https://file.aiquickdraw.com/mathbot/file/1736508956700y73cb6q9.jpg

Let's solve the problems from the provided Algebra review step by step. Below, I will start solving each problem, grouped into simplifications and solving equations.

---

### **Simplify:**
#### **34. \( \frac{x}{2x} \)**
Cancel \( x \) from numerator and denominator:
\[
\frac{x}{2x} = \frac{1}{2}.
\]

---

#### **35. \( \frac{a^2b}{ab^2} \)**
Divide \( a^2 \) by \( a \), leaving \( a \). Divide \( b \) by \( b^2 \), leaving \( \frac{1}{b} \):
\[
\frac{a^2b}{ab^2} = \frac{a}{b}.
\]

---

#### **36. \( \frac{x-2}{x^2-4} \)**
Factor \( x^2 - 4 \) as a difference of squares:
\[
x^2 - 4 = (x-2)(x+2).
\]
Cancel \( x-2 \) from numerator and denominator:
\[
\frac{x-2}{x^2-4} = \frac{1}{x+2}, \quad x \neq 2 \, \text{or} \, -2 \, (\text{to avoid zero in the denominator}).
\]

---

#### **37. \( \frac{3x + 6}{x^2 + 5x + 6} \)**
Factor numerator and denominator:
\[
3x + 6 = 3(x + 2), \quad x^2 + 5x + 6 = (x + 2)(x + 3).
\]
Cancel \( x + 2 \):
\[
\frac{3x + 6}{x^2 + 5x + 6} = \frac{3}{x + 3}, \quad x \neq -2 \, \text{or} \, -3.
\]

---

#### **38. \( \sqrt{81} \)**
The square root of 81 is:
\[
\sqrt{81} = 9.
\]

---

#### **39. \( \sqrt{50} \)**
Simplify \( \sqrt{50} \) by factoring out \( \sqrt{25} \):
\[
\sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2}.
\]

---

#### **40. \( \sqrt{9x^2} \)**
Simplify:
\[
\sqrt{9x^2} = 3|x|.
\]
(\( |x| \) represents the absolute value of \( x \), as square roots are non-negative.)

---

#### **41. \( \sqrt{45y^3} \)**
Simplify \( \sqrt{45y^3} \) by factoring \( \sqrt{9y^2} \):
\[
\sqrt{45y^3} = \sqrt{9y^2 \cdot 5y} = 3y\sqrt{5y}.
\]

---

### **Solve:**
#### **42. \( \frac{2}{3} = \frac{x}{9} \)**
Cross-multiply:
\[
2 \cdot 9 = 3 \cdot x \quad \Rightarrow \quad 18 = 3x \quad \Rightarrow \quad x = 6.
\]

---

#### **43. \( \frac{5}{8} = \frac{20}{x} \)**
Cross-multiply:
\[
5x = 8 \cdot 20 \quad \Rightarrow \quad 5x = 160 \quad \Rightarrow \quad x = 32.
\]

---

#### **44. \( \frac{3}{x} = \frac{7\frac{1}{2}}{25} \)**
Convert \( 7\frac{1}{2} \) to an improper fraction:
\[
7\frac{1}{2} = \frac{15}{2}.
\]
Rewrite the equation:
\[
\frac{3}{x} = \frac{\frac{15}{2}}{25}.
\]
Simplify \( \frac{\frac{15}{2}}{25} \) by multiplying:
\[
\frac{15}{2} \div 25 = \frac{15}{2} \cdot \frac{1}{25} = \frac{15}{50} = \frac{3}{10}.
\]
Now solve:
\[
\frac{3}{x} = \frac{3}{10} \quad \Rightarrow \quad x = 10.
\]

---

#### **45. \( \frac{x}{15} = \frac{28}{60} \)**
Simplify \( \frac{28}{60} \):
\[
\frac{28}{60} = \frac{7}{15}.
\]
Now solve:
\[
\frac{x}{15} = \frac{7}{15} \quad \Rightarrow \quad x = 7.
\]

---

### Let me know if you'd like further details or explanations for any of the above! 

Here are **5 related questions** to expand your understanding:
1. How do you determine when to factorize a quadratic expression?
2. What is the importance of absolute value when simplifying square roots?
3. How can you identify and cancel common terms in a rational expression?
4. How do you simplify complex fractions like \( \frac{\frac{15}{2}}{25} \)?
5. Why is cross-multiplication a valid method for solving proportions?

**Tip:** Always check for restrictions on variables (e.g., values that make the denominator zero) when simplifying rational expressions!

Simplify and solve the algebra problems from the provided image, including simplifying expressions and solving equations.

Simplify and solve algebraic expressions and equations step-by-step, including rational expressions, square roots, and proportions. Learn key techniques like factoring, simplifying radicals, and cross-multiplication. Perfect for students in Grades 8-10.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation