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The problem asks to determine the "Himpunan Penyelesaian" (Solution Set) for each inequality in the system. I will solve each one step by step.

### 1. \( 2 - 3x \leq 12 \)

Solve for \( x \):

\[
2 - 3x \leq 12
\]
\[
-3x \leq 12 - 2
\]
\[
-3x \leq 10
\]
\[
x \geq \frac{10}{-3}
\]
\[
x \geq -\frac{10}{3}
\]

So, the solution is \( x \geq -\frac{10}{3} \).

### 2. \( 3x - 5 < 4x - 6 \)

Solve for \( x \):

\[
3x - 5 < 4x - 6
\]
\[
-5 + 6 < 4x - 3x
\]
\[
1 < x
\]

So, the solution is \( x > 1 \).

### 3. \( -3 \leq 1 - 6x < 4 \)

This is a compound inequality, so we solve it step by step.

For the left part:
\[
-3 \leq 1 - 6x
\]
\[
-3 - 1 \leq -6x
\]
\[
-4 \leq -6x
\]
\[
x \leq \frac{2}{3}
\]

For the right part:
\[
1 - 6x < 4
\]
\[
-6x < 3
\]
\[
x > -\frac{1}{2}
\]

Combining both parts: \( -\frac{1}{2} < x \leq \frac{2}{3} \).

### 4. \( 2x^2 - 5x - 3 < 0 \)

We solve the quadratic inequality by first solving the equation \( 2x^2 - 5x - 3 = 0 \) using the quadratic formula:

\[
x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(-3)}}{2(2)}
\]
\[
x = \frac{5 \pm \sqrt{25 + 24}}{4}
\]
\[
x = \frac{5 \pm \sqrt{49}}{4}
\]
\[
x = \frac{5 \pm 7}{4}
\]
\[
x_1 = \frac{12}{4} = 3, \quad x_2 = \frac{-2}{4} = -\frac{1}{2}
\]

The roots are \( x = 3 \) and \( x = -\frac{1}{2} \). Since it's a quadratic inequality, the solution is between the roots:

\[
-\frac{1}{2} < x < 3
\]

### 5. \( x^2 - 5x - 6 \leq 0 \)

We solve \( x^2 - 5x - 6 = 0 \) using the quadratic formula:

\[
x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(-6)}}{2(1)}
\]
\[
x = \frac{5 \pm \sqrt{25 + 24}}{2}
\]
\[
x = \frac{5 \pm \sqrt{49}}{2}
\]
\[
x = \frac{5 \pm 7}{2}
\]
\[
x_1 = \frac{12}{2} = 6, \quad x_2 = \frac{-2}{2} = -1
\]

The roots are \( x = 6 \) and \( x = -1 \). Since the inequality is \( \leq \), the solution includes the endpoints:

\[
-1 \leq x \leq 6
\]

### 6. \( \frac{1}{x+1} < \frac{2}{3x - 1} \)

We solve by cross-multiplying:

\[
(1)(3x - 1) < (2)(x + 1)
\]
\[
3x - 1 < 2x + 2
\]
\[
3x - 2x < 2 + 1
\]
\[
x < 3
\]

However, we need to exclude values that make the denominators zero, so \( x \neq -1 \) and \( x \neq \frac{1}{3} \). Thus, the solution is:

\[
x < 3, \quad x \neq -1, \quad x \neq \frac{1}{3}
\]

### 7. \( \frac{x+2}{x+4} < \frac{x-1}{x-2} \)

We solve by cross-multiplying:

\[
(x + 2)(x - 2) < (x - 1)(x + 4)
\]
\[
x^2 - 4 < x^2 + 4x - x - 4
\]
\[
x^2 - 4 < x^2 + 3x - 4
\]
\[
-4 < 3x - 4
\]
\[
0 < 3x
\]
\[
x > 0
\]

However, we need to exclude values that make the denominators zero, so \( x \neq -4 \) and \( x \neq 2 \). Thus, the solution is:

\[
0 < x, \quad x \neq -4, \quad x \neq 2
\]

---

Let me know if you'd like more details or further explanation! Here are some related questions:

1. How do you solve compound inequalities involving multiple steps?
2. Can you explain why certain values are excluded when dealing with rational inequalities?
3. How does the quadratic formula work for inequalities?
4. What's the difference between solving linear and quadratic inequalities?
5. How do boundary conditions affect the solution of inequalities?

**Tip:** Always remember to check for values that make denominators zero when solving rational inequalities.

Tentukan Himpunan Penyelesaian (HP) untuk setiap pertidaksamaan berikut:

Solve various inequalities, including linear, quadratic, and rational inequalities. Learn step-by-step solutions using the quadratic formula, cross-multiplication, and other algebraic methods. Suitable for high school students.

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