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We are asked to solve the inequality \( |7x - 1| > |3x - 5| \).

### Step-by-step breakdown:

1. **Absolute value setup**:
   The expression involves two absolute values, which requires us to consider different cases depending on whether the expressions inside the absolute values are positive or negative.

2. **Case 1: \( 7x - 1 \geq 0 \) and \( 3x - 5 \geq 0 \)**:
   - In this case, both expressions inside the absolute values are non-negative, so we can remove the absolute value signs:
   \[
   7x - 1 > 3x - 5
   \]
   Simplifying this inequality:
   \[
   7x - 3x > -5 + 1
   \]
   \[
   4x > -4
   \]
   \[
   x > -1
   \]
   Additionally, the conditions for this case require:
   \[
   7x - 1 \geq 0 \implies x \geq \frac{1}{7}
   \quad \text{and} \quad 3x - 5 \geq 0 \implies x \geq \frac{5}{3}
   \]
   Therefore, from this case, we have \( x > \frac{5}{3} \).

3. **Case 2: \( 7x - 1 \geq 0 \) and \( 3x - 5 \leq 0 \)**:
   - Here, we have:
   \[
   7x - 1 > -(3x - 5)
   \]
   Simplifying:
   \[
   7x - 1 > -3x + 5
   \]
   \[
   7x + 3x > 5 + 1
   \]
   \[
   10x > 6
   \]
   \[
   x > \frac{3}{5}
   \]
   The conditions for this case are:
   \[
   7x - 1 \geq 0 \implies x \geq \frac{1}{7}
   \quad \text{and} \quad 3x - 5 \leq 0 \implies x \leq \frac{5}{3}
   \]
   Therefore, from this case, we have \( \frac{5}{3} \geq x > \frac{3}{5} \).

4. **Case 3: \( 7x - 1 \leq 0 \) and \( 3x - 5 \geq 0 \)**:
   - Here, we have:
   \[
   -(7x - 1) > 3x - 5
   \]
   Simplifying:
   \[
   -7x + 1 > 3x - 5
   \]
   \[
   -7x - 3x > -5 - 1
   \]
   \[
   -10x > -6
   \]
   \[
   x < \frac{3}{5}
   \]
   The conditions are:
   \[
   7x - 1 \leq 0 \implies x \leq \frac{1}{7}
   \quad \text{and} \quad 3x - 5 \geq 0 \implies x \geq \frac{5}{3}
   \]
   But since there is no overlap between these conditions, this case gives no solution.

5. **Case 4: \( 7x - 1 \leq 0 \) and \( 3x - 5 \leq 0 \)**:
   - Here, we have:
   \[
   -(7x - 1) > -(3x - 5)
   \]
   Simplifying:
   \[
   -7x + 1 > -3x + 5
   \]
   \[
   -7x + 3x > 5 - 1
   \]
   \[
   -4x > 4
   \]
   \[
   x < -1
   \]
   The conditions for this case are:
   \[
   7x - 1 \leq 0 \implies x \leq \frac{1}{7}
   \quad \text{and} \quad 3x - 5 \leq 0 \implies x \leq \frac{5}{3}
   \]
   Thus, from this case, we get \( x < -1 \).

### Combining all cases:
- From Case 1, we have \( x > \frac{5}{3} \).
- From Case 2, we have \( \frac{3}{5} < x \leq \frac{5}{3} \).
- From Case 4, we have \( x < -1 \).

Thus, the final solution is:
\[
x < -1 \quad \text{or} \quad \frac{3}{5} < x \leq \frac{5}{3} \quad \text{or} \quad x > \frac{5}{3}
\]

Let me know if you need more details or clarifications! Here are five related questions to deepen your understanding:

1. How do absolute value inequalities change when the expression inside is negative?
2. How would the solution process differ for \( |7x - 1| \geq |3x - 5| \)?
3. What are the geometric interpretations of absolute value inequalities on the number line?
4. Can you derive a general formula for solving inequalities involving absolute values?
5. How can we verify if our solution satisfies the original inequality?

**Tip**: Always double-check your case conditions when solving absolute value inequalities to ensure no possible solutions are missed.

This page explains how to solve the inequality |7x - 1| > |3x - 5| with a step-by-step solution. Learn how to handle absolute values and inequalities, with key concepts like absolute value rules and algebraic inequality manipulations.

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