Theater Revenues

Problem:

Solution:

This is not an easy question - it involves three variables, now let's set them up

Let $ x $ be the number of orchestra seating.
Let $ y $ be the main floor seating.
Let $ z $ be the balcony seating.

We know there are 600 seats, therefore we have

$ x + y + z = 600 $ (Equation 1)

Next, we know if they are all sold, the revenue would be 33,500, that means

$ 80x + 60y + 25z = 33500 $ (Equation 2)

Last but not least, we know on that evening, this equation holds.

$ 80x + \frac{3}{5}60y + \frac{4}{5}25z = 24640 $.

The simplify to

$ 80x + 36y + 20z = 24640 $.  (Equation 3)

These give us three equations with 3 unknowns, now we are all set to solve them.

Subtract equation 2 by the equation 3, we get

$ 24y + 5z = 8860 $.  (Equation 4)

Subtract 80 times equation 1 by the equation 2, we get

$ 20y + 55z  = 14500 $  (Equation 5)

Finally subtract the 5 times equation 4 by 6 times equation 5, we get

$ -305z = -42700 $, or simply $ z = 140.$

Substitute this result to equation 4, solving get $ y = 340 $.

Finally, substitute both values back to equation 1, get $ x = 120 $.