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Let’s solve these problems step-by-step by substitution. I will solve each problem individually. Let’s start:

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### Problem 8
\[
\begin{cases}
x + y = 5 \\
y + 1 = 3x^2 + 2x
\end{cases}
\]

1. From the first equation: \( y = 5 - x \).
2. Substitute \( y = 5 - x \) into the second equation:
   \[
   (5 - x) + 1 = 3x^2 + 2x
   \]
   Simplify:
   \[
   6 - x = 3x^2 + 2x
   \]
   Rearrange:
   \[
   3x^2 + 3x - 6 = 0
   \]
   Simplify further:
   \[
   x^2 + x - 2 = 0
   \]
   Factorize:
   \[
   (x + 2)(x - 1) = 0
   \]
   So, \( x = -2 \) or \( x = 1 \).

3. For \( x = -2 \): Substitute into \( y = 5 - x \):
   \[
   y = 5 - (-2) = 7
   \]
   Solution: \( (-2, 7) \).

4. For \( x = 1 \): Substitute into \( y = 5 - x \):
   \[
   y = 5 - 1 = 4
   \]
   Solution: \( (1, 4) \).

**Solutions for Problem 8**: \( (-2, 7) \) and \( (1, 4) \).

---

### Problem 9
\[
\begin{cases}
x^2 + y - 8 = 0 \\
x + y - 2 = 0
\end{cases}
\]

1. From the second equation: \( y = 2 - x \).
2. Substitute \( y = 2 - x \) into the first equation:
   \[
   x^2 + (2 - x) - 8 = 0
   \]
   Simplify:
   \[
   x^2 - x - 6 = 0
   \]
   Factorize:
   \[
   (x - 3)(x + 2) = 0
   \]
   So, \( x = 3 \) or \( x = -2 \).

3. For \( x = 3 \): Substitute into \( y = 2 - x \):
   \[
   y = 2 - 3 = -1
   \]
   Solution: \( (3, -1) \).

4. For \( x = -2 \): Substitute into \( y = 2 - x \):
   \[
   y = 2 - (-2) = 4
   \]
   Solution: \( (-2, 4) \).

**Solutions for Problem 9**: \( (3, -1) \) and \( (-2, 4) \).

---

### Problem 10
\[
\begin{cases}
5x + y = 2x^2 + 6 \\
y + 4x = 7x - 2
\end{cases}
\]

1. Simplify the second equation:
   \[
   y + 4x = 7x - 2 \implies y = 3x - 2
   \]

2. Substitute \( y = 3x - 2 \) into the first equation:
   \[
   5x + (3x - 2) = 2x^2 + 6
   \]
   Simplify:
   \[
   8x - 2 = 2x^2 + 6
   \]
   Rearrange:
   \[
   2x^2 - 8x + 8 = 0
   \]
   Simplify further:
   \[
   x^2 - 4x + 4 = 0
   \]
   Factorize:
   \[
   (x - 2)^2 = 0
   \]
   So, \( x = 2 \).

3. For \( x = 2 \): Substitute into \( y = 3x - 2 \):
   \[
   y = 3(2) - 2 = 6 - 2 = 4
   \]
   Solution: \( (2, 4) \).

**Solution for Problem 10**: \( (2, 4) \).

---

Do you want me to provide detailed work for any specific problem?  

Here are related questions:  

1. How can substitution be used in solving nonlinear systems?  
2. What are the steps to factor quadratic equations systematically?  
3. Can a nonlinear system have infinitely many solutions?  
4. How does substitution compare to elimination for solving systems?  
5. What happens if a system of equations has no solutions?  

**Tip**: Always simplify one equation first to express a variable in terms of another before substituting into the second.

Solve the given linear and quadratic systems by substitution and find the solutions.

Explore step-by-step solutions for solving linear and quadratic systems of equations using substitution. Problems involve simplifying, substituting, factoring, and identifying solutions. Suitable for Grades 9-11 students studying algebra.

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