`(x-3)/(x-5) > 0` then x belongs to -

Options

A. (-infty,3)cup(5,infty)`
B. (-infty,-3)cup(-5,infty)`
C. (-infty,3)cup(5,infty)`
D. (3,5)

Answer

To solve the inequality \((x-3)/(x-5) > 0\), we will follow these steps:

### Step 1: Identify the critical points
The critical points occur when the numerator and denominator are equal to zero.

1. Set the numerator \(x - 3 = 0\) which gives us:
\[ x = 3 \]
2. Set the denominator \(x - 5 = 0\) which gives us:
\[ x = 5 \]

### Step 2: Determine the intervals
The critical points divide the number line into intervals. The intervals are:
- \((-\infty, 3)\)
- \((3, 5)\)
- \((5, \infty)\)

### Step 3: Test each interval
We will test a point from each interval to determine where the inequality holds true.

1. **Interval \((-\infty, 3)\)**: Choose \(x = 0\)
\[ \frac{0 - 3}{0 - 5} = \frac{-3}{-5} = \frac{3}{5} > 0 \quad \text{(True)} \]

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

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

### Step 4: Combine the results
From our tests, the inequality \((x-3)/(x-5) > 0\) holds true in the intervals \((-\infty, 3)\) and \((5, \infty)\).

### Step 5: Write the solution set
Since the inequality does not include equality (i.e., it is strictly greater than), we do not include the critical points \(x = 3\) and \(x = 5\).

Thus, the solution set is:
\[ x \in (-\infty, 3) \cup (5, \infty) \]

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