To determine the correct solution within the context of this situation, let's evaluate each option based on the inequality $6x + 3y \leq 84$ and the context.

Step 1: Check the inequality condition $6x + 3y \leq 84$:

1. Option A: (9,10) $6(9) + 3(10) = 54 + 30 = 84$ Since $84 \leq 84$, this satisfies the inequality.

2. Option B: (8.5,11) $6(8.5) + 3(11) = 51 + 33 = 84$ Since $84 \leq 84$, this also satisfies the inequality.

3. Option C: (11, -2) $6(11) + 3(-2) = 66 - 6 = 60$ While this does satisfy the inequality $60 \leq 84$, the context of the situation indicates that $y$ (the number of exhibits) cannot be negative. Therefore, this option is not valid in this context.

4. Option D: (-3,15) $6(-3) + 3(15) = -18 + 45 = 27$ Although $27 \leq 84$, the context also requires that both $x$ (the number of rides) and $y$ be non-negative, so this option is also not valid.

Conclusion:

The valid options that satisfy both the inequality and the context of the problem are:

• A. (9,10)
• B. (8.5,11)

Since both A and B are valid solutions, the problem does not specify if we need integer values only. However, if integer solutions are preferred, Option A would be more appropriate.

Final Answer:

A. (9,10)

Would you like a further breakdown on any part of this answer, or do you have any questions?

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Here are 5 additional related questions:

1. What are the possible integer solutions for $x$ and $y$ that satisfy the inequality $6x + 3y \leq 84$?
2. How would the inequality change if each ride cost $5 and each exhibit cost $4?
3. What is the maximum number of rides Darren can go on if he views 5 exhibits?
4. How would the graph of this inequality change if Darren’s budget increased to $100?
5. If Darren wanted to spend exactly $84, what are the possible combinations of rides and exhibits he could choose?

Tip: When working with inequalities, always consider the context to ensure solutions make practical sense (e.g., non-negative values in real-world contexts).