Math homework help..

A broadway theater has 400 seats, divided into Orchestra, main, and balcony seating. Orchesta seats sell for $50, main sell for $40, and balcony sell for $25. If all the seats are sold, the gross revenue to the theater is $14,850. If all the main and balcony seats are sold, but only half of the orchestra seats are sold, the gross revenue is $12,850. How many are there of each kind of seat?

50% of the orchestra seats unsold makes the collection less by ( 14850 -12850) = 2000$

2000 /50 = 40 seats unsold

So total orchestra seats in the theater = 80 seats

So total collection for orchestra seats = 80X50 = 4000$

REMAINING NO OF SEATS = 400 -80 = 320 seats

Total collection for the 320 seats will be 14850 -4000 = 10850 $

Let the no of main seats be X

No of balcony seats will be 320 - X

Collection = 40X + 25 ( 320 - X) = 10850

15X = 10850 - 8000 = 2850

X = 190 seats

ANSWER ORCHESTRA = 80 seats
MAIN = 190 seats
BALCONY = 130 seats
CHECK
80X 50 = 4000 $
190X 40 = 7600 $
130 X 25 = 3250 $
TOTAL = 14850 $

Let:

x = number of orchestra seats
y = number of main seats
z = number of balcony seats

Our equations are:

400 = x + y + z
14850 = 50x + 40y + 25z
12850 = 50(x/2) + 40y + 25z

Using the elimination method with the last two equations shows us that x = 80.

Our first two equations can then be changed to:

320 = y + z
10850 = 40y + 25z

These two equations can be solved to show that y = 190 and z = 130

o+m+b=400
50o+40m+25b=14850
50/2o+40m+25b=12850

three equations, three unknowns. I think you can take it from here.