The solution of the inequality `((x+7)/(x-5)+(3x+1)/(2) ge 0 ` is

Options

A. `[1,3] cup (5,oo)`
B. `(1,3) cup (5,oo)`
C. `(-oo,1) cup (5,oo)`
D. None of these

Explanation

To solve the inequality

\[ \frac{x+7}{x-5} + \frac{3x+1}{2} \geq 0, \]

we will follow these steps:

### Step 1: Combine the fractions
First, we need to find a common denominator for the two fractions. The common denominator is \(2(x - 5)\). Therefore, we can rewrite the inequality as:

\[ \frac{2(x + 7) + (3x + 1)(x - 5)}{2(x - 5)} \geq 0. \]

### Step 2: Expand the numerator
Now, we will expand the numerator:

\[ 2(x + 7) = 2x + 14, \]
and
\[ (3x + 1)(x - 5) = 3x^2 - 15x + x - 5 = 3x^2 - 14x - 5. \]

So, the numerator becomes:

\[ 2x + 14 + 3x^2 - 14x - 5 = 3x^2 - 12x + 9. \]

### Step 3: Rewrite the inequality
Now, we can rewrite the inequality as:

\[ \frac{3x^2 - 12x + 9}{2(x - 5)} \geq 0. \]

### Step 4: Factor the numerator
Next, we will factor the quadratic expression in the numerator:

\[ 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x - 1)(x - 3). \]

Thus, the inequality becomes:

\[ \frac{3(x - 1)(x - 3)}{2(x - 5)} \geq 0. \]

### Step 5: Identify critical points
The critical points from the factors are \(x = 1\), \(x = 3\), and \(x = 5\). These points will divide the number line into intervals.

### Step 6: Test intervals
We will test the sign of the expression in each interval determined by the critical points:

1. **Interval \((-∞, 1)\)**: Choose \(x = 0\):
\[ \frac{3(0 - 1)(0 - 3)}{2(0 - 5)} = \frac{3(-1)(-3)}{2(-5)} = \frac{9}{-10} < 0. \]

2. **Interval \((1, 3)\)**: Choose \(x = 2\):
\[ \frac{3(2 - 1)(2 - 3)}{2(2 - 5)} = \frac{3(1)(-1)}{2(-3)} = \frac{-3}{-6} > 0. \]

3. **Interval \((3, 5)\)**: Choose \(x = 4\):
\[ \frac{3(4 - 1)(4 - 3)}{2(4 - 5)} = \frac{3(3)(1)}{2(-1)} = \frac{9}{-2} < 0. \]

4. **Interval \((5, ∞)\)**: Choose \(x = 6\):
\[ \frac{3(6 - 1)(6 - 3)}{2(6 - 5)} = \frac{3(5)(3)}{2(1)} = \frac{45}{2} > 0. \]

### Step 7: Determine the solution set
From the tests, we find that the expression is non-negative in the intervals \([1, 3]\) and \((5, ∞)\).

### Step 8: Include endpoints
Since the inequality is \(\geq 0\), we include the endpoints where the expression is zero:

- At \(x = 1\) and \(x = 3\), the expression equals zero.
- At \(x = 5\), the expression is undefined.

Thus, the solution set is:

\[ x \in [1, 3] \cup (5, \infty). \]

### Final Answer
The solution of the inequality is:

\[ x \in [1, 3] \cup (5, \infty). \]