

$y(1-2y-5x)=0$,

One of the solutions is $x=1/3, y=-1/3$, but I don't see how to find it. Some help would be appreciated. I'm obviously having a gap here.

Answer 1

$$ \begin{cases} x(1-4x-y)=0 \\ y(1-2y-5x)=0 \end{cases}\Longleftrightarrow $$ $$ \begin{cases} x-4x^2-xy=0 \\ y-2y^2-5xy=0 \end{cases}\Longleftrightarrow $$ $$ \begin{cases} x=0\space\space\vee\space\space x=\frac{1-y}{4} \\ y-2y^2-5xy=0 \end{cases}\Longleftrightarrow $$ $$ \begin{cases} x=0\space\space\vee\space\space x=\frac{1-y}{4} \\ y-2y^2-5\cdot 0\cdot y=0\space\space\vee\space\space y-2y^2-5\cdot \frac{1-y}{4}\cdot y=0 \end{cases}\Longleftrightarrow $$ $$ \begin{cases} x=0\space\space\vee\space\space x=\frac{1-y}{4} \\ y-2y^2=0\space\space\vee\space\space \left(-\frac{3y}{4}-\frac{1}{4}\right)y=0 \end{cases}\Longleftrightarrow $$ $$ \begin{cases} x=0\space\space\vee\space\space x=\frac{1-y}{4} \\ y=0\vee y=\frac{1}{2}\space\space\vee\space\space y=-\frac{1}{3}\vee y=0 \end{cases} $$

So your solutions are:

$$x=0,y=\frac{1}{2}$$ $$x=\frac{1}{4},y=0$$ $$x=\frac{1}{3},y=-\frac{1}{3}$$ $$x=0,y=0$$