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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?

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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.

Find the solution set of the system of non-linear equations: y = x^2 - 2x + 1 and x + y = 3

Learn how to solve the system of non-linear equations y = x^2 - 2x + 1 and x + y = 3. This step-by-step solution explains the use of the substitution method to find the solution set. Suitable for high school students (Grades 10-12).

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