Question

A theater owner wants to divide a 3600 seat theater into three​ sections, with tickets costing ​$30​, ​$80​, and ​$120​, depending on the section. He wants to have twice as many $30 tickets as the sum of the other​ tickets, and he wants to earn ​$176,000 from a full house. Find how many seats he should have in each section.

Answer 1

2400 $30 tickets

1000 $80 tickets

200 ​$120 tickets

Explanation:

a= $30 tickets

b = $80 tickets

c= ​$120tickets

a=2(b+c)

a+b+c = 3600

30a + 80b + 120c = 176000

a+b+c = 3600

2(b+c) + b+c = 3600. (substitute a=2(b+c) in)

2b + 2c + b+c = 3600 (calculate out 2(b+c))

3b + 3c = 3600 (combine like terms)

b + c = 1200 (divide 3 on both sides) *

a=2(b+c)

a = 2(1200) (substitute b + c = 1200 in)

a = 2400

30a + 80b + 120c = 176000

30(2400) + 80b + 120c = 176000 (substitute a = 2400 in)

72000 + 80b + 120c = 176000

80b + 120c = 104000 (subtract 72000 from both sides)

2b + 3c =2600 (divide 40 from both sides)

*b + c = 1200

2b + 2c = 2400 (mutiply by 2 on both sides)

2b + 3c =2600

-

2b + 2c = 2400

------------------------

c = 200

*b + c = 1200

b + 200 = 1200 (substitute c = 200 in)

b = 1000

a = 2400

b = 1000

c = 200