To model the situation with a two-variable inequality, we can let:
- \( x \) represent the number of adult tickets
- \( y \) represent the number of children's tickets
The cost for adult tickets is \( 15x \) and the cost for children's tickets is \( 12y \). The family wants to spend a maximum of $150, so the inequality representing this situation is:
\[ 15x + 12y \leq 150 \]
Contextual Constraints:
- \( x \geq 0 \) (number of adult tickets cannot be negative)
- \( y \geq 0 \) (number of children's tickets cannot be negative)
Now, we need to determine which of the given points is a viable solution within these constraints.
Let's evaluate each point:
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Point (2, 5): \[ 15(2) + 12(5) = 30 + 60 = 90 \] Since \( 90 \leq 150 \) and both \( 2 \geq 0 \) and \( 5 \geq 0 \) hold, this point is a viable solution.
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Point (3, 4.5): \[ 15(3) + 12(4.5) = 45 + 54 = 99 \] Since \( 99 \leq 150 \), and both \( 3 \geq 0 \) and \( 4.5 \geq 0 \) hold, this point is a viable solution as well. However, note that \( y \) must be a whole number as it represents the number of children's tickets, which makes this point less acceptable.
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Point (10, 5): \[ 15(10) + 12(5) = 150 + 60 = 210 \] Since \( 210 > 150 \), this point is NOT a viable solution.
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Point (-3, 5): This point has \( x = -3 \), which is negative. Thus it cannot be a viable solution.
Conclusion: The viable solution among the options based on the constraints provided is \( (2, 5) \).
Although \( (3, 4.5) \) mathematically satisfies the spending condition, it is not practical in the context of number of tickets. Hence, the best answer is:
Viable Solution: (2, 5)