We are given the equation:

\[
\frac{-10s}{s^2 - 25} - \frac{9}{s + 5} = \frac{1}{s - 5}
\]

### Step 1: Factor the denominator of the first term
Notice that \(s^2 - 25\) is a difference of squares, so we can factor it as:

\[
s^2 - 25 = (s + 5)(s - 5)
\]

Substituting this back into the equation, we get:

\[
\frac{-10s}{(s + 5)(s - 5)} - \frac{9}{s + 5} = \frac{1}{s - 5}
\]

### Step 2: Get a common denominator for the left-hand side
To combine the fractions on the left-hand side, we need to make the denominator the same. The second term \(\frac{9}{s + 5}\) has a denominator of \(s + 5\), so we need to multiply both the numerator and denominator of the second term by \(s - 5\) to make the denominators match.

\[
\frac{-10s}{(s + 5)(s - 5)} - \frac{9(s - 5)}{(s + 5)(s - 5)} = \frac{1}{s - 5}
\]

### Step 3: Combine the fractions on the left-hand side
Now that the denominators are the same, we can combine the two fractions on the left-hand side:

\[
\frac{-10s - 9(s - 5)}{(s + 5)(s - 5)} = \frac{1}{s - 5}
\]

Simplify the numerator:

\[
-10s - 9(s - 5) = -10s - 9s + 45 = -19s + 45
\]

So the equation becomes:

\[
\frac{-19s + 45}{(s + 5)(s - 5)} = \frac{1}{s - 5}
\]

### Step 4: Eliminate the denominators
To eliminate the denominators, multiply both sides of the equation by \((s + 5)(s - 5)\):

\[
-19s + 45 = 1(s + 5)
\]

### Step 5: Expand and solve for \(s\)
Now expand the right-hand side:

\[
-19s + 45 = s + 5
\]

Move all terms involving \(s\) to one side of the equation and constants to the other:

\[
-19s - s = 5 - 45
\]

Simplify:

\[
-20s = -40
\]

Solve for \(s\):

\[
s = \frac{-40}{-20} = 2
\]

### Solution:
The solution to the equation is:

\[
s = 2
\]

Would you like a more detailed explanation or have any questions about this solution?

Here are five related questions to explore:
1. How would you solve if the equation had an additional fraction on the right-hand side?
2. Can you solve similar equations involving higher degree polynomials in the denominator?
3. What happens if the denominator becomes a cubic polynomial?
4. How do you check for extraneous solutions when solving equations with fractions?
5. Can you derive alternative factorizations for more complex terms in the denominator?

**Tip:** Always check for extraneous solutions by substituting back into the original equation, especially when working with rational expressions.

This detailed solution explains how to solve the rational equation -10s/(s^2 - 25) - 9/(s + 5) = 1/(s - 5) by factoring, combining fractions, and solving linear equations. Learn step-by-step how to factor using the difference of squares and eliminate denominators. Suitable for Grades 10-12.

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