A theatre has 700 seats. The orchestra seats sell for $40 each, the main seats sell for $25 each, and the balcony seats seel for $20 each. When all orchestra, main, and balcony seats are sold, the theatre collects $18,450. When half the orchestra seats and all the main and balcony seats sell, then the theatre collects 15,650. How many orchestra, main, and balcony seats are available?

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Curlyben:

I definitely have attempted this. I am coming up in circles. I solved for o(orchestra) first to come up with o = 700 -m -b. Then I substituted o with the above equation to solve for b (trying to get the equation into two variables). My answer came to b=477.5 - .75m. Then, I solved for m (main seats) which came to m = 839.58. I AM LOST since there are only 700 seats to begin with (m cannot possibly be 839.58). So, as you can see you have no worries that I have not tried this one.

Quote:

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Originally Posted by wardas

A theatre has 700 seats. The orchestra seats sell for $40 each, the main seats sell for $25 each, and the balcony seats seel for $20 each. When all orchestra, main, and balcony seats are sold, the theatre collects $18,450. When half the orchestra seats and all the main and balcony seats sell, then the theatre collects 15,650. How many orchestra, main, and balcony seats are available?

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Let O=# of orchestra seats, M=# main seats. B=# balcony seats.

Then, we have three equations with three unknowns.

The third constraint says that when 'half' the orchestra seats are sold, That would be O/2.

Which leaves 40(O/2)=20O in the third equation.







Solve away.

The difference between equation 2 and 3 is
20o and 2800
so
20o=2800
so
o=140
take out of first equation
m+b=560
and 3rd equation
25m+20b=12850
which makes m=330 and b=230 and o=140