Answer:
no, Marcus is not correct
Marcus's numbers do not satisfy the problem statement.
Step-by-step explanation:
Marcus should have written equations like ...
3a +6s = 102 . . . . . . where a = the adult ticket price, s = student price
2a +7s = 95
If he were to graph these equations, he might see the result shown below.
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If he were to solve them by the "elimination" method, he might multiply the first equation by -2/3 and add it to the second.
-2/3(3a +6s) +(2a +7s) = -2/3(102) +(95)
-2a -4s +2a +7s = -68 +95
3s = 27
s = 9
Using this value in the first equation, Marcus would have ...
3a +6·9 = 102
3a = 48 . . . . . . . . . . subtract 54
a = 16 . . . . . . . . . . . . divide by 3
Each student ticket was $9; each adult ticket was $16.
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<em>Check</em>
3·16 +6·9 = 48 +54 = 102 . . . . first van price agrees
2·16 +7·9 = 32 +63 = 95 . . . . . second van price agrees
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If Marcus wrote his equations without defining the variables first, he could easily get the adult price and the student price mixed up. If he didn't know his math facts, he could make a mistake in arithmetic at any of a number of places along the line. His final mistake was that he did not check his answer.
