A theater owner wants to divide a 3300 seat theater into three sections, with tickets costing $ 60, $70, and $120, depending on the section. He wants to have twice as many $60 tickets as the sum of the other tickets, and he wants to earn $ 224,000 from a full house. Find how many seats he should have in each section.

Section A = 2,200 seats

Section B = 800 seats

Section C = 300 seats

Step-by-step explanation:

Three sections

Section A cost = $60

Section B cost = $70

Section C cost = $120

Total seat = 3300

Total earnings = $224,000

He wants to have twice as many $60 tickets as the sum of the other tickets

A = 2(B + C)

A + B + C = 3,300

60A + 70B + 120C = 224,000

substitute A = 2(B + C)

2(B + C) + B + C = 3,300

60(2B + 2C) + 70B + 120C = 224,000

2B + 2C + B + C = 3,300

120B + 120C + 70B + 120C = 224,000

3B + 3C = 3,300 (1)

190B + 240C = 224,000 (2)

Multiply (1) by 80

240B + 240C = 264,000 (3)

190B + 240C = 224,000 (2)

Subtract the equations

240B - 190B = 264,000 - 224,000

50B = 40,000

B = 40,000 / 50

= 800

B = 800

Substitute the value of B into (1)

3B + 3C = 3,300 (1)

3(800) + 3C = 3,300

2,400 + 3C = 3,300

3C = 3,300 - 2,400

3C = 900

C = 900 / 3

= 300

C = 300

Substitute values of B and C into

A + B + C = 3,300

A + 800 + 300 = 3,300

A + 1,100 = 3,300

A = 3,300 - 1,100

= 2,200

A = 2,200

Section A = 2,200 seats

Section B = 800 seats

Section C = 300 seats

y=sec x is the correct answer

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