|1-1/5x| = 3
|5 - 6x| = 7
|x + 7| > 3
|1/3x + 4| > 1
|4 - 1/2x| <_ 6 
|3 - x| <_ 3 
|2 - 3x| >_ 2/3 

Let's solve each of the equations and inequalities step by step.

### 1. \( |1 - \frac{1}{5}x| = 3 \)

This is an absolute value equation, so we need to split it into two cases:

#### Case 1: \( 1 - \frac{1}{5}x = 3 \)
\[
1 - \frac{1}{5}x = 3 \implies -\frac{1}{5}x = 3 - 1 = 2 \implies x = \frac{2}{-\frac{1}{5}} = -10
\]

#### Case 2: \( 1 - \frac{1}{5}x = -3 \)
\[
1 - \frac{1}{5}x = -3 \implies -\frac{1}{5}x = -3 - 1 = -4 \implies x = \frac{-4}{-\frac{1}{5}} = 20
\]

**Solution**: \( x = -10 \) or \( x = 20 \).

---

### 2. \( |5 - 6x| = 7 \)

Again, split into two cases:

#### Case 1: \( 5 - 6x = 7 \)
\[
5 - 6x = 7 \implies -6x = 7 - 5 = 2 \implies x = \frac{2}{-6} = -\frac{1}{3}
\]

#### Case 2: \( 5 - 6x = -7 \)
\[
5 - 6x = -7 \implies -6x = -7 - 5 = -12 \implies x = \frac{-12}{-6} = 2
\]

**Solution**: \( x = -\frac{1}{3} \) or \( x = 2 \).

---

### 3. \( |x + 7| > 3 \)

This is an absolute value inequality. We split it into two cases:

#### Case 1: \( x + 7 > 3 \)
\[
x + 7 > 3 \implies x > 3 - 7 = -4
\]

#### Case 2: \( x + 7 < -3 \)
\[
x + 7 < -3 \implies x < -3 - 7 = -10
\]

**Solution**: \( x > -4 \) or \( x < -10 \). So, the solution in interval notation is:
\[
x \in (-\infty, -10) \cup (-4, \infty)
\]

---

### 4. \( \left| \frac{1}{3}x + 4 \right| > 1 \)

Split into two cases:

#### Case 1: \( \frac{1}{3}x + 4 > 1 \)
\[
\frac{1}{3}x + 4 > 1 \implies \frac{1}{3}x > 1 - 4 = -3 \implies x > -9
\]

#### Case 2: \( \frac{1}{3}x + 4 < -1 \)
\[
\frac{1}{3}x + 4 < -1 \implies \frac{1}{3}x < -1 - 4 = -5 \implies x < -15
\]

**Solution**: \( x > -9 \) or \( x < -15 \), so in interval notation:
\[
x \in (-\infty, -15) \cup (-9, \infty)
\]

---

### 5. \( 4 - \frac{1}{2}x \leq 6 \)

Solve the inequality:

\[
4 - \frac{1}{2}x \leq 6 \implies -\frac{1}{2}x \leq 6 - 4 = 2 \implies x \geq \frac{2}{\frac{1}{2}} = 4
\]

**Solution**: \( x \geq 4 \).

---

### 6. \( 3 - x \leq 3 \)

Solve the inequality:

\[
3 - x \leq 3 \implies -x \leq 3 - 3 = 0 \implies x \geq 0
\]

**Solution**: \( x \geq 0 \).

---

### 7. \( 2 - 3x \geq \frac{2}{3} \)

Solve the inequality:

\[
2 - 3x \geq \frac{2}{3} \implies -3x \geq \frac{2}{3} - 2 = \frac{2}{3} - \frac{6}{3} = -\frac{4}{3}
\]
\[
x \leq \frac{-\frac{4}{3}}{-3} = \frac{4}{9}
\]

**Solution**: \( x \leq \frac{4}{9} \).

---

Would you like more details on any of these steps or have any questions? 

Here are some related questions to deepen your understanding:

1. How do you solve absolute value inequalities with multiple conditions?
2. Can you explain how interval notation works in inequalities?
3. How do we approach compound inequalities with absolute values?
4. What are the different cases in solving an absolute value equation?
5. How do you graph inequalities on a number line?

**Tip**: Always check both cases when solving absolute value equations or inequalities to ensure you've covered all possible solutions.

|1-1/5x| = 3 |5 - 6x| = 7 |x + 7| > 3 |1/3x + 4| > 1 |4 - 1/2x| <_ 6 |3 - x| <_ 3 |2 - 3x| >_ 2/3

Discover how to solve absolute value equations and inequalities step-by-step, covering |1 - 1/5x| = 3, |5 - 6x| = 7, and more. Learn key algebraic techniques like splitting cases for absolute value problems. Suitable for Grades 9-11 students.

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