Solve the following inequalities:
`(x^(2)-3x-18)/(13x-x^(2)-42)gt0`

Answer

To solve the inequality \((x^2 - 3x - 18) / (13x - x^2 - 42) > 0\), we will follow these steps:

### Step 1: Factor the numerator and the denominator

**Numerator:**
We need to factor \(x^2 - 3x - 18\).
To factor this, we look for two numbers that multiply to \(-18\) and add to \(-3\). The numbers \(-6\) and \(3\) work.

So, we can write:
\[ x^2 - 3x - 18 = (x - 6)(x + 3) \]

**Denominator:**
Now, we will factor \(13x - x^2 - 42\). We can rearrange it as:
\[ -x^2 + 13x - 42 = -(x^2 - 13x + 42) \]
Next, we need to factor \(x^2 - 13x + 42\). We look for two numbers that multiply to \(42\) and add to \(-13\). The numbers \(-6\) and \(-7\) work.

So, we can write:
\[ x^2 - 13x + 42 = (x - 6)(x - 7) \]
Thus, the denominator becomes:
\[ -(x - 6)(x - 7) \]

### Step 2: Rewrite the inequality

Now we can rewrite the inequality:
\[ \frac{(x - 6)(x + 3)}{-(x - 6)(x - 7)} > 0 \]
This simplifies to:
\[ \frac{(x - 6)(x + 3)}{-(x - 6)(x - 7)} > 0 \]

### Step 3: Determine the critical points

The critical points occur where the numerator or the denominator is zero:
1. From the numerator: \(x - 6 = 0\) gives \(x = 6\) and \(x + 3 = 0\) gives \(x = -3\).
2. From the denominator: \(x - 6 = 0\) gives \(x = 6\) and \(x - 7 = 0\) gives \(x = 7\).

Thus, the critical points are \(x = -3\), \(x = 6\), and \(x = 7\).

### Step 4: Test intervals

We will test the intervals determined by these critical points:
- \( (-\infty, -3) \)
- \( (-3, 6) \)
- \( (6, 7) \)
- \( (7, \infty) \)

**Interval 1: \( (-\infty, -3) \)**
Choose \(x = -4\):
\[ \frac{(-4 - 6)(-4 + 3)}{-(-4 - 6)(-4 - 7)} = \frac{(-10)(-1)}{-(-10)(-11)} = \frac{10}{110} > 0 \]

**Interval 2: \( (-3, 6) \)**
Choose \(x = 0\):
\[ \frac{(0 - 6)(0 + 3)}{- (0 - 6)(0 - 7)} = \frac{(-6)(3)}{-(-6)(-7)} = \frac{-18}{42} < 0 \]

**Interval 3: \( (6, 7) \)**
Choose \(x = 6.5\):
\[ \frac{(6.5 - 6)(6.5 + 3)}{- (6.5 - 6)(6.5 - 7)} = \frac{(0.5)(9.5)}{- (0.5)(-0.5)} = \frac{4.75}{0.25} > 0 \]

**Interval 4: \( (7, \infty) \)**
Choose \(x = 8\):
\[ \frac{(8 - 6)(8 + 3)}{- (8 - 6)(8 - 7)} = \frac{(2)(11)}{- (2)(1)} = \frac{22}{-2} < 0 \]

### Step 5: Combine results

From our testing:
- The inequality is satisfied in the intervals \( (-\infty, -3) \) and \( (6, 7) \).

### Step 6: Write the final solution

Thus, the solution to the inequality is:
\[ x \in (-\infty, -3] \cup (6, 7) \]