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The question is asking to determine the set of integer solutions (denoted as \( HP \)) for the inequality:

\[
4\left(\frac{x}{2} - 2\right) \leq 3x - 13
\]

### Step-by-Step Solution:

1. **Simplify the Left Side of the Inequality:**

   \[
   4\left(\frac{x}{2} - 2\right) = 4 \cdot \frac{x}{2} - 4 \cdot 2 = 2x - 8
   \]

   So the inequality becomes:

   \[
   2x - 8 \leq 3x - 13
   \]

2. **Move Terms Involving \( x \) to One Side:**

   Subtract \( 2x \) from both sides:

   \[
   -8 \leq x - 13
   \]

3. **Isolate \( x \):**

   Add 13 to both sides:

   \[
   5 \leq x
   \]

   or equivalently:

   \[
   x \geq 5
   \]

4. **Determine the Solution Set in Integers:**

   Since the question specifies \( x \in \mathbb{Z} \) (integers), the solution set for \( x \) is all integers greater than or equal to 5.

   Thus, the solution set \( HP \) is:

   \[
   HP = \{ x \in \mathbb{Z} \mid x \geq 5 \} = \{ 5, 6, 7, 8, \dots \}
   \]

### Final Answer:

\[
HP = \{ x \in \mathbb{Z} \mid x \geq 5 \}
\]

Would you like a detailed explanation of each step or have any questions?

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### Related Questions:

1. What would the solution be if the inequality symbol were \( > \) instead of \( \leq \)?
2. How would the solution set change if \( x \in \mathbb{R} \) (real numbers) instead of integers?
3. Can you show how to graph this inequality on a number line?
4. What happens if we have a different constant on the right side, like \( 3x - 10 \) instead of \( 3x - 13 \)?
5. How would we approach solving this if there were a quadratic term in \( x \)?

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### Tip:
When solving inequalities, always remember to check if you need to flip the inequality symbol—this happens if you multiply or divide by a negative number.

Tentukan HP bila 4(\frac{x}{2} - 2) \leq 3x - 13, x ∈ ℤ

Learn how to solve the inequality 4(x/2 - 2) ≤ 3x - 13 and determine the integer solution set HP. This problem covers algebraic concepts related to inequalities and integer solutions, suitable for students in Grades 10-11.

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