Let's define the variables in the context of the problem:
- Let \( x \) represent the number of adult tickets.
- Let \( y \) represent the number of children's tickets.
The cost of the adult tickets is $15 each, and the cost of the children's tickets is $12 each. The Ramirez family wants to spend a maximum of $150 on tickets. Therefore, the inequality representing this situation can be written as:
\[ 15x + 12y \leq 150 \]
Contextual Constraints
- \( x \geq 0 \): The number of adult tickets must be a non-negative integer.
- \( y \geq 0 \): The number of children's tickets must be a non-negative integer.
Now we need to check which of the provided points is a viable solution to the inequality \( 15x + 12y \leq 150 \) while also satisfying the non-negativity constraints.
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Check point (3, 4.5):
- \( x = 3 \), \( y = 4.5 \)
- \( 15(3) + 12(4.5) = 45 + 54 = 99 \) (This satisfies \( 99 \leq 150 \))
- However, \( y = 4.5 \) is not a valid number of children's tickets since it must be a whole number.
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Check point (10, 5):
- \( x = 10 \), \( y = 5 \)
- \( 15(10) + 12(5) = 150 + 60 = 210 \) (This does not satisfy \( 210 \leq 150 \))
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Check point (-3, 5):
- \( x = -3 \), \( y = 5 \)
- Point is invalid because \( x \) cannot be negative.
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Check point (2, 5):
- \( x = 2 \), \( y = 5 \)
- \( 15(2) + 12(5) = 30 + 60 = 90 \) (This satisfies \( 90 \leq 150 \))
- Both \( x \) and \( y \) are non-negative.
Viable Solution
The only viable solution among the options provided is (2, 5).