For the opening night of an opera house, a total of 1000 tickets were sold. Front orchestra seats cost $80 each, rear orchestra seats cost $60 each, and front balcony seats cost $50 each. The combined number of tickets sold for the front orchestra and rear orchestra exceeded twice the number of front balcony tickets sold by 400. The total receipts for the performance were $62,800. Determine how many tickets of each type were sold.
3 answers
80(240) + 60(560) + 50(200) = 62800
one third of the problem is just solved and one unknown is just found: the number of front balcony tickets is exactly 200.
Regarding the remaining part, you have now only two unknowns: the number of tickets for the front orchestra
and the number of tickets for rear orchestra.
Regarding the remaining part, you have now only two unknowns: the number of tickets for the front orchestra
and the number of tickets for rear orchestra.
number of front orchestra seats ---- x
number of rear orchestra seats ---- y
number of front balcony seats --- 1000-x-y
"The combined number of tickets sold for the front orchestra and rear orchestra exceeded twice the number of front balcony tickets sold by 400"
---> (x+y) - 2(1000-x-y) = 400
x+y-2000 + 2x + 2y = 400
3x + 3y = 2400
x + y = 800 or y = 800-x
cost equation:
80x + 60y + 50(1000-x-y) = 62800
80x + 60y + 50000 - 50x - 50y = 62800
30x + 10y = 12800
3x + y = 1280
use substitution:
3x + 800-x = 1280
You take over, we're almost there.
number of rear orchestra seats ---- y
number of front balcony seats --- 1000-x-y
"The combined number of tickets sold for the front orchestra and rear orchestra exceeded twice the number of front balcony tickets sold by 400"
---> (x+y) - 2(1000-x-y) = 400
x+y-2000 + 2x + 2y = 400
3x + 3y = 2400
x + y = 800 or y = 800-x
cost equation:
80x + 60y + 50(1000-x-y) = 62800
80x + 60y + 50000 - 50x - 50y = 62800
30x + 10y = 12800
3x + y = 1280
use substitution:
3x + 800-x = 1280
You take over, we're almost there.
1 answer