Adult tickets to a play cost 20.50, and student tickets are 16.00. Ms powers can spend no more than 120.00 on tickets. If she buys 3 student tickets, how many adult tickets is she able to buy? Use a linear inequality in two variables to explain your reasoning.
(3 * 16) + 20.50a ≤ 120
Solve for a.
Let's represent the number of adult tickets Ms Powers can buy as 'x'.
The cost of each adult ticket is $20.50, so the total cost of 'x' adult tickets will be 20.50x.
Ms Powers has already bought 3 student tickets, which cost $16.00 each. So, the cost of the student tickets she has bought is 3 * 16.00 = $48.00.
The total amount Ms Powers can spend on tickets is $120.00, so the sum of the cost of adult tickets and the cost of student tickets should not exceed $120.00.
Therefore, the inequality representing the situation is:
20.50x + 48.00 ≤ 120.00
Simplifying the inequality:
20.50x ≤ 120.00 - 48.00
20.50x ≤ 72.00
Dividing both sides of the inequality by 20.50:
x ≤ 72.00 / 20.50
x ≤ 3.51
Since Ms Powers cannot buy a fraction of a ticket, the maximum number of adult tickets she can buy is 3.
Therefore, she is able to buy a maximum of 3 adult tickets.
To solve this problem, we can set up a linear inequality in two variables:
Let's assume "x" represents the number of adult tickets that Ms. Powers can buy.
The cost of one adult ticket is $20.50. So the total cost of adult tickets she can buy is 20.50x.
Ms. Powers can buy 3 student tickets, and the cost of one student ticket is $16.00. So the total cost of student tickets she bought is 16.00 * 3 = $48.00.
The total amount Ms. Powers can spend is $120.00.
Therefore, we can write the inequality:
20.50x + 48.00 ≤ 120.00
Simplifying the inequality:
20.50x ≤ 120.00 - 48.00
20.50x ≤ 72.00
Dividing both sides of the inequality by 20.50 to solve for x:
x ≤ 72.00 / 20.50
x ≤ 3.51
Since the number of adult tickets cannot be a fraction, Ms. Powers can buy a maximum of 3 adult tickets to stay within her budget.
So, she is able to buy a maximum of 3 adult tickets.