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### Problem Summary:
You are given a system of inequalities:
\[
\begin{cases}
x - 6a \geq -1 \\
|x + a - 1| < 5
\end{cases}
\]
The goal is to determine which of the following statements are correct:

1. **For \(x = 1\)**, there are **6 integer solutions** satisfying the inequalities.
2. **For \(x = 2\)**, the sum of all integers \(x\) that satisfy the inequalities is \(-8\).
3. **If \(x = 0\)**, it lies within the interval \(-4 < x < 6\).
4. The solutions of \(x\) for both inequalities overlap when \(a < 1\).

We are asked to pick the correct combination of statements from options (a) to (e). Now, let’s solve the inequalities step-by-step.

---

### Step 1: Solve the Inequalities

1. **Inequality 1**:  
   \(x - 6a \geq -1\)  
   Rearranging:  
   \(x \geq 6a - 1\).

2. **Inequality 2**:  
   \(|x + a - 1| < 5\).  
   This absolute value inequality splits into two parts:  
   \(-5 < x + a - 1 < 5\).  
   Rearranging:  
   \(-4 < x + a < 6\).  
   Therefore:
   \[
   -4 - a < x < 6 - a.
   \]

---

### Step 2: Analyze Each Statement

1. **Statement (1):**  
   For \(x = 1\), the condition is whether there are 6 integer solutions that meet both inequalities.  
   - From \(x \geq 6a - 1\), if \(a = 0\), we get \(x \geq -1\).
   - From \(-4 - a < x < 6 - a\), if \(a = 0\), this becomes \(-4 < x < 6\), giving possible integer solutions: \(-3, -2, -1, 0, 1, 2, 3, 4, 5\).  
     That’s 9 integer solutions, not 6. So, **(1) is false**.

2. **Statement (2):**  
   For \(x = 2\), we want the sum of all integers satisfying both inequalities.  
   - If \(a = 0\), the valid range for \(x\) is \(-4 < x < 6\), i.e., the integers \(-3, -2, -1, 0, 1, 2, 3, 4, 5\).  
     The sum of these integers is:
     \[
     (-3) + (-2) + (-1) + 0 + 1 + 2 + 3 + 4 + 5 = 9.
     \]
   This contradicts the claim that the sum is \(-8\), so **(2) is false**.

3. **Statement (3):**  
   For \(x = 0\), we check whether it lies within the interval \(-4 < x < 6\).  
   Clearly, \(0\) is between \(-4\) and \(6\), so **(3) is true**.

4. **Statement (4):**  
   We need to determine if the solutions of \(x\) for both inequalities overlap when \(a < 1\).
   - If \(a < 1\), say \(a = 0\), the ranges are:
     - \(x \geq -1\),
     - \(-4 < x < 6\).
   The intersection is \(-1 \leq x < 6\), which is non-empty, so **(4) is true**.

---

### Step 3: Conclusion

- Statements **(3)** and **(4)** are correct.

Thus, the correct answer is:

**(c) (2) dan (4) benar**.

---

Do you have any questions or need further clarifications?  

Here are 5 related questions:
1. How do you solve absolute value inequalities?
2. How do you determine the intersection of two inequalities?
3. What are the methods to verify integer solutions in inequalities?
4. How does changing the parameter \(a\) affect the solutions?
5. What is the geometric interpretation of absolute value inequalities?

**Tip:** Always break down absolute value inequalities into two cases for easier solving.

Diketahui:
\[ \begin{cases} x - 6a \geq -1 \\ |x + a - 1| < 5 \end{cases} \]
Manakah dari pernyataan berikut yang benar?

1) untuk \(x = 1\), terdapat 6 bilangan bulat yang memenuhi pertidaksamaan
2) untuk \(x = 2\), penjumlahan dari seluruh bilangan bulat \(x\) yang memenuhi pertidaksamaan adalah \(-8\)
3) Jika \(x = 0\), \(x\) memenuhi interval \(-4 < x < 6\)
4) solusi \(x\) yang memenuhi masing-masing kedua pertidaksamaan beririsan ketika \(a < 1\)

Pilih jawaban yang benar dari opsi:

a. (1), (2), dan (3) benar
b. (1) dan (3) benar
c. (2) dan (4) benar
d. (4) saja yang benar
e. (1), (2), (3), dan (4) yang benar

Explore a problem involving a system of inequalities and determine which statements about integer solutions, intervals, and intersections are true. This problem involves absolute values, inequalities, and solution analysis, suitable for high school students.

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