The inequality provided is:

$| |x - 1| - 5 | < 2x - 5$

Let's break this down step by step.

Step 1: Solve the Inner Absolute Value

The expression inside the double absolute value is $|x - 1|$. To analyze this, we consider two cases:

• Case 1: $x - 1 \geq 0$ so $|x - 1| = x - 1$
• Case 2: $x - 1 < 0$ so $|x - 1| = -(x - 1) = 1 - x$

We need to deal with both cases separately.

Case 1: $x - 1 \geq 0$ or $x \geq 1$

For $x \geq 1$, the expression simplifies to $|x - 1| = x - 1$, so we have: $| x - 1 | - 5 = x - 1 - 5 = x - 6$ Now the inequality becomes: $| x - 6 | < 2x - 5$ We again apply the definition of absolute value:

• Subcase 1.1: If $x - 6 \geq 0$, this simplifies to: $x - 6 < 2x - 5$ Solving this: $x - 6 < 2x - 5 \quad \Rightarrow \quad -6 + 5 < x \quad \Rightarrow \quad x > -1$ Since this subcase applies when $x \geq 6$, we now have $x > -1$ and $x \geq 6$, meaning $x \geq 6$.

• Subcase 1.2: If $x - 6 < 0$, this simplifies to: $-(x - 6) < 2x - 5 \quad \Rightarrow \quad -x + 6 < 2x - 5$ Solving this: $6 + 5 < 3x \quad \Rightarrow \quad x > 11/3$ Since this applies for $x < 6$, we get $11/3 < x < 6$.

Thus, the solution for Case 1 is $x > 11/3$.

Case 2: $x - 1 < 0$ or $x < 1$

For $x < 1$, we have $|x - 1| = -(x - 1) = 1 - x$. So the inequality becomes: $| 1 - x - 5 | < 2x - 5$ Simplifying: $| 1 - x - 5 | = | -x - 4 | = | -(x + 4) | = | x + 4 |$ Thus, the inequality is: $| x + 4 | < 2x - 5$ Now we again have two subcases:

• Subcase 2.1: $x + 4 \geq 0$, so: $x + 4 < 2x - 5 \quad \Rightarrow \quad 4 + 5 < x \quad \Rightarrow \quad x > 9$ This is not possible as it contradicts the condition $x < 1$.

• Subcase 2.2: $x + 4 < 0$, so: $-(x + 4) < 2x - 5 \quad \Rightarrow \quad -x - 4 < 2x - 5$ Solving this: $-4 + 5 < 3x \quad \Rightarrow \quad 1 < 3x \quad \Rightarrow \quad x > 1/3$ However, this subcase requires $x < -4$, which is not possible since $x > 1/3$.

Thus, Case 2 doesn't give any valid solutions.

Final Answer:

The solution is $x \in (11/3, \infty)$, which corresponds to option (2) in the provided list.

Would you like more details or have any questions?

Here are five questions related to this concept:

1. What is the definition of absolute value and how is it applied in inequalities?
2. How do you handle cases when solving inequalities involving absolute values?
3. What is the difference between solving single and double absolute value inequalities?
4. How do we interpret the intersection of solutions from different cases?
5. How can graphing inequalities help visualize the solution set?

Tip: When dealing with absolute value inequalities, always split them into cases based on the definition of absolute value to handle each possibility correctly.