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

### 4. \( 12(b + 2) = 8(b + 5) \)
1. Distribute the constants on both sides:
   \[
   12b + 24 = 8b + 40
   \]
2. Move the terms involving \( b \) to one side:
   \[
   12b - 8b = 40 - 24
   \]
   \[
   4b = 16
   \]
3. Solve for \( b \):
   \[
   b = 4
   \]

### 5. \( 6(y + 8) = 3(2y - 7) \)
1. Distribute the constants:
   \[
   6y + 48 = 6y - 21
   \]
2. Subtract \( 6y \) from both sides:
   \[
   48 = -21
   \]
   This is a contradiction, so there is **no solution**.

### 6. \( \frac{3}{4}(12c - 4) = 15c + 15 \)
1. Distribute \( \frac{3}{4} \):
   \[
   9c - 3 = 15c + 15
   \]
2. Move the terms involving \( c \) to one side:
   \[
   9c - 15c = 15 + 3
   \]
   \[
   -6c = 18
   \]
3. Solve for \( c \):
   \[
   c = -3
   \]

### 7. \( 11(4p + 4) - 4p = 4(7p - 7) \)
1. Distribute on both sides:
   \[
   44p + 44 - 4p = 28p - 28
   \]
2. Combine like terms:
   \[
   40p + 44 = 28p - 28
   \]
3. Move the terms involving \( p \) to one side:
   \[
   40p - 28p = -28 - 44
   \]
   \[
   12p = -72
   \]
4. Solve for \( p \):
   \[
   p = -6
   \]

### 8. \( 3(2d - 8) = 11d - 18(d - 3) \)
1. Distribute on both sides:
   \[
   6d - 24 = 11d - 18d + 54
   \]
   \[
   6d - 24 = -7d + 54
   \]
2. Move the terms involving \( d \) to one side:
   \[
   6d + 7d = 54 + 24
   \]
   \[
   13d = 78
   \]
3. Solve for \( d \):
   \[
   d = 6
   \]

### 9. \( 5(4 + r) = \frac{1}{2}(40 + 10r) \)
1. Distribute on both sides:
   \[
   20 + 5r = 20 + 5r
   \]
2. The equation simplifies to:
   \[
   20 + 5r = 20 + 5r
   \]
   This is an identity, so **any value** of \( r \) will satisfy the equation.

### 10. \( \frac{3}{5}e - 6 = \frac{2}{5}(e - 10) - 7 \)
1. Distribute the fractions:
   \[
   \frac{3}{5}e - 6 = \frac{2}{5}e - 4 - 7
   \]
   \[
   \frac{3}{5}e - 6 = \frac{2}{5}e - 11
   \]
2. Move the terms involving \( e \) to one side:
   \[
   \frac{3}{5}e - \frac{2}{5}e = -11 + 6
   \]
   \[
   \frac{1}{5}e = -5
   \]
3. Solve for \( e \):
   \[
   e = -25
   \]

### 11. **Three consecutive integers are \( n \), \( n + 1 \), and \( n + 2 \). Four times the sum of the least and greatest integers is 12 less than three times the least integer. What is the least integer?**
1. Express the sum of the least and greatest integers:
   \[
   \text{Sum} = n + (n + 2) = 2n + 2
   \]
2. Four times the sum:
   \[
   4(2n + 2) = 8n + 8
   \]
3. Set up the equation based on the problem:
   \[
   8n + 8 = 3n - 12
   \]
4. Move the terms involving \( n \) to one side:
   \[
   8n - 3n = -12 - 8
   \]
   \[
   5n = -20
   \]
5. Solve for \( n \):
   \[
   n = -4
   \]

Would you like a detailed explanation for any of these solutions or have any questions?

### Related Questions:
1. How do you solve systems of linear equations?
2. What are the different types of equations encountered in algebra?
3. Can you solve quadratic equations by factoring?
4. What is the method of substitution in solving linear equations?
5. How do you interpret the solution of a linear equation graphically?

### Tip:
Always verify your solutions by substituting them back into the original equations to ensure they satisfy all given conditions.

Learn how to solve linear equations step-by-step with detailed solutions. This problem covers key algebraic concepts such as linear equations and the distributive property. Suitable for students in Grades 7-9.

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