I'll solve each of the problems in your image. Here's the breakdown:

Zad. 8 (Solve the equations)

(a)

$\frac{3x - 5}{2} + 3 = \frac{3x + 5}{6}$

1. Eliminate the fractions by multiplying through by the least common denominator (LCD = 6): $6 \cdot \frac{3x - 5}{2} + 6 \cdot 3 = 6 \cdot \frac{3x + 5}{6}$ $3(3x - 5) + 18 = 3x + 5$

2. Expand and simplify: $9x - 15 + 18 = 3x + 5$ $9x + 3 = 3x + 5$

3. Solve for $x$: $9x - 3x = 5 - 3$ $6x = 2 \quad \Rightarrow \quad x = \frac{1}{3}$

(b)

$|x + 1| - 4 = 0$

1. Rearrange: $|x + 1| = 4$

2. Solve the absolute value equation: $x + 1 = 4 \quad \text{or} \quad x + 1 = -4$ $x = 3 \quad \text{or} \quad x = -5$

Zad. 9 (Solve the inequalities)

(a)

$\frac{2 - 3(x - 1)}{4} \geq \frac{1}{2} \cdot \frac{2x + 1}{8}$

1. Simplify each term and multiply through by the LCD (8): $8 \cdot \frac{2 - 3(x - 1)}{4} \geq 8 \cdot \frac{1}{2} \cdot \frac{2x + 1}{8}$ $2(2 - 3(x - 1)) \geq 2x + 1$

2. Expand: $4 - 6(x - 1) \geq 2x + 1$ $4 - 6x + 6 \geq 2x + 1$

3. Simplify: $10 - 6x \geq 2x + 1$

4. Solve for $x$: $10 - 1 \geq 6x + 2x$ $9 \geq 8x \quad \Rightarrow \quad x \leq \frac{9}{8}$

(b)

$|x - 7| < 4$

1. Solve the absolute value inequality: $-4 < x - 7 < 4$

2. Add 7 to all parts: $3 < x < 11$

Zad. 10 (Solve the system of equations)