`(x^(2)-y^(2))=16` and `xy=-15`. Find the value of `(x+y)` if `(x+y)` is a positive number.

Options

A. 3
B. 5
C. 2
D. none of these

Answer

To solve the equations \(x^2 - y^2 = 16\) and \(xy = -15\) and find the value of \(x + y\) (where \(x + y\) is a positive number), we can follow these steps:

### Step 1: Rewrite the equations
We have two equations:
1. \(x^2 - y^2 = 16\)
2. \(xy = -15\)

### Step 2: Express \(y\) in terms of \(x\)
From the second equation, we can express \(y\) in terms of \(x\):
\[ y = \frac{-15}{x} \]

### Step 3: Substitute \(y\) into the first equation
Now, substitute \(y\) into the first equation:
\[ x^2 - \left(\frac{-15}{x}\right)^2 = 16 \]
This simplifies to:
\[ x^2 - \frac{225}{x^2} = 16 \]

### Step 4: Eliminate the fraction
Multiply through by \(x^2\) to eliminate the fraction:
\[ x^4 - 225 = 16x^2 \]

### Step 5: Rearrange the equation
Rearranging gives us:
\[ x^4 - 16x^2 - 225 = 0 \]

### Step 6: Let \(u = x^2\)
Let \(u = x^2\). The equation becomes:
\[ u^2 - 16u - 225 = 0 \]

### Step 7: Factor or use the quadratic formula
We can factor this quadratic equation. We look for two numbers that multiply to \(-225\) and add to \(-16\). The factors are \(-25\) and \(9\):
\[ (u - 25)(u + 9) = 0 \]

### Step 8: Solve for \(u\)
Setting each factor to zero gives:
\[ u - 25 = 0 \quad \Rightarrow \quad u = 25 \]
\[ u + 9 = 0 \quad \Rightarrow \quad u = -9 \quad (\text{not valid since } u = x^2 \geq 0) \]

### Step 9: Find \(x\)
Since \(u = x^2\), we have:
\[ x^2 = 25 \quad \Rightarrow \quad x = 5 \quad \text{or} \quad x = -5 \]

### Step 10: Find corresponding \(y\) values
Using \(xy = -15\):
1. If \(x = 5\):
\[ 5y = -15 \quad \Rightarrow \quad y = -3 \]
2. If \(x = -5\):
\[ -5y = -15 \quad \Rightarrow \quad y = 3 \]

### Step 11: Calculate \(x + y\)
Now calculate \(x + y\) for both cases:
1. \(x = 5\) and \(y = -3\):
\[ x + y = 5 - 3 = 2 \quad (\text{positive}) \]
2. \(x = -5\) and \(y = 3\):
\[ x + y = -5 + 3 = -2 \quad (\text{not positive}) \]

### Conclusion
The only valid solution where \(x + y\) is positive is:
\[ \boxed{2} \]