Find the smallest integral x satisfying the inequality `(x-5)/(x^(2) + 5x -14) gt 0`.

Answer

To solve the inequality \(\frac{x-5}{x^2 + 5x - 14} > 0\), we will follow these steps:

### Step 1: Factor the Denominator
First, we need to factor the quadratic expression in the denominator \(x^2 + 5x - 14\).

To factor \(x^2 + 5x - 14\), we look for two numbers that multiply to \(-14\) (the constant term) and add to \(5\) (the coefficient of \(x\)). The numbers \(7\) and \(-2\) satisfy this condition.

Thus, we can write:
\[ x^2 + 5x - 14 = (x + 7)(x - 2) \]

### Step 2: Rewrite the Inequality
Now we can rewrite the inequality:
\[ \frac{x-5}{(x + 7)(x - 2)} > 0 \]

### Step 3: Identify Critical Points
Next, we find the points where the expression is equal to zero or undefined. This occurs when the numerator or denominator is zero.

1. **Numerator**: \(x - 5 = 0 \Rightarrow x = 5\)
2. **Denominator**:
- \(x + 7 = 0 \Rightarrow x = -7\)
- \(x - 2 = 0 \Rightarrow x = 2\)

So, the critical points are \(x = -7\), \(x = 2\), and \(x = 5\).

### Step 4: Test Intervals
We will test the sign of the expression in the intervals defined by these critical points:
- \((- \infty, -7)\)
- \((-7, 2)\)
- \((2, 5)\)
- \((5, \infty)\)

1. **Interval \((- \infty, -7)\)**: Choose \(x = -8\)
\[ \frac{-8-5}{(-8+7)(-8-2)} = \frac{-13}{(-1)(-10)} = \frac{-13}{10} < 0 \]

2. **Interval \((-7, 2)\)**: Choose \(x = 0\)
\[ \frac{0-5}{(0+7)(0-2)} = \frac{-5}{(7)(-2)} = \frac{-5}{-14} > 0 \]

3. **Interval \((2, 5)\)**: Choose \(x = 3\)
\[ \frac{3-5}{(3+7)(3-2)} = \frac{-2}{(10)(1)} = \frac{-2}{10} < 0 \]

4. **Interval \((5, \infty)\)**: Choose \(x = 6\)
\[ \frac{6-5}{(6+7)(6-2)} = \frac{1}{(13)(4)} > 0 \]

### Step 5: Combine Results
From our tests, the expression is positive in the intervals:
- \((-7, 2)\)
- \((5, \infty)\)

### Step 6: Determine Valid Intervals
Since we are looking for values where the expression is greater than zero, we exclude the critical points where the expression is zero or undefined:
- We cannot include \(x = -7\), \(x = 2\), or \(x = 5\).

Thus, the solution set is:
\[ (-7, 2) \cup (5, \infty) \]

### Step 7: Find the Smallest Integral Value
The smallest integer in the interval \((-7, 2)\) is \(-6\).

### Final Answer
The smallest integral \(x\) satisfying the inequality is:
\[ \boxed{-6} \]