Answer :

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]