Solve for the roots in the following equations I have two equalities:
$$ \alpha x^{2} + \alpha y^{2} - y = 0 $$
$$ \beta  x^{2} + \beta y^{2} - x = 0 $$

Where $$ \alpha, \beta $$ are both known constants.

How can I solve for $x$ and $y$?
The two curves are circles through the origin and they will intersect in at most one other point.

By eliminating $x^2+y^2$, we have

$$\beta y=\alpha x,$$

then

$$\beta(\beta x^2+\beta y^2-x)=(\beta^2 + \alpha^2)x^2 -\beta x = 0. $$

The rest is immediate.


  $$\left(\frac\beta{\beta^2+\alpha^2},\frac\alpha{\beta^2+\alpha^2}\right).$$


We also have the special case $\alpha=\beta=0$, which is trivial.
We have


  $$ \alpha x^{2}+\alpha y^{2}-y=0 \longrightarrow(1) $$ and $\beta
 x^{2}+\beta y^{2}-x=0 \quad \longrightarrow \quad(2)$


Both equations can be written as:


  $\alpha\left(x^{2}+y^{2}\right)=y$ and $\quad \beta\left(x^{2}+y^{2}\right)=x$


Dividing first equation with later one, we get:


  $$
\begin{aligned}
&amp; \frac{\alpha}{\beta}=\frac{y}{x} \\
\Rightarrow &amp; \frac{\alpha x}{\beta}=y
\end{aligned}
$$


Putting the value of $y$ in $(1),$ we get:


  \begin{aligned}
&amp; \alpha x^{2}+\frac{\alpha^{3} x^{2}}{\beta^{2}}-\frac{\alpha x}{\beta}=0 \\
\Rightarrow &amp; \alpha x\left(x+\frac{\alpha^{2} x}{\beta^{2}}-\frac{1}{\beta}\right)=0 \\
\Rightarrow &amp; x+\frac{\alpha^{2} x}{\beta^{2}}-\frac{1}{\beta}=0 \\
\Rightarrow &amp; x\left(1+\frac{\alpha 2}{\beta^{2}}\right)=\frac{1}{\beta} \\
\Rightarrow &amp; x\left(\frac{\beta^{2}+\alpha^{2}}{\beta^{2}}\right)=\frac{1}{\beta} \\
\Rightarrow &amp; x=\frac{\beta}{\beta^{2}+\alpha^{2}}
\end{aligned}


Similarly find the value of $y$, yourself. $^_^$

Note: In this solution it is clear that I have assumed$\alpha \neq0, \beta \neq0,  x \neq0 y, \neq 0$.. For if this was not the case, as Eric Towers states in his answer, solution will be trivial.