$$ \begin{vmatrix}2x-1\\ \end{vmatrix} $$$$ \ge\left\vert x-1\right\vert $$

To solve the inequality $ \begin{vmatrix}2x-1\\ \end{vmatrix} \ge \left\vert x-1 \right\vert $, let's break it down by considering the absolute values separately.

### 1. Consider the two cases for $ |2x - 1| $:

\[
|2x - 1| = \begin{cases}
2x - 1 & \text{if } 2x - 1 \geq 0 \\
-(2x - 1) & \text{if } 2x - 1 < 0
\end{cases}
\]

This leads to two scenarios:
1. $ 2x - 1 \geq 0 $ implies $ x \geq \frac{1}{2} $.
2. $ 2x - 1 < 0 $ implies $ x < \frac{1}{2} $.

### 2. Consider the two cases for $ |x - 1| $:

\[
|x - 1| = \begin{cases}
x - 1 & \text{if } x - 1 \geq 0 \\
-(x - 1) & \text{if } x - 1 < 0
\end{cases}
\]

This leads to two scenarios:
1. $ x - 1 \geq 0 $ implies $ x \geq 1 $.
2. $ x - 1 < 0 $ implies $ x < 1 $.

### 3. Analyzing each case:

#### **Case 1: $ x \geq 1 $**

Here, $ |2x - 1| = 2x - 1 $ and $ |x - 1| = x - 1 $.

The inequality becomes:
\[
2x - 1 \geq x - 1
\]
Simplifying:
\[
x \geq 0
\]
Since $ x \geq 1 $ already, this inequality is satisfied.

#### **Case 2: $ \frac{1}{2} \leq x < 1 $**

Here, $ |2x - 1| = 2x - 1 $ and $ |x - 1| = -(x - 1) = 1 - x $.

The inequality becomes:
\[
2x - 1 \geq 1 - x
\]
Simplifying:
\[
3x \geq 2 \implies x \geq \frac{2}{3}
\]
So, $ \frac{2}{3} \leq x < 1 $ satisfies this case.

#### **Case 3: $ x < \frac{1}{2} $**

Here, $ |2x - 1| = -(2x - 1) = 1 - 2x $ and $ |x - 1| = 1 - x $.

The inequality becomes:
\[
1 - 2x \geq 1 - x
\]
Simplifying:
\[
-x \geq 0 \implies x \leq 0
\]
So, $ 0 \leq x < \frac{1}{2} $ satisfies this case.

### 4. Combine the intervals:

- Case 1 gives $ x \geq 1 $.
- Case 2 gives $ \frac{2}{3} \leq x < 1 $.
- Case 3 gives $ 0 \leq x < \frac{1}{2} $.

### Final Solution:

The solution set is the union of all valid intervals:
\[
x \in [0, \frac{1}{2}) \cup [\frac{2}{3}, \infty)
\]

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

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**Related Questions:**
1. How do absolute value inequalities differ from standard inequalities?
2. Can you provide examples of other types of absolute value inequalities?
3. What are the practical applications of solving absolute value inequalities?
4. How would the solution change if the inequality was a strict inequality ($>$ instead of $ \geq $)?
5. How do we solve absolute value inequalities with more than two terms?

**Tip:** When dealing with absolute value inequalities, always break them into cases based on the definition of the absolute value function. This approach helps in solving the inequality correctly for different intervals.

Learn how to solve the absolute value inequality |2x - 1| >= |x - 1| with a detailed step-by-step solution. This problem covers absolute value inequalities and involves breaking down the inequality into cases based on the definitions of absolute values. Suitable for advanced high school students and anyone preparing for college-level mathematics.

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