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The problem asks us to find \( g(f(x)) \), where:

\[
f(x) = \frac{x + 5}{2x + 1}
\]
and
\[
g(x) = \frac{5 - x}{2x - 1}
\]

### Steps to solve:

1. Start by substituting \( f(x) \) into \( g(x) \).
   We are asked to find \( g(f(x)) \), which means we substitute \( f(x) \) wherever \( x \) appears in \( g(x) \).

2. Substituting \( f(x) = \frac{x + 5}{2x + 1} \) into \( g(x) = \frac{5 - x}{2x - 1} \):

\[
g(f(x)) = \frac{5 - \frac{x + 5}{2x + 1}}{2\left( \frac{x + 5}{2x + 1} \right) - 1}
\]

3. Simplify the numerator:
\[
5 - \frac{x + 5}{2x + 1} = \frac{5(2x + 1)}{2x + 1} - \frac{x + 5}{2x + 1} = \frac{10x + 5 - x - 5}{2x + 1} = \frac{9x}{2x + 1}
\]

4. Simplify the denominator:
\[
2\left( \frac{x + 5}{2x + 1} \right) - 1 = \frac{2(x + 5)}{2x + 1} - 1 = \frac{2(x + 5) - (2x + 1)}{2x + 1}
\]
\[
= \frac{2x + 10 - 2x - 1}{2x + 1} = \frac{9}{2x + 1}
\]

5. Therefore, the expression for \( g(f(x)) \) becomes:

\[
g(f(x)) = \frac{\frac{9x}{2x + 1}}{\frac{9}{2x + 1}} = x
\]

### Final Answer:
\[
g(f(x)) = x
\]

Would you like more details or clarifications on any step?

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Here are 5 questions for further exploration:
1. How can you interpret the composition of functions graphically?
2. What is the domain of the function \( f(x) \)?
3. What is the domain of the function \( g(f(x)) \)?
4. How would you verify if \( f \) and \( g \) are inverses of each other?
5. Can you generalize this process to find \( g(f(h(x))) \) for another function \( h(x) \)?

**Tip**: Always check the domain when composing functions, especially if the denominator could cause division by zero.

What is g(f(x)) where f(x) = (x + 5) / (2x + 1) and g(x) = (5 - x) / (2x - 1)?

Learn how to compute the composition of two functions, f(x) = (x + 5) / (2x + 1) and g(x) = (5 - x) / (2x - 1), step by step. This algebraic concept is commonly taught in Grades 10-12 and is part of understanding function transformations and inverses.

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