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.