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To solve for \( f(g(x)) \) and \( g(f(x)) \), we can follow these steps with the provided functions:

1. **Given Functions:**
   - \( f(x) = \frac{1}{x - 3} \)
   - \( g(x) = \frac{8}{x} + 3 \)

### Step 1: Finding \( f(g(x)) \)

To find \( f(g(x)) \), we substitute \( g(x) \) into \( f(x) \):
\[
f(g(x)) = f\left(\frac{8}{x} + 3\right)
\]
Substitute \( \frac{8}{x} + 3 \) for \( x \) in \( f(x) = \frac{1}{x - 3} \):
\[
f(g(x)) = \frac{1}{\left(\frac{8}{x} + 3\right) - 3}
\]
Simplify inside the denominator:
\[
f(g(x)) = \frac{1}{\frac{8}{x} + 3 - 3} = \frac{1}{\frac{8}{x}} = \frac{x}{8}
\]
So,
\[
f(g(x)) = \frac{x}{8}
\]

### Step 2: Finding \( g(f(x)) \)

To find \( g(f(x)) \), we substitute \( f(x) \) into \( g(x) \):
\[
g(f(x)) = g\left(\frac{1}{x - 3}\right)
\]
Substitute \( \frac{1}{x - 3} \) for \( x \) in \( g(x) = \frac{8}{x} + 3 \):
\[
g(f(x)) = \frac{8}{\frac{1}{x - 3}} + 3
\]
Simplify the fraction:
\[
g(f(x)) = 8(x - 3) + 3
\]
Expanding,
\[
g(f(x)) = 8x - 24 + 3 = 8x - 21
\]
So,
\[
g(f(x)) = 8x - 21
\]

### Final Answers:
- \( f(g(x)) = \frac{x}{8} \)
- \( g(f(x)) = 8x - 21 \)

Would you like more detailed steps or further explanations? 

### Related Questions:
1. How do we find compositions for more complex functions?
2. What are some common applications of function compositions in real-world problems?
3. How would we verify if two functions are inverses by using compositions?
4. Can you explain the concept of domain and range in the context of function compositions?
5. What happens if we reverse the order in a function composition?

### Tip:
When working with compositions, always substitute carefully and simplify step by step to avoid errors.

Use the pair of functions to find f(g(x)) and g(f(x)). Simplify your answers. Given f(x) = 1/(x - 3) and g(x) = 8/x + 3.

Step-by-step solution for finding f(g(x)) and g(f(x)) for the functions f(x) = 1/(x - 3) and g(x) = 8/x + 3. This problem covers key algebraic concepts and is suitable for high school students.

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