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
The denominator is \(x^2 + 5x - 14\). We can factor this quadratic expression.

To factor \(x^2 + 5x - 14\), we look for two numbers that multiply to \(-14\) and add to \(5\). These numbers are \(7\) and \(-2\).

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 critical points by setting the numerator and denominator to zero:
1. \(x - 5 = 0 \Rightarrow x = 5\)
2. \(x + 7 = 0 \Rightarrow x = -7\)
3. \(x - 2 = 0 \Rightarrow x = 2\)

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

### Step 4: Determine the intervals
The critical points divide the number line into the following intervals:
1. \((-∞, -7)\)
2. \((-7, 2)\)
3. \((2, 5)\)
4. \((5, ∞)\)

### Step 5: Test each interval
We will test a point from each interval to see where the inequality holds.

1. **Interval \((-∞, -7)\)**: Choose \(x = -8\)
\[ \frac{-8 - 5}{(-8 + 7)(-8 - 2)} = \frac{-13}{(-1)(-10)} = \frac{-13}{10} < 0 \quad \text{(not satisfied)} \]

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

3. **Interval \((2, 5)\)**: Choose \(x = 3\)
\[ \frac{3 - 5}{(3 + 7)(3 - 2)} = \frac{-2}{10 \cdot 1} = \frac{-2}{10} < 0 \quad \text{(not satisfied)} \]

4. **Interval \((5, ∞)\)**: Choose \(x = 6\)
\[ \frac{6 - 5}{(6 + 7)(6 - 2)} = \frac{1}{13 \cdot 4} = \frac{1}{52} > 0 \quad \text{(satisfied)} \]

### Step 6: Combine results
From our tests, the inequality \(\frac{x - 5}{(x + 7)(x - 2)} > 0\) is satisfied in the intervals \((-7, 2)\) and \((5, ∞)\).

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

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