Question 545110: A vendor sold 200 tickets for an upcoming rock concert. Floor seats were $36 and stadium seats were $28. The vendor sold $6080 in tickets. How many $36 and $28 tickets did the vendor sell?
Answer by lmeeks54(111)   (Show Source):
You can put this solution on YOUR website! Let f = # of floor seat tickets sold
Let s = # stadium seat tickets sold
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Given:
total of 200 tickets sold
floor tickets cost $36
stadium tickets cost $28
value of all tickets sold = $6,080
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f + s = 200
36f + 28s = 6080
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Two equations and two unknowns. Let's rewrite the 1st equation solving for one variable in terms of the other and then substitute that identity into the 2nd equation:
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subtract s from both sides:
f = 200 - s
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substitute this into the 2nd equation:
36(200 - s) + 28s = 6080
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no we have one equation and one unknown, which we can easily solve:
7200 - 36s + 28s = 6080
7200 - 8s = 6080
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add 8s and subtract 6080 from both sides to simplify:
7200 - 6080 - 8s + 8s = 6080 - 6080 + 8s
1120 = 8s
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divide both sides by 8:
s = 140, which is the # of stadium seats sold.
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Let's go back to the f = 200 - s identity to find out how many floor tickets were sold:
f = 200 - s
f = 200 - 140
f = 60, which is the # of floor tickets sold.
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Let's check our work with the value equation:
36f + 28s = 6080
36(60) + 28(140) = 6080
2160 + 3920 = 6080
6080 = 6080 checks
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The vendor sold 60 floor tix @ $36 ea + 140 stadium tix @ $28 ea.
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cheers,
Lee
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