A stadium has 45,000 seats. Seats sell for $28 in Section A, $24 in Section B, and $20 in Section C. The number of seats in Section A equals the total number of seats in Sections B and C. Suppose the stadium takes in $1,139,600 from each sold-out event. How many seats does each section hold?

section A = 22,500 seatssection B = 14,900 seatssection C = 7,600 seats

Step-by-step 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:

45,000 = A + B + C                    eq. 11,139,600 = 28A + 24B + 20C   eq. 2

since A = B + C, substitute A into eq. 1

45,000 = A + B + C45,000 = (B + C) + B + C45,000 = 2B + 2C45,000 - 2C = 2BB = 45,000 - 2C

                       2

B = 22,500 - C                             eq. 3

since A = 22,500

         B = 22,500 - C

substitute A and B into eq. 2 to get C

1,139,600 = 28A + 24B + 20C1,139,600 = 28(22,500) + 24(22,500 - C) + 20C1,139,600 = 630,000 + 540,000 - 24C + 20C1,139,600 - 630,000 - 540,000 = -24C + 20C-30,400 = -2CC = 30,400

                 4

C = 7,600

now plugin the value of C=15,200 into eq. 3

B = 22,500 - CB = 22,500 - 7,600B = 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 + C45,000 = 22,500 + 14,900 + 7,600

45,000 = 45,000 √

1,139,600 = 28A + 24B + 20C1,139,600 = 28(22,500) + 24(14,900) + 20(7,600)1,139,600 = 630,000 + 357,600 + 152,0001,139,600 = 1,139,600 √

Idon’t exactly understand but i think dbc=rst