Abasketball fieldhouse seats 15,000. courtside seats sell for $9, end zone for $6 and balcony for $4. a full house earns $79000in ticket revenue. if half the courtside and balcony seats and all the endzone seats are sold, the total ticket revenue is $45,500. how many of each type of seat are there?

Let's assume number of Courtside seats = x

number of endzone seats = y

number of balcony seats = z

It says that a basketball fieldhouse seats 15,000.

⇒ x + y + z = 15000

Courtside seats sell for $9, end zone for $6 and balcony for $4. A full house earns $79000in ticket revenue.

⇒ 9x + 6y + 4z = 79000

If half the courtside and balcony seats and all the endzone seats are sold, the total ticket revenue is $45,500.

⇒ 9() + 6y + 4() = 45500

⇒ 9x + 12y + 4z = 91000

Writing all equations together;

1x + 1y + 1z = 15000 ..............(equation 1)

9x + 6y + 4z = 79000 ..............(equation 2)

9x + 12y + 4z = 91000 ..............(equation 3)

Subtracting equation(2) from equation(3):-

6y = 91000 - 79000 = 12000

y = 2000

Plugging y=2000 into equation(1) and equation(2) :-

x + 2000 + z = 15000 and 9x + 6*2000 + 4z = 79000

⇒ x + z = 13000 and 9x + 4z = 67000

x + z = 13000 ...............(equation 4)

and 9x + 4z = 67000 ................(equation 5)

Performing equation(5) - 4*equation(4) :-

(9x + 4z) - 4·(x + z) = 67000 - 4×13000

9x - 4x + 4z - 4z = 67000 - 52000

5x = 15000

x = 3000

Plugging x=3000 into equation(4) :-

3000 + z = 13000

z = 13000 - 3000

z = 10000

So, final answers are :-

number of Courtside seats = 3000

number of endzone seats = 2000

number of balcony seats = 10000

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