To solve the system of equations:
[tex]\[
\begin{cases}
x = x^2 + y^2 \\
y = 2xy
\end{cases}
\][/tex]
we will find the values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] that satisfy both equations simultaneously. Let's go through the process step by step.
### Step 1: Analyze the Second Equation
First, consider the second equation:
[tex]\[
y = 2xy
\][/tex]
We can rewrite this equation as:
[tex]\[
y - 2xy = 0 \quad \Rightarrow \quad y(1 - 2x) = 0
\][/tex]
This gives us two possible scenarios:
1. [tex]\( y = 0 \)[/tex]
2. [tex]\( 1 - 2x = 0 \)[/tex] or [tex]\( x = \frac{1}{2} \)[/tex]
### Step 2: Solve for [tex]\( y = 0 \)[/tex]
When [tex]\( y = 0 \)[/tex], substitute this into the first equation:
[tex]\[
x = x^2 + 0^2 \quad \Rightarrow \quad x = x^2
\][/tex]
This can be rewritten as:
[tex]\[
x^2 - x = 0 \quad \Rightarrow \quad x(x - 1) = 0
\][/tex]
So, we get two solutions for [tex]\( x \)[/tex]:
[tex]\[
x = 0 \quad \text{or} \quad x = 1
\][/tex]
Thus, two solutions from this scenario are:
1. [tex]\( (x, y) = (0, 0) \)[/tex]
2. [tex]\( (x, y) = (1, 0) \)[/tex]
### Step 3: Solve for [tex]\( x = \frac{1}{2} \)[/tex]
When [tex]\( x = \frac{1}{2} \)[/tex], substitute this into the first equation:
[tex]\[
\frac{1}{2} = \left( \frac{1}{2} \right)^2 + y^2 \quad \Rightarrow \quad \frac{1}{2} = \frac{1}{4} + y^2
\][/tex]
Subtract [tex]\(\frac{1}{4}\)[/tex] from both sides:
[tex]\[
\frac{1}{2} - \frac{1}{4} = y^2 \quad \Rightarrow \quad \frac{1}{4} = y^2
\][/tex]
This gives us two solutions for [tex]\( y \)[/tex]:
[tex]\[
y = \frac{1}{2} \quad \text{or} \quad y = -\frac{1}{2}
\][/tex]
Thus, two additional solutions from this scenario are:
1. [tex]\( (x, y) = \left( \frac{1}{2}, \frac{1}{2} \right) \)[/tex]
2. [tex]\( (x, y) = \left( \frac{1}{2}, -\frac{1}{2} \right) \)[/tex]
### Summary of Solutions
Combining all the solutions, we have the following four solutions that satisfy the system of equations:
[tex]\[
(x, y) = (0, 0), \quad (1, 0), \quad \left( \frac{1}{2}, \frac{1}{2} \right), \quad \left( \frac{1}{2}, -\frac{1}{2} \right)
\][/tex]
