Question

A stadium has 10,000 seats, divided into box seats, lower deck seats, and upper deck seats. Box seats sell for $10, lower deck seats sell for $8, and upper deck seats sell for $5. When all the seats for a game are sold, the total revenue is $70,000. The stadium has four times as many upper deck seats as box seats. Find the number of lower deck seats in the stadium using elimination.​

Answer 1

the number of box seats x = 1000

the number of lower deck seats y = 5000

the number of upper deck seats z = 4000

Explanation:

Let´s call

"x" the number of box seats

"y" the number of lower deck seats and

"z" the number of upper deck seats

Then

x + y + z = 10000 ( the total number of seats) (1)

10*x + 8*y + 5*z = 70000 (total revenue) (2)

And z = 4*x

Plugging the value z = 4*x in equation (1)

x + y + 4*x = 10000

5*x + y = 10000 ⇒ y = 10000 - 5*x

And plugging this value in equation (2)

10*x + 8 * (10000 - 5*x ) + 20*x = 70000

10*x + 80000 - 40*x + 20*x = 70000

- 10*x = -10000

x = 1000

Then

y = 10000 - 5*x ⇒ y = 10000 - 5 * ( 1000 )

y = 5000

And z = 4*x ⇒ z = 4000