A stadium has 53,000 seats. Seats sell for $28 in Section A, $16 in Section B, and $12 in Section C. Th number of seats in Section A equals the total number of seast in Sections B and C. Suppose the stadium takes in $1,118,000 from each sold-out event. How many seats does each section hold?

Section A holds:
Section B holds:
Section C holds:

2 answers

28a + 16b + 12c = 1,118,000
but a = b+c
28(b+c) + 16b + 12c = 1,118,000
44b + 40c = 1,118,000
11b+10c = 111,800

but we also know that a + b + c = 53000
b+c + b + c = 53000
b+c = 26500
c = 26500 - b

sub in that value of c into 11b+10c = 111,800
and you can find b
after that you can find c
and finally find a in a = b+c
A equals B plus C ... half of the seats are in A ... 53,000 / 2 = ?

value of A seats ... 28 * (53000 / 2) = 742,000

number of B and C seats ... B + C = 53,000 / 2 ... 12B + 12C = 6 * 53,000

value of B and C seats ... 16B + 12C = 1,118,000 - 742,000

subtract the two equations to eliminate C
... 4B = 1,118,000 - 742,000 - (6 * 53,000)

solve for B , then substitute back to find C