The image shows a system of two equations:

$1. \quad x + y = 2$ $2. \quad y = x^2 - 4x + 4$

To solve this system, we can use substitution. Since $y$ is already expressed in terms of $x$ in the second equation, we can substitute this expression for $y$ into the first equation:

Step-by-step solution:

1. From the first equation: $x + y = 2$ Substitute $y = x^2 - 4x + 4$: $x + (x^2 - 4x + 4) = 2$

2. Simplify the equation: $x^2 - 3x + 4 = 2$

3. Move all terms to one side: $x^2 - 3x + 4 - 2 = 0$ $x^2 - 3x + 2 = 0$

4. Solve this quadratic equation using the quadratic formula: $x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(2)}}{2(1)}$ $x = \frac{3 \pm \sqrt{9 - 8}}{2}$ $x = \frac{3 \pm \sqrt{1}}{2}$ $x = \frac{3 \pm 1}{2}$

5. This gives two solutions for $x$: $x = \frac{3 + 1}{2} = 2 \quad \text{or} \quad x = \frac{3 - 1}{2} = 1$

6. Now substitute these values of $x$ back into the equation $y = x^2 - 4x + 4$ to find $y$.

   • When $x = 2$: $y = 2^2 - 4(2) + 4 = 4 - 8 + 4 = 0$

   • When $x = 1$: $y = 1^2 - 4(1) + 4 = 1 - 4 + 4 = 1$

Final solution:

• $(x, y) = (2, 0)$
• $(x, y) = (1, 1)$

Would you like more details on this solution, or have any questions?

Here are 5 related questions you could explore:

1. How do we solve quadratic equations using the quadratic formula?
2. Can we solve the system graphically and verify the solutions?
3. How would the solution change if the second equation were different?
4. What happens if the quadratic discriminant is negative?
5. Can we use another method, like elimination, to solve this system?

Tip: Always simplify your equations as much as possible before solving. It reduces errors and makes solving faster.