Suppose `xy-5x+2y=30,` where x and y are positive integers. Find the number of possible values of x.

Answer

To solve the equation \( xy - 5x + 2y = 30 \) for positive integer values of \( x \) and \( y \), we can follow these steps:

### Step 1: Rearranging the Equation
Start with the original equation:
\[ xy - 5x + 2y = 30 \]
Rearranging gives:
\[ xy + 2y - 5x = 30 \]

### Step 2: Factoring
We can factor out \( x \) from the first two terms:
\[ y(x + 2) - 5x = 30 \]
Now, isolate \( y \):
\[ y(x + 2) = 30 + 5x \]
Thus,
\[ y = \frac{30 + 5x}{x + 2} \]

### Step 3: Finding Integer Solutions
For \( y \) to be a positive integer, \( 30 + 5x \) must be divisible by \( x + 2 \). We can express this condition mathematically:
\[ 30 + 5x \equiv 0 \mod (x + 2) \]

### Step 4: Analyzing Divisibility
Let’s simplify \( 30 + 5x \):
\[ 30 + 5x = 5(x + 6) \]
This means \( x + 2 \) must be a divisor of \( 5(x + 6) \).

### Step 5: Finding Divisors
The divisors of \( 30 \) (since \( 5 \) is a constant factor) are \( 1, 2, 3, 5, 6, 10, 15, 30 \). We can set \( x + 2 \) equal to each divisor and solve for \( x \).

### Step 6: Setting Up the Equations
1. \( x + 2 = 1 \) → \( x = -1 \) (not positive)
2. \( x + 2 = 2 \) → \( x = 0 \) (not positive)
3. \( x + 2 = 3 \) → \( x = 1 \) (positive)
4. \( x + 2 = 5 \) → \( x = 3 \) (positive)
5. \( x + 2 = 6 \) → \( x = 4 \) (positive)
6. \( x + 2 = 10 \) → \( x = 8 \) (positive)
7. \( x + 2 = 15 \) → \( x = 13 \) (positive)
8. \( x + 2 = 30 \) → \( x = 28 \) (positive)

### Step 7: Valid Positive Integer Values
The valid positive integer values of \( x \) are \( 1, 3, 4, 8, 13, 28 \).

### Step 8: Counting the Values
Now, we count the valid values:
- \( x = 1 \)
- \( x = 3 \)
- \( x = 4 \)
- \( x = 8 \)
- \( x = 13 \)
- \( x = 28 \)

Thus, there are **6 possible values of \( x \)**.

### Final Answer
The number of possible values of \( x \) is **6**.