Question

The student Fun Club plans to go to the movies. At the matinee, tickets cost $6 and popcorn is $3. At evening shows, tickets cost $9 and popcorn is $4. The Fun Club attends a matinee and spends less than $60, and then attends an evening show and spends more than $36. If they purchased the same number of tickets and popcorns at each show, which of the following is a possible solution for the number of tickets and popcorns purchased?

Answer 1

Matinee

Cost of each ticket: $6

Cost of popcorn: $3

Evening:

Ticket: $9

Popcorn: $4

Number of tickets: x

Number of popcorns : y

The Fun Club attends a matinee and spends less than $60

6x + 3y < 60

Then attends an evening show and spends more than $36

9x+ 4y < 36

We have the system:

6x + 3y < 60 (a)

9x+ 4y >36 (b)

Graph each inequality:

The intersection of red and blue is the solution.

7 tickets and 5 popcorns (7,5) is inside the intersection, So, it is the solution.