The number of integral solutions of `1/x+1/y=1/6 " with " x le y " is " 'alpha'`. The value of `'alpha-6'` is ……….. .

Answer

To solve the equation \( \frac{1}{x} + \frac{1}{y} = \frac{1}{6} \) under the condition \( x \leq y \), we will follow these steps:

### Step 1: Rewrite the Equation
We start with the equation:
\[ \frac{1}{x} + \frac{1}{y} = \frac{1}{6} \]
This can be rewritten as:
\[ \frac{y + x}{xy} = \frac{1}{6} \]
Cross-multiplying gives:
\[ 6(y + x) = xy \]
Rearranging this, we have:
\[ xy - 6x - 6y = 0 \]

### Step 2: Factor the Equation
To factor the equation, we can rewrite it as:
\[ xy - 6x - 6y + 36 = 36 \]
This can be factored as:
\[ (x - 6)(y - 6) = 36 \]

### Step 3: Find Factor Pairs of 36
Next, we need to find the integer factor pairs of 36. The pairs are:
\[ (1, 36), (2, 18), (3, 12), (4, 9), (6, 6), (-1, -36), (-2, -18), (-3, -12), (-4, -9), (-6, -6) \]

### Step 4: Solve for \(x\) and \(y\)
For each factor pair \((a, b)\), we can set:
\[ x - 6 = a \quad \text{and} \quad y - 6 = b \]
Thus:
\[ x = a + 6 \quad \text{and} \quad y = b + 6 \]

Now we will calculate \(x\) and \(y\) for each factor pair:
1. For \( (1, 36) \): \( x = 1 + 6 = 7, y = 36 + 6 = 42 \) → \( (7, 42) \)
2. For \( (2, 18) \): \( x = 2 + 6 = 8, y = 18 + 6 = 24 \) → \( (8, 24) \)
3. For \( (3, 12) \): \( x = 3 + 6 = 9, y = 12 + 6 = 18 \) → \( (9, 18) \)
4. For \( (4, 9) \): \( x = 4 + 6 = 10, y = 9 + 6 = 15 \) → \( (10, 15) \)
5. For \( (6, 6) \): \( x = 6 + 6 = 12, y = 6 + 6 = 12 \) → \( (12, 12) \)
6. For \( (-1, -36) \): \( x = -1 + 6 = 5, y = -36 + 6 = -30 \) → \( (5, -30) \)
7. For \( (-2, -18) \): \( x = -2 + 6 = 4, y = -18 + 6 = -12 \) → \( (4, -12) \)
8. For \( (-3, -12) \): \( x = -3 + 6 = 3, y = -12 + 6 = -6 \) → \( (3, -6) \)
9. For \( (-4, -9) \): \( x = -4 + 6 = 2, y = -9 + 6 = -3 \) → \( (2, -3) \)
10. For \( (-6, -6) \): \( x = -6 + 6 = 0, y = -6 + 6 = 0 \) → \( (0, 0) \)

### Step 5: Filter Valid Solutions
Now we need to filter these solutions to find those that satisfy \( x \leq y \):
1. \( (7, 42) \)
2. \( (8, 24) \)
3. \( (9, 18) \)
4. \( (10, 15) \)
5. \( (12, 12) \)

The negative pairs do not satisfy \( x \leq y \) as they yield negative or zero values.

### Step 6: Count the Integral Solutions
We have found 5 valid solutions:
1. \( (7, 42) \)
2. \( (8, 24) \)
3. \( (9, 18) \)
4. \( (10, 15) \)
5. \( (12, 12) \)

Thus, the total number of integral solutions, denoted as \( \alpha \), is 5.

### Step 7: Calculate \( \alpha - 6 \)
Finally, we need to find \( \alpha - 6 \):
\[ \alpha - 6 = 5 - 6 = -1 \]

### Final Answer
The value of \( \alpha - 6 \) is \( -1 \).

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