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The question asks us to simplify the following expression given that \( x \neq 3 \), \( y \neq -5 \), and \( xy = 15 \):

\[
\frac{1}{x - 3} + \frac{1}{y + 5}
\]

### Step 1: Finding a common denominator

To add the fractions, we need to find a common denominator. The denominators are \( x - 3 \) and \( y + 5 \), so the common denominator will be their product: 

\[
(x - 3)(y + 5)
\]

Thus, we rewrite each fraction:

\[
\frac{1}{x - 3} = \frac{y + 5}{(x - 3)(y + 5)}
\]
\[
\frac{1}{y + 5} = \frac{x - 3}{(x - 3)(y + 5)}
\]

### Step 2: Add the fractions

Now we add the numerators while keeping the common denominator:

\[
\frac{1}{x - 3} + \frac{1}{y + 5} = \frac{(y + 5) + (x - 3)}{(x - 3)(y + 5)}
\]

Simplifying the numerator:

\[
(y + 5) + (x - 3) = x + y + 2
\]

So the expression becomes:

\[
\frac{x + y + 2}{(x - 3)(y + 5)}
\]

### Step 3: Match the result with the options

This matches **option B**, which is:

\[
\boxed{B} \quad \frac{x + y + 2}{(x - 3)(y + 5)}
\]

Would you like more details on any step or have any other questions?

Here are some related questions you might explore:
1. How do you handle expressions with more than two fractions?
2. What are the properties of rational expressions?
3. How would the process change if \(xy\) were not provided?
4. What would happen if one of the restrictions \(x \neq 3\) or \(y \neq -5\) were not applied?
5. Can you explain how the Least Common Denominator works in more complex examples?

**Tip**: Always check for restrictions on variables to avoid invalid expressions (such as dividing by zero).

If x ≠ 3, y ≠ -5, and xy = 15, which of the following is equivalent to 1 / (1 / (x - 3) + 1 / (y + 5))?

Learn how to simplify the expression 1 / (1 / (x - 3) + 1 / (y + 5)) with given conditions x ≠ 3, y ≠ -5, and xy = 15. This problem covers rational expressions and finding common denominators, suitable for students in Grades 9-12.

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