Answer:
section A = 22,500 seats
section B = 14,900 seats
section C = 7,600 seats
Explanation:
given:
stadium has 45,000 seats.
section A = $28 per ticket
section B = $24 per ticket
section C = $20 per ticket
$1,139,600 from each sold out event
Seats on A = Seats B + Seats C
find:
How many seats does each section hold?
solution:
- 1,139,600 = 28A + 24B + 20C eq. 2
since A = B + C, substitute A into eq. 1
- 45,000 = A + B + C
- 45,000 = (B + C) + B + C
- 45,000 = 2B + 2C
- 45,000 - 2C = 2B
- B = 45,000 - 2C
2
since A = 22,500
B = 22,500 - C
substitute A and B into eq. 2 to get C
- 1,139,600 = 28A + 24B + 20C
- 1,139,600 = 28(22,500) + 24(22,500 - C) + 20C
- 1,139,600 = 630,000 + 540,000 - 24C + 20C
- 1,139,600 - 630,000 - 540,000 = -24C + 20C
- -30,400 = -2C
- C = 30,400
4
now plugin the value of C=15,200 into eq. 3
- B = 22,500 - C
- B = 22,500 - 7,600
- B = 14,900
therefore, the number of seats on each sections are:
seats A = 22,500
seats B = 14,900
seats C = 7,600
Proof:
- 45,000 = A + B + C
- 45,000 = 22,500 + 14,900 + 7,600
45,000 = 45,000 √
- 1,139,600 = 28A + 24B + 20C
- 1,139,600 = 28(22,500) + 24(14,900) + 20(7,600)
- 1,139,600 = 630,000 + 357,600 + 152,000
- 1,139,600 = 1,139,600 √