Let's define the variables:
- Let \( x \) be the number of adult tickets.
- Let \( y \) be the number of children's tickets.
The cost for adult tickets is \( 15x \) and the cost for children's tickets is \( 12y \). The goal is to keep the total spending under or equal to $150. Therefore, the inequality that models this situation can be written as:
\[ 15x + 12y \leq 150 \]
Next, we need to consider the contextual constraints for \( x \) and \( y \). Since the number of tickets cannot be negative, we have the constraints:
\[ x \geq 0 \] \[ y \geq 0 \]
Thus, the complete model consists of the inequality \( 15x + 12y \leq 150 \) along with \( x \geq 0 \) and \( y \geq 0 \).
Now, let's analyze the provided points to determine which one is a viable solution:
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Point (2, 5): \[ 15(2) + 12(5) = 30 + 60 = 90 \leq 150 \] This point is a viable solution.
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Point (10, 5): \[ 15(10) + 12(5) = 150 + 60 = 210 \nleq 150 \] This point is not a viable solution.
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Point (3, 4.5): \[ 15(3) + 12(4.5) = 45 + 54 = 99 \leq 150 \] This point is a viable solution.
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Point (-3, 5): [ x ) cannot be negative, so this point is not allowed.
From the analyses, both points (2, 5) and (3, 4.5) are viable solutions. However, it is important to note that point (3, 4.5) involves a non-integer number for children's tickets, which is unrealistic in this context depending on how ticket sales are conducted (usually they sell whole tickets).
Thus, while both may technically satisfy the inequality, the most practical solution in a real-world context would be:
Point (2, 5) is a simple and clear viable answer in the context of counting whole tickets.
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