If `(2x+1)/(x-2) le 1 , x in R ` , then the solution set is

Options

A. `( -oo,1/2)`
B. `(- oo,-1/2]`
C. `[-3,2]`
D. `[-3,2)`

Answer

To solve the inequality \(\frac{2x + 1}{x - 2} \leq 1\), we will follow these steps:

### Step 1: Rewrite the Inequality
We start with the inequality:
\[ \frac{2x + 1}{x - 2} \leq 1 \]

### Step 2: Move All Terms to One Side
We can subtract 1 from both sides:
\[ \frac{2x + 1}{x - 2} - 1 \leq 0 \]

### Step 3: Combine the Fractions
To combine the fractions, we need a common denominator:
\[ \frac{2x + 1 - (x - 2)}{x - 2} \leq 0 \]
This simplifies to:
\[ \frac{2x + 1 - x + 2}{x - 2} \leq 0 \]
\[ \frac{x + 3}{x - 2} \leq 0 \]

### Step 4: Find Critical Points
Now, we need to find the critical points where the expression is equal to zero or undefined. The critical points are:
1. \(x + 3 = 0 \Rightarrow x = -3\)
2. \(x - 2 = 0 \Rightarrow x = 2\) (undefined)

### Step 5: Test Intervals
We will test the intervals determined by the critical points \(-3\) and \(2\):
- Interval 1: \( (-\infty, -3) \)
- Interval 2: \( (-3, 2) \)
- Interval 3: \( (2, \infty) \)

**Testing Interval 1: \(x = -4\)**
\[ \frac{-4 + 3}{-4 - 2} = \frac{-1}{-6} > 0 \quad \text{(not a solution)} \]

**Testing Interval 2: \(x = 0\)**
\[ \frac{0 + 3}{0 - 2} = \frac{3}{-2} < 0 \quad \text{(solution)} \]

**Testing Interval 3: \(x = 3\)**
\[ \frac{3 + 3}{3 - 2} = \frac{6}{1} > 0 \quad \text{(not a solution)} \]

### Step 6: Include Critical Points
The inequality is less than or equal to zero, so we include the point where the expression equals zero:
- At \(x = -3\), \(\frac{-3 + 3}{-3 - 2} = 0\) (included)
- At \(x = 2\), the expression is undefined (not included)

### Final Solution Set
Thus, the solution set is:
\[ x \in [-3, 2) \]

### Summary
The final answer is:
\[ \text{Solution set: } [-3, 2) \]