The solution set of the inequation `|2x-3| lt x-1`, is

Options

A. `(4//3, 3//2) cup (3//2, 2)`
B. `(4//3, 2)`
C. `[3//2, 2)`
D. none of these

Explanation

To solve the inequality \( |2x - 3| < x - 1 \), we will break it down into different cases based on the properties of absolute values.

### Step 1: Identify the critical point
The expression inside the absolute value is \( 2x - 3 \). We first find the critical point where \( 2x - 3 = 0 \):
\[ 2x - 3 = 0 \implies 2x = 3 \implies x = \frac{3}{2} \]
**Hint:** The critical point helps us determine the intervals to test.

### Step 2: Consider the intervals
We will consider three intervals based on the critical point \( x = \frac{3}{2} \):
1. \( x < \frac{3}{2} \)
2. \( x = \frac{3}{2} \)
3. \( x > \frac{3}{2} \)

### Step 3: Case 1: \( x < \frac{3}{2} \)
In this case, \( 2x - 3 < 0 \), so \( |2x - 3| = -(2x - 3) = -2x + 3 \). The inequality becomes:
\[ -2x + 3 < x - 1 \]
Rearranging gives:
\[ 3 + 1 < x + 2x \implies 4 < 3x \implies x > \frac{4}{3} \]
Now, we combine this with the condition \( x < \frac{3}{2} \):
\[ \frac{4}{3} < x < \frac{3}{2} \]
**Hint:** Check if the range \( \frac{4}{3} \) to \( \frac{3}{2} \) is valid.

### Step 4: Case 2: \( x = \frac{3}{2} \)
Substituting \( x = \frac{3}{2} \) into the original inequality:
\[ |2(\frac{3}{2}) - 3| < \frac{3}{2} - 1 \]
This simplifies to:
\[ |3 - 3| < \frac{1}{2} \implies 0 < \frac{1}{2} \]
This condition is true, so \( x = \frac{3}{2} \) is included in the solution set.
**Hint:** Verify if the equality holds true.

### Step 5: Case 3: \( x > \frac{3}{2} \)
In this case, \( 2x - 3 > 0 \), so \( |2x - 3| = 2x - 3 \). The inequality becomes:
\[ 2x - 3 < x - 1 \]
Rearranging gives:
\[ 2x - x < -1 + 3 \implies x < 2 \]
Now, we combine this with the condition \( x > \frac{3}{2} \):
\[ \frac{3}{2} < x < 2 \]
**Hint:** Ensure the range \( \frac{3}{2} \) to \( 2 \) is valid.

### Step 6: Combine all intervals
From the three cases, we have:
1. From Case 1: \( \frac{4}{3} < x < \frac{3}{2} \)
2. From Case 2: \( x = \frac{3}{2} \)
3. From Case 3: \( \frac{3}{2} < x < 2 \)

Combining these intervals, we get:
\[ x \in \left(\frac{4}{3}, 2\right) \]
**Final Answer:** The solution set is \( x \in \left(\frac{4}{3}, 2\right) \).