https://file.aiquickdraw.com/mathbot/file/17263109740526669om5u.jpg

The problem in the image asks to "find the set of real number solutions" for the inequality:

\[
\left| |x - 1| - |x + 1| \right| \leq 1
\]

### Step-by-step solution:

1. **Breaking the absolute values**:
   We need to first handle the absolute values by breaking them into cases based on the critical points for \(x\), which are \(x = -1\) and \(x = 1\) (because these are the points where the expressions inside the absolute values change signs).

2. **Case 1: \(x \geq 1\)**:
   - When \(x \geq 1\), we have:
     \[
     |x - 1| = x - 1 \quad \text{and} \quad |x + 1| = x + 1
     \]
     So the inequality becomes:
     \[
     \left| (x - 1) - (x + 1) \right| \leq 1
     \]
     Simplifying inside the absolute value:
     \[
     \left| x - 1 - x - 1 \right| = \left| -2 \right| = 2
     \]
     Now we check the inequality:
     \[
     2 \leq 1
     \]
     This is false. Hence, **there are no solutions** in the region \(x \geq 1\).

3. **Case 2: \(-1 \leq x < 1\)**:
   - In this interval, \(x - 1 \leq 0\) and \(x + 1 \geq 0\), so:
     \[
     |x - 1| = 1 - x \quad \text{and} \quad |x + 1| = x + 1
     \]
     The inequality becomes:
     \[
     \left| (1 - x) - (x + 1) \right| \leq 1
     \]
     Simplifying inside the absolute value:
     \[
     \left| 1 - x - x - 1 \right| = \left| -2x \right| = 2|x|
     \]
     The inequality now becomes:
     \[
     2|x| \leq 1
     \]
     Dividing both sides by 2:
     \[
     |x| \leq \frac{1}{2}
     \]
     This implies:
     \[
     -\frac{1}{2} \leq x \leq \frac{1}{2}
     \]
     Therefore, in this region, the solution is \(x \in \left[ -\frac{1}{2}, \frac{1}{2} \right]\).

4. **Case 3: \(x < -1\)**:
   - When \(x < -1\), both \(x - 1 < 0\) and \(x + 1 < 0\), so:
     \[
     |x - 1| = 1 - x \quad \text{and} \quad |x + 1| = -(x + 1) = -x - 1
     \]
     The inequality becomes:
     \[
     \left| (1 - x) - (-x - 1) \right| \leq 1
     \]
     Simplifying inside the absolute value:
     \[
     \left| 1 - x + x + 1 \right| = \left| 2 \right| = 2
     \]
     Checking the inequality:
     \[
     2 \leq 1
     \]
     This is false. Hence, **there are no solutions** in the region \(x < -1\).

### Final solution:
The solution to the inequality is the interval:

\[
x \in \left[ -\frac{1}{2}, \frac{1}{2} \right]
\]

Would you like more details, or do you have any further questions?

Here are five related questions you could explore:
1. What happens if we replace the inequality \(\leq 1\) with \(< 1\)?
2. How would the solution change if the constants inside the absolute values were different?
3. What is the geometric interpretation of absolute value inequalities?
4. How would we solve \(\left| |x - 2| - |x + 3| \right| \leq 2\)?
5. Can this method be extended to solve inequalities with more absolute values?

**Tip:** Breaking absolute value expressions into cases based on critical points is a powerful technique for solving complex inequalities.

Encontrar el conjunto de números reales solución: | |x - 1| - |x + 1| | ≤ 1

Learn how to solve the absolute value inequality | |x - 1| - |x + 1| | ≤ 1 with a detailed step-by-step solution. This problem covers key algebraic concepts, including absolute value inequalities and case analysis. Suitable for students in Grades 9-12.

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