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To solve this problem, let's break down the information given:

1. Let $ x $ represent the number of twelve-dollar tickets.
2. Then, the number of fifteen-dollar tickets is $ \frac{3}{8}x - 17 $ (since there are 17 fewer fifteen-dollar tickets than three-eighths of the number of twelve-dollar tickets).

3. The total cost of the tickets is given as $2001. Thus, we can set up the following equation based on the costs:

\[
12x + 15\left(\frac{3}{8}x - 17\right) = 2001
\]

### Step-by-Step Solution

1. Expand the equation:

   \[
   12x + 15 \times \frac{3}{8}x - 15 \times 17 = 2001
   \]

2. Simplify each term:

   \[
   12x + \frac{45}{8}x - 255 = 2001
   \]

3. Move $ -255 $ to the other side:

   \[
   12x + \frac{45}{8}x = 2256
   \]

4. To clear the fraction, multiply through by 8:

   \[
   8 \cdot 12x + 8 \cdot \frac{45}{8}x = 8 \cdot 2256
   \]

   \[
   96x + 45x = 18048
   \]

5. Combine terms:

   \[
   141x = 18048
   \]

6. Solve for $ x $:

   \[
   x = \frac{18048}{141} = 128
   \]

So, the number of twelve-dollar tickets is $ x = 128 $.

7. Now, substitute $ x = 128 $ into the expression for the fifteen-dollar tickets:

   \[
   \frac{3}{8} \times 128 - 17 = 48 - 17 = 31
   \]

### Final Answer
- The number of twelve-dollar tickets is **128**.
- The number of fifteen-dollar tickets is **31**.

Would you like a detailed explanation of any specific step? 

Here are 5 related questions to expand on this problem:
1. How would the solution change if the ticket prices were different?
2. What if the total spending was a different amount, say $2500?
3. How can we use a similar approach for a problem with three different ticket types?
4. Can this be solved using matrices or another method?
5. How does this problem change if the 17 fewer tickets condition was instead "10 more tickets"?

**Tip:** When solving word problems, translating each part of the description into mathematical expressions is key to setting up the correct equations.

The local amateur football club spent $2001 on tickets to a professional football game. If the club bought 17 fewer fifteen-dollar tickets than three-eighths the number of twelve-dollar tickets, how many tickets of each type did the club buy?

Solve this algebra problem where a football club spent $2001 on tickets. Find out the number of twelve-dollar and fifteen-dollar tickets, given that the club bought 17 fewer fifteen-dollar tickets than three-eighths the number of twelve-dollar tickets. Ideal for students in Grades 9-10.

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