You can
put this solution on YOUR website!"A concert sold 1,068 tickets" __ A+B+C=1068
"bringing in total revenue of $15,060" __ 10A+15B+20C=15060
"4 times as many "A" tickets sold as there were "B" tickets" __ A=4B
substituting __ (4B)+B+C=1068 __ 5B+C=1068 __ multiplying by 11 __ 55B+11C=11748
substituting __ 10(4B)+15B+20C=15060 __ 55B+20C=15060
subtracting __ [55B+20C=15060]-[55B+11C=11748] __ 9C=3312 __ C=368
substituting __ 5B+(368)=1068 __ B=140
substituting __ A=4(140) __ A=560
You can
put this solution on YOUR website!A concert sold 1,068 tickets bringing in total revenue of $15,060. Three categories of tickets were sold. Type "A" sold for $10, type "B" sold for $15 and type "C" sold for $20. There were 4 times as many "A" tickets sold as there were "B" tickets. How many of each type of ticket generate
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Total tickets equation:
A + B + C = 1068
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Total Revenue equation
10A + 15B + 20C = 15060
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"There were 4 times as many "A" tickets sold as there were "B" tickets."
A = 4B
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Substitute 4B for A in both total equations
2B + B + C = 1068
3B + C = 1068
C = (1068-3B)
:
10(2B) + 15B + 20C = 15060
20B + 15B + 20C = 15060
35B + 20C = 15060
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Substitute (1068-3B) for C in the above equation:
35B + 20(1068-3B) = 15060
35B + 21360 - 60B = 15060
35B - 60B = 15060 - 21360
-25B = -6300
B =

B = 252 ea $15 tickets
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Using equation C = 1068 - 3B)
C = 1068 - 3(252)
C = 1068 - 756
C = 312 ea $20 tickets
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Find A
A + 252 + 312 = 1068
A = 1068 - 564
A = 504 ea $10 tickets
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You can check our solution:
10(504) + 15(252) + 20(312) =