To solve this system of equations:

1. $y = x^2 - 2x + 1$
2. $x + y = 3$

Step 1: Substitute $y$ from the second equation into the first.

From equation (2), solve for $y$: $y = 3 - x$

Substitute this into the first equation: $3 - x = x^2 - 2x + 1$

Step 2: Simplify the resulting quadratic equation.

Move all terms to one side: $x^2 - 2x + 1 + x - 3 = 0$ $x^2 - x - 2 = 0$

Step 3: Factor the quadratic equation.

The equation $x^2 - x - 2 = 0$ can be factored as: $(x - 2)(x + 1) = 0$

Step 4: Solve for $x$.

Set each factor equal to zero:

1. $x - 2 = 0 \implies x = 2$
2. $x + 1 = 0 \implies x = -1$

Step 5: Find the corresponding $y$-values.

For $x = 2$: $y = 3 - 2 = 1$

For $x = -1$: $y = 3 - (-1) = 4$

Final Solution:

The solution set for the system is: $(2, 1) \quad \text{and} \quad (-1, 4)$

Would you like more details, or do you have any questions?

Here are 5 related questions to expand your understanding:

1. What methods can be used to solve non-linear systems in general?
2. How do you determine when to use substitution versus elimination in solving systems of equations?
3. Can all non-linear systems be solved algebraically, or are numerical methods sometimes required?
4. How does graphing the system help visualize the solution of non-linear equations?
5. What are some practical applications of solving systems of non-linear equations?

Tip: Always check the solutions by substituting them back into both original equations to ensure accuracy.