I'll now extract and solve the problems from the uploaded image. Let's start by addressing each exercise one by one.

Left Column (Exercises 9–22: Solve the equation)

1. 11. |m + 3| = 7 $m + 3 = 7 \quad \text{or} \quad m + 3 = -7$ $m = 4 \quad \text{or} \quad m = -10$ Answer: 4, -10

2. 13. |x + 3| = 13.4 $x + 3 = 13.4 \quad \text{or} \quad x + 3 = -13.4$ $x = 10.4 \quad \text{or} \quad x = -16.4$ Answer: 10.4, -16.4

3. 15. |3.5d| = 15.4 $3.5d = 15.4 \quad \text{or} \quad 3.5d = -15.4$ $d = 4.4 \quad \text{or} \quad d = -4.4$ Answer: 4.4, -4.4

4. 17. |-8w + 6| = 76 $-8w + 6 = 76 \quad \text{or} \quad -8w + 6 = -76$ $-8w = 70 \quad \text{or} \quad -8w = -82$ $w = -8.75 \quad \text{or} \quad w = 10.25$ Answer: -8.75, 10.25

5. 19. |-4b - 5| = 13 $-4b - 5 = 13 \quad \text{or} \quad -4b - 5 = -13$ $-4b = 18 \quad \text{or} \quad -4b = -8$ $b = -4.5 \quad \text{or} \quad b = 2$ Answer: -4.5, 2

6. 21. |x - 1| + 5 = 2 Subtracting 5 from both sides: $|x - 1| = -3$ Since the absolute value cannot be negative, no solution.

7. 23. 9|4p + 2| + 8 = 35 Simplifying: $9|4p + 2| = 27 \quad \Rightarrow \quad |4p + 2| = 3$ $4p + 2 = 3 \quad \text{or} \quad 4p + 2 = -3$ $4p = 1 \quad \text{or} \quad 4p = -5$ $p = 0.25 \quad \text{or} \quad p = -1.25$ Answer: 0.25, -1.25

Right Column (Exercises 29–38: Solve the equation)

1. 29. |4n - 15| = 1 $4n - 15 = 1 \quad \text{or} \quad 4n - 15 = -1$ $4n = 16 \quad \text{or} \quad 4n = 14$ $n = 4 \quad \text{or} \quad n = 3.5$ Answer: 4, 3.5

2. 30. |2c + 8| = |10c| Two cases: $2c + 8 = 10c \quad \text{or} \quad 2c + 8 = -10c$ First case: $8c = 8 \quad \Rightarrow \quad c = 1$ Second case: $12c = -8 \quad \Rightarrow \quad c = -\frac{2}{3}$ Answer: 1, -$\frac{2}{3}$

3. 32. |4p - 2| = |2p + 8| Two cases: $4p - 2 = 2p + 8 \quad \text{or} \quad 4p - 2 = -(2p + 8)$ First case: $2p = 10 \quad \Rightarrow \quad p = 5$ Second case: $6p = -6 \quad \Rightarrow \quad p = -1$ Answer: 5, -1

4. 34. |4m - 1| = 3|4v + 1| Simplify: $4m - 1 = 12v + 3$ Solve