To solve the system of equations:
[tex]\[
\begin{cases}
x = x^2 + y^2 \\
y = 2xy
\end{cases}
\][/tex]
we will analyze and solve each equation step by step.
### Step 1: Analyze the second equation, [tex]\( y = 2xy \)[/tex]
Rewrite the equation:
[tex]\[ y - 2xy = 0 \][/tex]
Factor out [tex]\( y \)[/tex]:
[tex]\[ y(1 - 2x) = 0 \][/tex]
This gives us two possible solutions:
1. [tex]\( y = 0 \)[/tex]
2. [tex]\( 1 - 2x = 0 \)[/tex]
### Step 2: Solve for [tex]\( x \)[/tex] when [tex]\( y = 0 \)[/tex]
Substitute [tex]\( y = 0 \)[/tex] into the first equation:
[tex]\[ x = x^2 + 0^2 \][/tex]
[tex]\[ x = x^2 \][/tex]
Rearrange the equation:
[tex]\[ x^2 - x = 0 \][/tex]
Factor out [tex]\( x \)[/tex]:
[tex]\[ x(x - 1) = 0 \][/tex]
This gives us two solutions:
[tex]\[ x = 0 \quad \text{or} \quad x = 1 \][/tex]
So, the solutions in this case are:
[tex]\[ (x, y) = (0, 0) \quad \text{or} \quad (1, 0) \][/tex]
### Step 3: Solve for [tex]\( x \)[/tex] when [tex]\( 1 - 2x = 0 \)[/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[ 1 - 2x = 0 \][/tex]
[tex]\[ 2x = 1 \][/tex]
[tex]\[ x = \frac{1}{2} \][/tex]
Substitute [tex]\( x = \frac{1}{2} \)[/tex] into the first equation:
[tex]\[ x = x^2 + y^2 \][/tex]
[tex]\[ \frac{1}{2} = \left(\frac{1}{2}\right)^2 + y^2 \][/tex]
[tex]\[ \frac{1}{2} = \frac{1}{4} + y^2 \][/tex]
Subtract [tex]\( \frac{1}{4} \)[/tex] from both sides:
[tex]\[ \frac{1}{2} - \frac{1}{4} = y^2 \][/tex]
[tex]\[ \frac{1}{4} = y^2 \][/tex]
Solve for [tex]\( y \)[/tex]:
[tex]\[ y = \pm \frac{1}{2} \][/tex]
So, the solutions in this case are:
[tex]\[ \left( \frac{1}{2}, \frac{1}{2} \right) \quad \text{and} \quad \left( \frac{1}{2}, -\frac{1}{2} \right) \][/tex]
### Conclusion
Combining all the solutions, we have four pairs [tex]\((x, y)\)[/tex] that satisfy the system of equations:
[tex]\[
\left\{ \left( 0, 0 \right), \left( 1, 0 \right), \left( \frac{1}{2}, \frac{1}{2} \right), \left( \frac{1}{2}, -\frac{1}{2} \right) \right\}
\][/tex]
