To raise money, the student council at a local middle school is hosting a talent show. Tickets are sold for $3 for students (x) and $7 for adults (y). The goal is to raise a total of $1905. Which of the following is a viable solution for the number of students and adult tickets that can be sold? (1 point)

Question

To raise money, the student council at a local middle school is hosting a talent show. Tickets are sold for $3 for students (x) and $7 for adults (y). The goal is to raise a total of $1905. Which of the following is a viable solution for the number of students and adult tickets that can be sold? (1 point)

Responses

(153, 278)
(153, 278)

(163, 202.3)
(163, 202.3)

(656,−9)
open paren 656 comma negative 9 close paren

(278, 153)

Answer 1

To solve this problem, we can set up a system of equations.

Let x be the number of student tickets sold, and y be the number of adult tickets sold.

The total amount raised from student tickets is 3x, and the total amount raised from adult tickets is 7y. The total amount raised is $1905, so:

3x + 7y = 1905

Now we need to find a feasible solution for the number of student and adult tickets sold. The number of tickets sold cannot be negative, so x and y must be positive integers.

Let's try the option (278, 153):

3(278) + 7(153) = 834 + 1071 = 1905

So, the option (278, 153) is a viable solution for the number of student and adult tickets that can be sold to raise a total of $1,905.