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.