Solve the inequations
(i) `-3le4-7xlt18`
(ii) `-12lt3x-5le-4`

Answer

Let's solve the given inequalities step by step.

### Part (i): Solve the inequality \(-3 \leq 4 - 7x < 18\)

1. **Break the compound inequality into two parts:**
\[ -3 \leq 4 - 7x \quad \text{and} \quad 4 - 7x < 18 \]

2. **Solve the first part:**
\[ -3 \leq 4 - 7x \]
Subtract 4 from both sides:
\[ -3 - 4 \leq -7x \implies -7 \leq -7x \]
Now, divide by -7 (remember to reverse the inequality sign):
\[ 1 \geq x \quad \text{or} \quad x \leq 1 \]

3. **Solve the second part:**
\[ 4 - 7x < 18 \]
Subtract 4 from both sides:
\[ -7x < 18 - 4 \implies -7x < 14 \]
Now, divide by -7 (again, reverse the inequality sign):
\[ x > -2 \]

4. **Combine the results:**
From the two parts, we have:
\[ -2 < x \leq 1 \]
Thus, the solution for part (i) is:
\[ x \in (-2, 1] \]

### Part (ii): Solve the inequality \(-12 < 3x - 5 \leq -4\)

1. **Break the compound inequality into two parts:**
\[ -12 < 3x - 5 \quad \text{and} \quad 3x - 5 \leq -4 \]

2. **Solve the first part:**
\[ -12 < 3x - 5 \]
Add 5 to both sides:
\[ -12 + 5 < 3x \implies -7 < 3x \]
Now, divide by 3:
\[ -\frac{7}{3} < x \quad \text{or} \quad x > -\frac{7}{3} \]

3. **Solve the second part:**
\[ 3x - 5 \leq -4 \]
Add 5 to both sides:
\[ 3x \leq -4 + 5 \implies 3x \leq 1 \]
Now, divide by 3:
\[ x \leq \frac{1}{3} \]

4. **Combine the results:**
From the two parts, we have:
\[ -\frac{7}{3} < x \leq \frac{1}{3} \]
Thus, the solution for part (ii) is:
\[ x \in \left(-\frac{7}{3}, \frac{1}{3}\right] \]

### Summary of Solutions:
- For part (i): \( x \in (-2, 1] \)
- For part (ii): \( x \in \left(-\frac{7}{3}, \frac{1}{3}\right] \)