To solve the system of equations
[tex]\[
\left\{
\begin{array}{l}
x = x^2 + y^2 \\
y = 2xy
\end{array}
\right.
\][/tex]
we'll solve it step by step:
1. Step 1: Analyze the second equation.
[tex]\[
y = 2xy
\][/tex]
Rearrange this equation to:
[tex]\[
y - 2xy = 0
\][/tex]
Factor out [tex]\( y \)[/tex]:
[tex]\[
y(1 - 2x) = 0
\][/tex]
This gives us two potential solutions:
[tex]\[
y = 0 \quad \text{or} \quad 1 - 2x = 0 \implies x = \frac{1}{2}
\][/tex]
2. Step 2: Solve for [tex]\( y \)[/tex] when [tex]\( y = 0 \)[/tex].
Substituting [tex]\( y = 0 \)[/tex] into the first equation:
[tex]\[
x = x^2 + 0^2 \implies x = x^2
\][/tex]
Rearrange to:
[tex]\[
x^2 - x = 0 \implies x(x - 1) = 0
\][/tex]
This gives us:
[tex]\[
x = 0 \quad \text{or} \quad x = 1
\][/tex]
So, we have two solutions from this scenario:
[tex]\[
(x, y) = (0, 0) \quad \text{and} \quad (x, y) = (1, 0)
\][/tex]
3. Step 3: Solve for [tex]\( y \)[/tex] when [tex]\( x = \frac{1}{2} \)[/tex].
Substitute [tex]\( x = \frac{1}{2} \)[/tex] into the first equation:
[tex]\[
\frac{1}{2} = \left(\frac{1}{2}\right)^2 + y^2
\][/tex]
Simplify the equation:
[tex]\[
\frac{1}{2} = \frac{1}{4} + y^2
\][/tex]
Subtract [tex]\(\frac{1}{4}\)[/tex] from both sides:
[tex]\[
\frac{1}{2} - \frac{1}{4} = y^2 \implies \frac{1}{4} = y^2
\][/tex]
Solving for [tex]\( y \)[/tex]:
[tex]\[
y = \pm \frac{1}{2}
\][/tex]
This gives us:
[tex]\[
(x, y) = \left(\frac{1}{2}, \frac{1}{2}\right) \quad \text{and} \quad (x, y) = \left(\frac{1}{2}, -\frac{1}{2}\right)
\][/tex]
Combining all these solutions, we get the final set of solutions for the given system of equations:
[tex]\[
\boxed{(0, 0), \left(\frac{1}{2}, -\frac{1}{2}\right), \left(\frac{1}{2}, \frac{1}{2}\right), (1, 0)}
\][/tex]
