To solve the inequality \( |2x - 1| \geq |x - 1| \), we can approach it by considering the different cases for the absolute values, based on the expressions inside them.

### Step 1: Identify critical points
The expressions inside the absolute values, \( 2x - 1 \) and \( x - 1 \), become zero when:
- \( 2x - 1 = 0 \) implies \( x = \frac{1}{2} \)
- \( x - 1 = 0 \) implies \( x = 1 \)

These points divide the number line into three intervals:
1. \( x < \frac{1}{2} \)
2. \( \frac{1}{2} \leq x < 1 \)
3. \( x \geq 1 \)

### Step 2: Analyze each interval
Let's consider the behavior of the absolute value expressions in each interval.

#### Case 1: \( x < \frac{1}{2} \)
In this interval:
- \( 2x - 1 < 0 \), so \( |2x - 1| = -(2x - 1) = -2x + 1 \)
- \( x - 1 < 0 \), so \( |x - 1| = -(x - 1) = -x + 1 \)

The inequality becomes:
\[
-2x + 1 \geq -x + 1
\]
Subtracting 1 from both sides:
\[
-2x \geq -x
\]
Adding \( 2x \) to both sides:
\[
0 \geq x
\]

Since \( x < \frac{1}{2} \), this inequality holds for all \( x \leq 0 \).

#### Case 2: \( \frac{1}{2} \leq x < 1 \)
In this interval:
- \( 2x - 1 \geq 0 \), so \( |2x - 1| = 2x - 1 \)
- \( x - 1 < 0 \), so \( |x - 1| = -x + 1 \)

The inequality becomes:
\[
2x - 1 \geq -x + 1
\]
Adding \( x \) to both sides:
\[
3x - 1 \geq 1
\]
Adding 1 to both sides:
\[
3x \geq 2
\]
Dividing by 3:
\[
x \geq \frac{2}{3}
\]

Since \( x \geq \frac{2}{3} \) and \( x < 1 \), the inequality holds for \( \frac{2}{3} \leq x < 1 \).

#### Case 3: \( x \geq 1 \)
In this interval:
- \( 2x - 1 \geq 0 \), so \( |2x - 1| = 2x - 1 \)
- \( x - 1 \geq 0 \), so \( |x - 1| = x - 1 \)

The inequality becomes:
\[
2x - 1 \geq x - 1
\]
Subtracting \( x \) from both sides:
\[
x \geq 0
\]

Since \( x \geq 1 \), this inequality holds for all \( x \geq 1 \).

### Step 3: Combine the intervals
- In \( x \leq 0 \)
- In \( \frac{2}{3} \leq x < 1 \)
- In \( x \geq 1 \)

Thus, the solution to the inequality \( |2x - 1| \geq |x - 1| \) is:
\[
x \leq 0 \quad \text{or} \quad x \geq \frac{2}{3}
\]

**Final Answer:** \( (-\infty, 0] \cup \left[\frac{2}{3}, \infty\right) \)

Would you like more details on this solution or have any other questions? 

Here are 5 related questions:
1. How do you solve inequalities involving absolute values?
2. What happens if you square both sides of an inequality with absolute values?
3. How do you graph the solution of \( |2x - 1| \geq |x - 1| \)?
4. Can absolute value inequalities have no solution?
5. How do absolute value inequalities differ from standard inequalities?

**Tip:** When dealing with absolute value inequalities, always remember to consider the critical points where the expressions inside the absolute values are zero; this helps to break the problem into manageable cases.

Learn how to solve the inequality |2x - 1| ≥ |x - 1| with a detailed step-by-step solution. This problem covers absolute value inequalities and critical point analysis, suitable for high school students.

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