Question 153283This question is from textbook Algebra and Trigonometry
: Tickets for the homecoming dance cost $20 for a single ticket or $35 for a couple. Ticket sales totaled $2280, and 128 people attended. How many tickets of each type were sold?
This question is from textbook Algebra and Trigonometry
Answer by orca(409)   (Show Source):
You can put this solution on YOUR website! Let x be the number of single tickets sold.
Let y be the number of couple tickets sold.
So
The total number of people attended is x + 2y.
The total money earned from selling the tickets is 20x + 35y.
As 128 people attended, setting x + 2y equal to 128, we have:
x + 2y = 128 ....................(1)
As the total revenue from selling tickets is $2,280, setting 20x + 35y equal to 2280, we have:
20x + 35y = 2280 ................(2)
Next solve the simultaneous equations (1) and (2).
From equation (1), we get x = 128 - 2y.
Substituting it into equation (2), we have:
20(128 - 2y) + 35y = 2280
Solving for y, we obtain:
2560 - 40y + 35y = 2280
-40y + 35y = 2280 - 2560
-5y = -280
y = (-280)/(-5)
y = 56
Substituting y = 56 into x = 128 - 2y, we have
x = 128 - 2*56 = 16
So 16 single and 56 couple tickets were sold.
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