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The image you provided shows a set of equations and inequalities. Let's solve them step by step.

### Question 6: Solve the following equations:

**(1) \( |3x| = 6 \)**

For absolute value equations, we consider two cases:
1. \( 3x = 6 \)
2. \( 3x = -6 \)

**Case 1**:
\[
3x = 6 \implies x = \frac{6}{3} = 2
\]

**Case 2**:
\[
3x = -6 \implies x = \frac{-6}{3} = -2
\]

So, the solution to \( |3x| = 6 \) is:
\[
x = 2 \quad \text{or} \quad x = -2
\]

---

**(2) \( |2x + 5| = 1 \)**

Similarly, we consider two cases:
1. \( 2x + 5 = 1 \)
2. \( 2x + 5 = -1 \)

**Case 1**:
\[
2x + 5 = 1 \implies 2x = 1 - 5 = -4 \implies x = \frac{-4}{2} = -2
\]

**Case 2**:
\[
2x + 5 = -1 \implies 2x = -1 - 5 = -6 \implies x = \frac{-6}{2} = -3
\]

So, the solution to \( |2x + 5| = 1 \) is:
\[
x = -2 \quad \text{or} \quad x = -3
\]

---

### Solve the following inequalities:

**(1) \( |2x - 3| \geq 5 \)**

We again split into two cases:
1. \( 2x - 3 \geq 5 \)
2. \( 2x - 3 \leq -5 \)

**Case 1**:
\[
2x - 3 \geq 5 \implies 2x \geq 8 \implies x \geq 4
\]

**Case 2**:
\[
2x - 3 \leq -5 \implies 2x \leq -2 \implies x \leq -1
\]

So, the solution to \( |2x - 3| \geq 5 \) is:
\[
x \geq 4 \quad \text{or} \quad x \leq -1
\]

---

**(2) \( 1 \leq |3x + 1| < 5 \)**

This inequality involves a compound inequality, and we treat it as two parts: \( 1 \leq |3x + 1| \) and \( |3x + 1| < 5 \).

For the first part, \( 1 \leq |3x + 1| \), we again split into two cases:
1. \( 3x + 1 \geq 1 \)
2. \( 3x + 1 \leq -1 \)

**Case 1**:
\[
3x + 1 \geq 1 \implies 3x \geq 0 \implies x \geq 0
\]

**Case 2**:
\[
3x + 1 \leq -1 \implies 3x \leq -2 \implies x \leq \frac{-2}{3}
\]

For the second part, \( |3x + 1| < 5 \), we split it again into two cases:
1. \( 3x + 1 < 5 \)
2. \( 3x + 1 > -5 \)

**Case 1**:
\[
3x + 1 < 5 \implies 3x < 4 \implies x < \frac{4}{3}
\]

**Case 2**:
\[
3x + 1 > -5 \implies 3x > -6 \implies x > -2
\]

Now, combining the results:

- From \( 1 \leq |3x + 1| \), we have \( x \geq 0 \) or \( x \leq \frac{-2}{3} \).
- From \( |3x + 1| < 5 \), we have \( -2 < x < \frac{4}{3} \).

Thus, the final solution to \( 1 \leq |3x + 1| < 5 \) is:
\[
-2 < x \leq \frac{-2}{3} \quad \text{or} \quad 0 \leq x < \frac{4}{3}
\]

---

Would you like more details or further clarification on any of these steps?

Here are some related questions to explore further:

1. What is the general method to solve absolute value inequalities?
2. How can compound inequalities be solved more efficiently?
3. What are the key differences between absolute value equations and inequalities?
4. How would the solutions change if the inequality symbols were reversed?
5. Can you express absolute value inequalities graphically?

**Tip**: When solving absolute value equations, always consider both the positive and negative cases for the expression inside the absolute value!

Solve the following absolute value equations and inequalities:
1. |3x| = 6 
2. |2x + 5| = 1 

Solve the following inequalities:
1. |2x - 3| ≥ 5 
2. 1 ≤ |3x + 1| < 5

This guide explains how to solve absolute value equations and inequalities step by step, including |3x| = 6, |2x + 5| = 1, and inequalities such as |2x - 3| ≥ 5. Learn the core mathematical concepts and apply formulas related to absolute values and inequalities. Suitable for students in Grades 9-10.

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