Ayouth group and their leaders visited a museum. three adults and six students in one van paid $102 for their admission. two adults and seven students paid $95 for their admission. marcus wrote a system of linear equations to find the price for each adult and student. he found that each student ticket was $12 and each adult ticket was $9. is marcus correct? explain.

  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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Check

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.

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