SOLUTION: at a college prodution of streetcar named desire 400tickets were sold. the ticket prices were $8, $10, and $12 and the total income from the ticket sales was 3700. how many tickets
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-> SOLUTION: at a college prodution of streetcar named desire 400tickets were sold. the ticket prices were $8, $10, and $12 and the total income from the ticket sales was 3700. how many tickets
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Question 127717: at a college prodution of streetcar named desire 400tickets were sold. the ticket prices were $8, $10, and $12 and the total income from the ticket sales was 3700. how many tickets of each type were sold if the combined number of $8 and $10 tickets sold was 7 times the number of $12 tickets sold? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! prices were $8, $10, and $12 and the total income from the ticket sales was $3700. how many tickets of each type were sold if the combined number of $8 and $10 tickets sold was 7 times the number of $12 tickets sold?
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It says 400 tickets sold:
x + y + z = 400
:
It says:
"prices were $8, $10, and $12 and the total income from the ticket sales was 3700.
8x + 10y + 12z = 3700
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It says:
"the combined number of $8 and $10 tickets sold was 7 times the number of $12 tickets sold?"
x + y = 7z
x = (7z-y)
:
Using the 1st equation, substitute (7z-y) for x
(7z-y) + y + z = 400
7z + z - y + y = 400
8z = 400
z =
z = 50 ea $12 ticket sold
;
Substitute 50 for z in the 1st equation:
x + y + 50 = 400
x + y = 400 - 50
x + y = 350
x = (350-y)
:
Subsitute 50 for z in the 2nd equation:
8x + 10y + 12(50) = 3700
8x + 10y + 600 = 3700
8x + 10y = 3700 - 600
8x + 10y = 3100
:
Substitute (350-y) for x in the above equation
8(350-y) + 10y = 3100
2800 - 8y + 10y = 3100
2y = 3100 - 2800
2y = 300
y =
y = 150 ea $10 tickets
:
I'll let you find the number of $8 tickets (x), there are several ways to do it.
