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12 June, 21:21

Tickets for the homecoming dance cost $20 for a single ticket or $35 for a couple. ticket sales totaled $2280, and 128 people attended. how many tickets of each type were sold?

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  1. 12 June, 23:13
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    Let x be the number of single tickets sold.

    Let y be the number of couple tickets sold.

    So

    The total number of people attended is x + 2y.

    The total money earned from selling the tickets is 20x + 35y.

    As 128 people attended, setting x + 2y equal to 128, we have:

    x + 2y = 128 ... (1)

    As the total revenue from selling tickets is $2,280, setting 20x + 35y equal to 2280, we have:

    20x + 35y = 2280 ... (2)

    Next solve the simultaneous equations (1) and (2).

    From equation (1), we get x = 128 - 2y.

    Substituting it into equation (2), we have:

    20 (128 - 2y) + 35y = 2280

    Solving for y, we obtain:

    2560 - 40y + 35y = 2280

    -40y + 35y = 2280 - 2560

    -5y = - 280

    y = (-280) / (-5)

    y = 56

    Substituting y = 56 into x = 128 - 2y, we have

    x = 128 - 2*56 = 16

    So 16 single and 56 couple tickets were sold.
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