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The image contains a list of problems labeled "Simplifying Rational Algebraic Expression." I'll work through each problem to provide the solutions. 

### 1. \(\frac{45}{10x - 10}\)
Factor out the greatest common factor (GCF) from the denominator:
\[
\frac{45}{10(x - 1)} = \frac{45 \div 5}{10 \div 5} = \frac{9}{2(x - 1)}
\]
Final answer: \(\frac{9}{2(x - 1)}\)

---

### 2. \(\frac{16p}{28p}\)
Simplify by canceling the common factors between numerator and denominator:
\[
\frac{16p}{28p} = \frac{16}{28} = \frac{4}{7}
\]
Final answer: \(\frac{4}{7}\)

---

### 3. \(\frac{28}{7 - 2}\)
First, simplify the denominator:
\[
7 - 2 = 5
\]
So:
\[
\frac{28}{5}
\]
Final answer: \(\frac{28}{5}\)

---

### 4. \(\frac{x - 4}{3x - 12x}\)
Simplify the denominator:
\[
3x - 12x = -9x
\]
So the expression becomes:
\[
\frac{x - 4}{-9x}
\]
Final answer: \(\frac{x - 4}{-9x}\)

---

### 5. \(\frac{x + 6}{x^2 + 5x - 6}\)
Factor the denominator:
\[
x^2 + 5x - 6 = (x + 6)(x - 1)
\]
So the expression becomes:
\[
\frac{x + 6}{(x + 6)(x - 1)}
\]
Cancel the common factor \(x + 6\):
\[
\frac{1}{x - 1}
\]
Final answer: \(\frac{1}{x - 1}\)

---

### 6. \(\frac{b^2 - 49}{b^2 + 3b - 28}\)
Factor both numerator and denominator:
\[
b^2 - 49 = (b - 7)(b + 7)
\]
\[
b^2 + 3b - 28 = (b - 7)(b + 4)
\]
So the expression becomes:
\[
\frac{(b - 7)(b + 7)}{(b - 7)(b + 4)}
\]
Cancel the common factor \(b - 7\):
\[
\frac{b + 7}{b + 4}
\]
Final answer: \(\frac{b + 7}{b + 4}\)

---

### 7. \(\frac{2x^2 - x^2 - 42x}{2x^2 - 20x + 42}\)
Simplify the numerator:
\[
2x^2 - x^2 - 42x = x^2 - 42x
\]
Factor both the numerator and denominator:
\[
x^2 - 42x = x(x - 42)
\]
\[
2x^2 - 20x + 42 = 2(x^2 - 10x + 21) = 2(x - 7)(x - 3)
\]
So the expression becomes:
\[
\frac{x(x - 42)}{2(x - 7)(x - 3)}
\]
Final answer: \(\frac{x(x - 42)}{2(x - 7)(x - 3)}\)

---

### 8. \(\frac{4}{60 - 20}\)
Simplify the denominator:
\[
60 - 20 = 40
\]
So the expression becomes:
\[
\frac{4}{40} = \frac{1}{10}
\]
Final answer: \(\frac{1}{10}\)

---

### 9. \(\frac{3r^2 - 39r + 90}{r^2 - 3r - 70}\)
Factor both the numerator and denominator:
\[
3r^2 - 39r + 90 = 3(r^2 - 13r + 30) = 3(r - 3)(r - 10)
\]
\[
r^2 - 3r - 70 = (r - 10)(r + 7)
\]
So the expression becomes:
\[
\frac{3(r - 3)(r - 10)}{(r - 10)(r + 7)}
\]
Cancel the common factor \(r - 10\):
\[
\frac{3(r - 3)}{r + 7}
\]
Final answer: \(\frac{3(r - 3)}{r + 7}\)

---

### 10. \(\frac{x^2 + 2x - 80}{2x^2 - 24x + 64}\)
Factor both the numerator and denominator:
\[
x^2 + 2x - 80 = (x - 8)(x + 10)
\]
\[
2x^2 - 24x + 64 = 2(x^2 - 12x + 32) = 2(x - 8)(x - 4)
\]
So the expression becomes:
\[
\frac{(x - 8)(x + 10)}{2(x - 8)(x - 4)}
\]
Cancel the common factor \(x - 8\):
\[
\frac{x + 10}{2(x - 4)}
\]
Final answer: \(\frac{x + 10}{2(x - 4)}\)

---

These are the simplified answers to the expressions. Let me know if you'd like any of them explained in more detail!

Here are 5 relative questions you might find interesting:

1. How do you factor quadratic expressions efficiently?
2. What are common mistakes when simplifying rational expressions?
3. How do you handle cases where the numerator and denominator share no common factors?
4. Why is it important to cancel factors carefully in algebraic fractions?
5. Can all rational expressions be simplified?

**Tip:** Always check for factors that can be canceled before multiplying or simplifying a rational expression!

Can you solve and answer the Simplifying Rational Algebraic Expressions problems?

Solve and simplify rational algebraic expressions through factoring and canceling common factors. This problem set covers key algebraic concepts such as factoring quadratics, canceling factors, and simplifying fractions. Suitable for students in Grades 9-12.

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