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Let's solve each of the equations step-by-step.

### First Equation:
\[
4(v + 1) + v = 5(v - 1) + 9
\]
1. Expand both sides:
   \[
   4v + 4 + v = 5v - 5 + 9
   \]
   Simplifies to:
   \[
   5v + 4 = 5v + 4
   \]

2. Subtract \(5v\) from both sides:
   \[
   4 = 4
   \]
   This is always true, which means that **all real numbers** are solutions for \(v\).

### Second Equation:
\[
3(u - 2) + 5 = 2(3u - 5)
\]
1. Expand both sides:
   \[
   3u - 6 + 5 = 6u - 10
   \]
   Simplifies to:
   \[
   3u - 1 = 6u - 10
   \]

2. Move terms involving \(u\) to one side:
   \[
   3u - 6u = -10 + 1
   \]
   Simplifies to:
   \[
   -3u = -9
   \]

3. Solve for \(u\):
   \[
   u = 3
   \]

### Final Answer:
- For the first equation, all real numbers are solutions.
- For the second equation, \( u = 3 \).

Would you like further clarification or more details?

Here are 5 related questions:
1. How can we verify if an equation has infinite solutions or no solution?
2. What are the conditions for a linear equation to have a unique solution?
3. How do you handle equations with variables on both sides?
4. What is the method to solve quadratic equations?
5. Can an equation with fractions have no solution?

**Tip:** When solving equations, always start by simplifying both sides before moving terms across the equal sign!

Solve the following equations: 4(v + 1) + v = 5(v - 1) + 9 and 3(u - 2) + 5 = 2(3u - 5).

Learn how to solve linear equations with variables on both sides. This example includes two equations: 4(v + 1) + v = 5(v - 1) + 9 and 3(u - 2) + 5 = 2(3u - 5). Suitable for students in Grades 8-10.

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