At a college production of a play, 420 tickets were sold. The ticket prices were $8, $10, and $12, and the total income from ticket sales was $3900. How many tickets of each type were sold if the combined number of $8 and $10 tickets sold was 5 times the number of $12 tickets sold?
+23245 Let x represent the number of $8.00 tickets sold
y represent the number of $10.00 tickets sold
z represent the number of $12.00 tickets sold
Since the combined number of $8.00 tickets and $10.00 tickets sold was 5 times the number of $12.00 tickets sold:
x + y = 5z ---> z = (x + y)/5
Value of the tickets sold: 8x + 10y + 12(x + y)/5 = 3900
(multiply by 5) 40x + 50y + 12x + 12y = 19 500
52x + 62y = 19 500
Number of tickets sold: x + y + (x + y)/5 = 420
(multiply by 5) 5x + 5y + x + y = 2100
6x + 6y = 2100
(divide by 6) x + y = 350
Combining the two equations: 52x + 62y = 19500
x + y = 350
(multiplying the bottom equation by -52:
52x + 62y = 19500
-52x - 52y = -18200
(adding down) 10y = 1300
y = 130
x + y = 350 ---> x + 130 = 350 ---> x = 220
z = (x + y)/5 ---> z = (220 + 130) / 5 ---> z = 70
