https://r2.erweima.ai/mathbot/image/1725439274824bostrijf.jpg

I'll analyze the image and solve the problems you requested. The problems are mainly algebraic and involve operations with fractions and solving equations. Here's the breakdown:

### Problems and Solutions:

1. **Problem 1:**
   \[
   \frac{a}{2} + \frac{3a}{4} = \frac{7a}{8}
   \]
   **Solution:**
   \[
   \text{Combine like terms: } \frac{a}{2} + \frac{3a}{4} = \frac{4a}{8} + \frac{6a}{8} = \frac{10a}{8} = \frac{5a}{4}
   \]
   So, the equation is incorrect as \( \frac{5a}{4} \neq \frac{7a}{8} \).

2. **Problem 2:**
   \[
   \frac{5}{2} + \frac{1}{4} = \frac{11}{4}
   \]
   **Solution:**
   \[
   \frac{5}{2} = \frac{10}{4} \quad \text{(same denominator)} \\
   \frac{10}{4} + \frac{1}{4} = \frac{11}{4} \quad \text{(correct)}
   \]
   The equation is correct.

3. **Problem 3:**
   \[
   3x - \frac{2x}{3} = 5 + \frac{1}{3}
   \]
   **Solution:**
   Multiply everything by 3 to eliminate the fraction:
   \[
   9x - 2x = 15 + 1 \\
   7x = 16 \\
   x = \frac{16}{7}
   \]
   So, \( x = \frac{16}{7} \) is the correct solution.

4. **Problem 4:**
   \[
   \frac{2a + b}{3} = \frac{a}{4} + \frac{b}{6}
   \]
   **Solution:**
   Multiply by 12 (LCM of 3, 4, and 6):
   \[
   4(2a + b) = 3a + 2b \\
   8a + 4b = 3a + 2b \\
   5a + 2b = 0 \quad \text{(rearrange terms)}
   \]
   Therefore, the equation can be simplified to \( 5a = -2b \).

5. **Problem 5:**
   \[
   \frac{2x}{5} - \frac{y}{3} = \frac{x}{4} + \frac{y}{2}
   \]
   **Solution:**
   Multiply everything by 60 (LCM of 5, 3, 4, 2):
   \[
   24x - 20y = 15x + 30y \\
   9x = 50y \quad \text{(rearrange terms)}
   \]
   Therefore, the solution can be written as \( x = \frac{50}{9}y \).

6. **Problem 6:**
   \[
   \frac{4x - 2}{x + 3} = 1 - \frac{2}{x + 3}
   \]
   **Solution:**
   Multiply by \( x + 3 \) to eliminate the denominator:
   \[
   4x - 2 = (x + 3) - 2 \\
   4x - 2 = x + 1 \\
   3x = 3 \\
   x = 1
   \]
   The solution is \( x = 1 \).

7. **Problem 7:**
   \[
   \frac{3x + 5}{x - 2} = \frac{x + 5}{x - 2}
   \]
   **Solution:**
   Since the denominators are the same, equate the numerators:
   \[
   3x + 5 = x + 5 \\
   2x = 0 \\
   x = 0
   \]
   So, \( x = 0 \).

### Summary
These are the detailed solutions for the algebraic problems in the provided image.

Would you like more detailed explanations or have any questions about these solutions?

#### Follow-Up Questions:
1. What happens when the denominators are not the same, and how do you solve such equations?
2. Can you explain how to deal with equations involving variables on both sides?
3. What is the method to simplify fractions before solving equations?
4. How does multiplying by the LCM help in solving equations with fractions?
5. Could you solve more complex

The user asked for solving a set of algebraic problems involving fractions, equations, and simplifications.

Step-by-step solutions for algebraic equations involving fractions, simplification techniques, and solving linear equations. This content covers key algebraic concepts suitable for students in Grades 8-10.

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