The inequality provided is:

$0 < \frac{5}{|x - 1|} < 3.$

We need to solve this compound inequality. Let’s break it into steps:

Step 1: Analyze the compound inequality

The inequality consists of two parts: $\frac{5}{|x - 1|} > 0 \quad \text{and} \quad \frac{5}{|x - 1|} < 3.$

The first part, $\frac{5}{|x - 1|} > 0$, is always true as long as $x \neq 1$, since the denominator $x - 1$ cannot be zero.

So, we focus on the second part: $\frac{5}{|x - 1|} < 3.$

Step 2: Solve $\frac{5}{|x - 1|} < 3$

Rewrite this inequality: $\frac{5}{|x - 1|} < 3 \implies |x - 1| > \frac{5}{3}.$

Step 3: Express the absolute value inequality

The inequality $|x - 1| > \frac{5}{3}$ implies two cases:

1. $x - 1 > \frac{5}{3} \implies x > 1 + \frac{5}{3} = \frac{8}{3}$,
2. $x - 1 < -\frac{5}{3} \implies x < 1 - \frac{5}{3} = -\frac{2}{3}$.

Thus, the solution to $\frac{5}{|x - 1|} < 3$ is: $x \in (-\infty, -\frac{2}{3}) \cup (\frac{8}{3}, \infty).$

Step 4: Combine with the condition $x \neq 1$

Since $x \neq 1$ is already true in the intervals $(-\infty, -\frac{2}{3}) \cup (\frac{8}{3}, \infty)$, no additional restrictions are needed.

Final Solution

The solution set is: $x \in (-\infty, -\frac{2}{3}) \cup (\frac{8}{3}, \infty).$

Would you like further clarification or additional examples?

5 Related Questions:

1. How do we handle inequalities involving absolute values in general?
2. Why does the inequality $\frac{5}{|x - 1|} > 0$ hold for all $x \neq 1$?
3. Can you explain why the inequality flips when solving $|x - 1| > \frac{5}{3}$?
4. How do compound inequalities differ from single inequalities in their solutions?
5. What happens if the inequality sign is reversed, e.g., $0 < \frac{5}{|x - 1|} \leq 3$?

Tip:

When solving inequalities with absolute values, always consider the definition of absolute value and split the problem into multiple cases!