Let's solve each part of the question.

Part 1: Finding the Inequality

1. Ticket Prices:

   • Lower Deck: $42 each
   • Upper Deck: 75% of $42, which is $0.75 \times 42 = 31.5$ dollars each.

2. Cost Constraint:

   • We want to find an inequality that represents spending no more than $800 on tickets. So, the total cost of $x$ lower deck tickets and $y$ upper deck tickets must satisfy: $42x + 31.5y \leq 800$

3. Matching the Inequality with Answer Choices:

   • The correct inequality is $42x + 31.5y \leq 800$.
   • Answer: B. $42x + 31.5y \leq 800$.

Part 2: Checking Possible Combinations

Using the inequality $42x + 31.5y \leq 800$, we substitute each combination of $x$ (lower deck tickets) and $y$ (upper deck tickets) to see if the cost is within the $800 budget.

1. Option A: 22 lower deck and 5 upper deck $42(22) + 31.5(5) = 924 + 157.5 = 1081.5 \quad (\text{exceeds 800})$

2. Option B: 16 lower deck and 6 upper deck $42(16) + 31.5(6) = 672 + 189 = 861 \quad (\text{exceeds 800})$

3. Option C: 8 lower deck and 14 upper deck $42(8) + 31.5(14) = 336 + 441 = 777 \quad (\text{within 800})$

4. Option D: 10 lower deck and 18 upper deck $42(10) + 31.5(18) = 420 + 567 = 987 \quad (\text{exceeds 800})$

Answer for possible combinations: C. 8 lower deck and 14 upper deck.

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Let me know if you would like further details or have any questions!

Additional Practice Questions:

1. If the budget was $1,000, which combinations would satisfy the inequality $42x + 31.5y \leq 1000$?
2. Calculate the maximum number of lower deck tickets possible within an $800 budget if no upper deck tickets are bought.
3. How would the inequality change if upper deck tickets were only 50% of the cost of lower deck tickets?
4. If someone buys 5 upper deck tickets, what is the maximum number of lower deck tickets they can buy while staying under $800?
5. Suppose each lower deck ticket also has a $5 service fee; how would this affect the inequality?

Tip:

When working with inequalities, always check the boundary cases by substituting values that could maximize one variable while minimizing the other. This helps understand the limits fully.