You have a system of two equations:

1. \( x + y = 2 \)
2. \( y = x^2 - 4x + 4 \)

To solve this system, we can substitute the expression for \( y \) from the second equation into the first equation.

### Step 1: Substitute \( y = x^2 - 4x + 4 \) into \( x + y = 2 \)

\[
x + (x^2 - 4x + 4) = 2
\]

### Step 2: Simplify the equation

\[
x + x^2 - 4x + 4 = 2
\]

\[
x^2 - 3x + 4 = 2
\]

### Step 3: Move all terms to one side

\[
x^2 - 3x + 4 - 2 = 0
\]

\[
x^2 - 3x + 2 = 0
\]

### Step 4: Factor the quadratic equation

\[
(x - 1)(x - 2) = 0
\]

### Step 5: Solve for \( x \)

Set each factor equal to zero:

\[
x - 1 = 0 \quad \text{or} \quad x - 2 = 0
\]

\[
x = 1 \quad \text{or} \quad x = 2
\]

### Step 6: Solve for \( y \)

Now, substitute each value of \( x \) back into the second equation \( y = x^2 - 4x + 4 \).

- When \( x = 1 \):

\[
y = 1^2 - 4(1) + 4 = 1 - 4 + 4 = 1
\]

- When \( x = 2 \):

\[
y = 2^2 - 4(2) + 4 = 4 - 8 + 4 = 0
\]

### Final Solution:

The solutions are:
1. \( x = 1 \), \( y = 1 \)
2. \( x = 2 \), \( y = 0 \)

Let me know if you'd like a more detailed explanation or have any further questions!

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### Further Exploration:

1. What are the graphical representations of these equations?
2. How does changing the constants in the equations affect the solution?
3. Can you solve a system where one equation is linear and the other is quadratic in a different way?
4. What happens to the solution when we add another constraint to the system?
5. Can this system be solved using matrices?

### Tip: 
When solving systems of equations, substitution and elimination are powerful methods, especially when one equation is already solved for a variable.

Learn how to solve the system of equations x + y = 2 and y = x^2 - 4x + 4 with detailed step-by-step solutions. The problem combines linear and quadratic equations and is suitable for students in Grades 9-11. It demonstrates the substitution method, factoring quadratic equations, and solving for two variables.

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