The inequality given is:

$|x - 1| \geq 13$

To solve this algebraically, we use the fact that for any absolute value inequality of the form $|A| \geq B$, where $B \geq 0$, the inequality can be rewritten as:

$A \geq B \quad \text{or} \quad A \leq -B$

In this case, $A = x - 1$ and $B = 13$. Applying this rule, we break the absolute value inequality into two cases:

Case 1: $x - 1 \geq 13$

$x - 1 \geq 13$ Add 1 to both sides: $x \geq 14$

Case 2: $x - 1 \leq -13$

$x - 1 \leq -13$ Add 1 to both sides: $x \leq -12$

Final Solution

Combining both cases, the solution is:

$x \geq 14 \quad \text{or} \quad x \leq -12$

In interval notation, this can be written as: $x \in (-\infty, -12] \cup [14, \infty)$

Would you like more details or have any questions?

Here are 5 related questions to expand your understanding:

1. How would the solution change if the inequality were strict ($|x - 1| > 13$) instead of non-strict?
2. How do you solve inequalities involving absolute values that are less than a number (e.g., $|x - 1| < 13$)?
3. Can you explain the geometric interpretation of absolute value inequalities on a number line?
4. How would this inequality change if we had $|x + 3| \geq 13$?
5. What happens when the inequality involves a quadratic expression inside the absolute value?

Tip: When solving absolute value inequalities, always remember to break the problem into two separate inequalities!