Solve the following inequality *algebraically*.

vertical bar, x, minus, 1, vertical bar, is greater than or equal to, 13

∣*x*−1∣≥13

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!

Solve the following inequality algebraically: ∣x−1∣≥13

Discover how to solve the absolute value inequality ∣x−1∣≥13 algebraically. This step-by-step guide breaks down the process using absolute value inequality theorems and key algebraic concepts. Perfect for students in Grades 8-10.

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