To solve the system of equations, we start by analyzing and solving the given equations simultaneously. The system of equations is:
[tex]\[
\begin{cases}
x = x^2 + y^2 \\
y = 2xy
\end{cases}
\][/tex]
Step 1: Solve the second equation for [tex]\( y \)[/tex]:
Let's focus on the second equation:
[tex]\[ y = 2xy \][/tex]
If we rearrange this equation, we get:
[tex]\[ y - 2xy = 0 \][/tex]
Factor out [tex]\( y \)[/tex]:
[tex]\[ y(1 - 2x) = 0 \][/tex]
This gives us two possible solutions:
[tex]\[ y = 0 \quad \text{or} \quad 1 - 2x = 0 \Rightarrow x = \frac{1}{2} \][/tex]
Step 2: Substitute [tex]\( y = 0 \)[/tex] into the first equation:
If [tex]\( y = 0 \)[/tex]:
[tex]\[ x = x^2 + 0^2 \Rightarrow x = x^2 \][/tex]
This simplifies to:
[tex]\[ x^2 - x = 0 \][/tex]
Factor out [tex]\( x \)[/tex]:
[tex]\[ x(x - 1) = 0 \][/tex]
This gives two possible solutions for [tex]\( x \)[/tex]:
[tex]\[ x = 0 \quad \text{or} \quad x = 1 \][/tex]
So, the pairs [tex]\((x, y)\)[/tex] we have so far are [tex]\((0, 0)\)[/tex] and [tex]\((1, 0)\)[/tex].
Step 3: Substitute [tex]\( x = \frac{1}{2} \)[/tex] into the first equation:
If [tex]\( x = \frac{1}{2} \)[/tex]:
[tex]\[ x = x^2 + y^2 \Rightarrow \frac{1}{2} = \left(\frac{1}{2}\right)^2 + y^2 \Rightarrow \frac{1}{2} = \frac{1}{4} + y^2 \][/tex]
Subtract [tex]\(\frac{1}{4}\)[/tex] from both sides:
[tex]\[ \frac{1}{2} - \frac{1}{4} = y^2 \Rightarrow \frac{1}{4} = y^2 \][/tex]
Taking the square root of both sides gives:
[tex]\[ y = \pm \frac{1}{2} \][/tex]
So, the pairs [tex]\((x, y)\)[/tex] we have here are [tex]\(\left(\frac{1}{2}, -\frac{1}{2}\right)\)[/tex] and [tex]\(\left(\frac{1}{2}, \frac{1}{2}\right)\)[/tex].
Conclusion:
By solving both equations simultaneously, we have found the complete set of solutions for [tex]\( (x, y) \)[/tex]:
[tex]\[ (0, 0), \left(\frac{1}{2}, -\frac{1}{2}\right), \left(\frac{1}{2}, \frac{1}{2}\right), (1, 0) \][/tex]
These are the solutions to the given system of equations.
