Let's determine all solutions of the system of equations $x^{2}+y^{2}=x, 2 x y=y$.

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Find all solutions of the system of equations $x^{2}+y^{2}=x, 2 x y=y$.
Combine the two equations:

$$
x^{2}+y^{2}=x, \quad (1) \quad 2xy=y, \quad (2)
$$

and we get

$$
(x+y)^{2}=x+y
$$

If $x+y \neq 0$, we can simplify by dividing by $x+y$, leading to the equation $x+y=1$. From this, substituting $x=1-y$ we get

$$
2y(1-y)=y
$$

If $y \neq 0$, simplifying by dividing by $y$ gives the equation $2(1-y)=1$. Solving this, we find $y=\frac{1}{2}$ and from (2) $x=\frac{1}{2}$.

If $y=0$, then from (1) $x=1$ (since we assumed $x+y \neq 0$).

Now, let's return to the case where $x+y=0$, or $x=-y$. From (2), we get $-2y^2=y$ (for $y \neq 0$), and thus $y=-\frac{1}{2}$ and $x=\frac{1}{2}$, or $y=0=x$.

The system of equations has 4 pairs of solutions:

$$
x_{1}=\frac{1}{2}, y_{1}=\frac{1}{2} ; \quad x_{2}=0, y_{2}=0 ; \quad x_{3}=\frac{1}{2}, y_{3}=-\frac{1}{2} ; \quad x_{4}=1, y_{4}=0
$$

By substitution, it is easy to verify that all 4 pairs are indeed solutions to the system of equations.

Remark. The system of equations can be visualized geometrically, showing that we get four solutions, and the solutions themselves are almost immediately apparent: (1) can be rewritten as $\left(x-\frac{1}{2}\right)^{2}+y^{2}=\left(\frac{1}{2}\right)^{2}$, which represents a circle with center $C\left(\frac{1}{2} ; 0\right)$ and radius $\frac{1}{2}$. (2) can be factored as $2y\left(x-\frac{1}{2}\right)=0$, which is the equation of a pair of intersecting lines. (The intersection point is the center of the circle.)

One pair of solutions is obtained as the intersection points of the x-axis and the circle: $(0,0)$ and $(1,0)$; the other pair is obtained as the intersection points of the line $x=\frac{1}{2}$ and the circle: $\left(\frac{1}{2},-\frac{1}{2}\right)$ and $\left(\frac{1}{2} ; \frac{1}{2}\right)$.

![](https://cdn.mathpix.com/cropped/2024_05_02_442fda23af3e66eeafbcg-1.jpg?height=423&width=449&top_left_y=1434&top_left_x=838)