The solution set of the inequation `(4x + 3)/(2x-5) lt 6`, is

Options

A. `(5//2, 33//8)`
B. `(-oo, 5//2)cup (33//8, oo)`
C. `(5//2, oo)`
D. `(33//8, oo)`

Answer

To solve the inequality \(\frac{4x + 3}{2x - 5} < 6\), we will follow these steps:

### Step 1: Rearranging the Inequality
First, we will rearrange the inequality to bring all terms to one side:
\[ \frac{4x + 3}{2x - 5} - 6 < 0 \]

### Step 2: Finding a Common Denominator
Next, we need to combine the terms on the left-hand side. We will express \(6\) as a fraction with the same denominator:
\[ \frac{4x + 3 - 6(2x - 5)}{2x - 5} < 0 \]
This simplifies to:
\[ \frac{4x + 3 - (12x - 30)}{2x - 5} < 0 \]

### Step 3: Simplifying the Numerator
Now, we will simplify the numerator:
\[ 4x + 3 - 12x + 30 = -8x + 33 \]
So, we have:
\[ \frac{-8x + 33}{2x - 5} < 0 \]

### Step 4: Rewriting the Inequality
We can rewrite the inequality as:
\[ \frac{8x - 33}{2x - 5} > 0 \]

### Step 5: Finding Critical Points
Next, we will find the critical points by setting the numerator and denominator to zero:
1. For the numerator: \(8x - 33 = 0 \Rightarrow x = \frac{33}{8}\)
2. For the denominator: \(2x - 5 = 0 \Rightarrow x = \frac{5}{2}\)

### Step 6: Testing Intervals
Now we will test the intervals determined by these critical points:
- \( (-\infty, \frac{5}{2}) \)
- \( (\frac{5}{2}, \frac{33}{8}) \)
- \( (\frac{33}{8}, \infty) \)

1. **Interval \( (-\infty, \frac{5}{2}) \)**: Choose \(x = 0\):
\[ \frac{8(0) - 33}{2(0) - 5} = \frac{-33}{-5} > 0 \quad \text{(True)} \]

2. **Interval \( (\frac{5}{2}, \frac{33}{8}) \)**: Choose \(x = 3\):
\[ \frac{8(3) - 33}{2(3) - 5} = \frac{24 - 33}{6 - 5} = \frac{-9}{1} < 0 \quad \text{(False)} \]

3. **Interval \( (\frac{33}{8}, \infty) \)**: Choose \(x = 5\):
\[ \frac{8(5) - 33}{2(5) - 5} = \frac{40 - 33}{10 - 5} = \frac{7}{5} > 0 \quad \text{(True)} \]

### Step 7: Conclusion
The solution set where the inequality holds true is:
\[ (-\infty, \frac{5}{2}) \cup (\frac{33}{8}, \infty) \]

### Final Answer
Thus, the solution set of the inequality \(\frac{4x + 3}{2x - 5} < 6\) is:
\[ (-\infty, \frac{5}{2}) \cup (\frac{33}{8}, \infty) \]