https://file.aiquickdraw.com/mathbot/file/1727312674680fdjowlyh.jpg

Let's solve these absolute value inequalities step by step:

### 1. \(|2x - 7| > 3\)

This inequality breaks down into two cases:
1. \( 2x - 7 > 3 \)
2. \( 2x - 7 < -3 \)

#### Case 1:
\[
2x - 7 > 3
\]
\[
2x > 10 \quad \Rightarrow \quad x > 5
\]

#### Case 2:
\[
2x - 7 < -3
\]
\[
2x < 4 \quad \Rightarrow \quad x < 2
\]

Thus, the solution is:
\[
x < 2 \quad \text{or} \quad x > 5
\]
This is the interval \( (-\infty, 2) \cup (5, \infty) \).

---

### 2. \(\left|\frac{x}{3} - 2\right| \leq 6\)

Again, break this into two cases:
1. \( \frac{x}{3} - 2 \leq 6 \)
2. \( \frac{x}{3} - 2 \geq -6 \)

#### Case 1:
\[
\frac{x}{3} - 2 \leq 6 \quad \Rightarrow \quad \frac{x}{3} \leq 8 \quad \Rightarrow \quad x \leq 24
\]

#### Case 2:
\[
\frac{x}{3} - 2 \geq -6 \quad \Rightarrow \quad \frac{x}{3} \geq -4 \quad \Rightarrow \quad x \geq -12
\]

Thus, the solution is:
\[
-12 \leq x \leq 24
\]
This is the interval \( [-12, 24] \).

---

### 3. \(|x - 1| < 2|x - 2|\)

We will consider the two cases based on the absolute values of \(x - 1\) and \(x - 2\):

- If \(x \geq 2\), then the absolute values drop:
  \[
  x - 1 < 2(x - 2)
  \]
  \[
  x - 1 < 2x - 4 \quad \Rightarrow \quad -x < -3 \quad \Rightarrow \quad x > 3
  \]
  So, in this case, \(x > 3\).

- If \(x < 2\), the absolute values become:
  \[
  -(x - 1) < 2(-(x - 2)) \quad \Rightarrow \quad -x + 1 < -2x + 4
  \]
  \[
  x > 3
  \]
  But \(x < 2\), so there's no solution in this range.

Thus, the solution is \(x > 3\).

---

### 4. \(|x - 1| < 2(x - 2)\)

Similar to problem 3, break into cases based on the sign of \(x - 1\):

- For \(x \geq 2\), we get:
  \[
  x - 1 < 2(x - 2) \quad \Rightarrow \quad x - 1 < 2x - 4 \quad \Rightarrow \quad -x < -3 \quad \Rightarrow \quad x > 3
  \]

- For \(x < 2\), we handle the absolute values and solve:
  \[
  -(x - 1) < 2(x - 2) \quad \Rightarrow \quad -x + 1 < 2x - 4 \quad \Rightarrow \quad -3x < -5 \quad \Rightarrow \quad x > \frac{5}{3}
  \]

Thus, the solution is \(x > \frac{5}{3}\).

---

### 5. \(x|x + 2| < 2|2x + 3|\)

This one will require splitting into more cases based on where the expressions inside the absolute value change sign.

---

### 6. \(x - 1 \leq \frac{2}{|x|}\)

We need to analyze the behavior when \(x = 0\) or when \(x\) changes sign.

---

### 7. \(\left|x + x\right| \leq 2\)

This simplifies to \(|2x| \leq 2\), which leads to:
\[
|x| \leq 1 \quad \Rightarrow \quad -1 \leq x \leq 1
\]

---

Would you like more details on any of these? Here are five follow-up questions for you to explore:

1. How do we solve compound absolute value inequalities with multiple expressions?
2. Can you explain how to graph the solutions to these inequalities?
3. What happens when there's a division by an absolute value term?
4. How does solving absolute value inequalities differ from solving regular inequalities?
5. Can you derive the interval solution for the inequality in question 5?

**Tip**: Always check if the values obtained satisfy all parts of an inequality, especially for more complex absolute value equations.

Solve the following absolute value inequalities:
1. |2x - 7| > 3
2. |x/3 - 2| ≤ 6
3. |x - 1| < 2|x - 2|
4. |x - 1| < 2(x - 2)
5. x|x + 2| < 2|2x + 3|
6. x - 1 ≤ 2/|x|
7. |x + x| ≤ 2

Learn how to solve various absolute value inequalities like |2x - 7| > 3 and |x/3 - 2| ≤ 6 step-by-step. This problem covers key algebraic concepts related to solving absolute value equations, inequalities, and applying piecewise functions. Suitable for high school students.

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