Let x be the number of adult tickets and y be the number of children's tickets.
The cost of x adult tickets is 15x and the cost of y children's tickets is 12y.
The total cost of the tickets should not exceed $150, so the two-variable inequality is:
15x + 12y ≤ 150
Now, let's evaluate the given points to determine which one is a viable solution:
(2, 5):
Substituting x = 2 and y = 5 into the inequality:
15(2) + 12(5) ≤ 150
30 + 60 ≤ 150
90 ≤ 150
The inequality holds true, so (2, 5) is a viable solution.
(10, 5):
Substituting x = 10 and y = 5 into the inequality:
15(10) + 12(5) ≤ 150
150 + 60 ≤ 150
210 ≤ 150
The inequality does not hold true, so (10, 5) is not a viable solution.
(-3, 5):
Substituting x = -3 and y = 5 into the inequality:
15(-3) + 12(5) ≤ 150
-45 + 60 ≤ 150
15 ≤ 150
The inequality holds true, so (-3, 5) is a viable solution.
(3, 4.5):
Substituting x = 3 and y = 4.5 into the inequality:
15(3) + 12(4.5) ≤ 150
45 + 54 ≤ 150
99 ≤ 150
The inequality holds true, so (3, 4.5) is a viable solution.
Therefore, the viable solutions are (2, 5), (-3, 5), and (3, 4.5).
The Ramirez family is going to the space museum. Adult tickets are $15 and children's tickets are $12. They want to spend a maximum of $150. Write a two-variable inequality to model this situation and determine the contextual constraints. Which of the following points is a viable solution ? (1 point ) (2, 5); (10, 5); (- 3, 5); (3, 4.5)
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Apologies for the mistake in my previous response. You are correct, and I apologize for the confusion.
The given points are (2, 5), (10, 5), (-3, 5), and (3, 4.5). We need to determine which of these points is a viable solution.
Let's evaluate each point in the inequality: 15x + 12y ≤ 150.
1. For (2, 5):
15(2) + 12(5) = 30 + 60 = 90
Since 90 is less than or equal to 150, (2, 5) is a viable solution.
2. For (10, 5):
15(10) + 12(5) = 150 + 60 = 210
Since 210 is greater than 150, (10, 5) is not a viable solution.
3. For (-3, 5):
15(-3) + 12(5) = -45 + 60 = 15
Since 15 is less than or equal to 150, (-3, 5) is a viable solution.
4. For (3, 4.5):
15(3) + 12(4.5) = 45 + 54 = 99
Since 99 is less than or equal to 150, (3, 4.5) is a viable solution.
To summarize:
(2, 5) and (-3, 5) are viable solutions, while (10, 5) and (3, 4.5) are not viable solutions.
The given points are (2, 5), (10, 5), (-3, 5), and (3, 4.5). We need to determine which of these points is a viable solution.
Let's evaluate each point in the inequality: 15x + 12y ≤ 150.
1. For (2, 5):
15(2) + 12(5) = 30 + 60 = 90
Since 90 is less than or equal to 150, (2, 5) is a viable solution.
2. For (10, 5):
15(10) + 12(5) = 150 + 60 = 210
Since 210 is greater than 150, (10, 5) is not a viable solution.
3. For (-3, 5):
15(-3) + 12(5) = -45 + 60 = 15
Since 15 is less than or equal to 150, (-3, 5) is a viable solution.
4. For (3, 4.5):
15(3) + 12(4.5) = 45 + 54 = 99
Since 99 is less than or equal to 150, (3, 4.5) is a viable solution.
To summarize:
(2, 5) and (-3, 5) are viable solutions, while (10, 5) and (3, 4.5) are not viable solutions.
1 answer