Finite math

A theater has a seating capacity of 900 and charges $5 for children, $7 for students, and $9 for adults. At a certain screening with full attendance, there were half as many adults as children and students combined. The receipts totaled $6100. How many children attended the show?

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  1. 5 c + 7 s + 9 a = 6100

    c + s + a = 900

    a = .5 (c+s) or 2a - c - s = 0

    ============================
    add two and three
    3 a = 900
    so
    a = 300

    c+s = 600 so s = 600 - c

    3 c + 7 s + 2700 = 6100

    3 c + 7(600-c) = 3400
    3 c + 4200 - 7 c = 3400
    4 c = 800
    c = 200

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  2. a = adults

    c = childrens

    s = students

    With a full audience:

    a + c + s = 900

    There were half as many adults as children and students combined mean:

    a = ( s + c ) / 2

    Replace this in equation:

    a + c + s = 900

    ( s + c ) / 2 + c + s = 900 Multiply both sides by 2

    s + c + 2 c + 2 s = 1,800

    3 c + 3 s = 1,800 Divide both sides by 3

    c + s = 600

    The receipts totaled $6,100 mean:

    $9 * adults + $7 * students + $5 * chidrens= $6,100

    $9 * a + $7 * s + $5 * c = $6,100

    9 a + 7 s + 5 c = 6,100

    Replace a = ( s + c ) / 2 in this equation.

    9 * ( s + c ) / 2 + 7 s + 5 c = 6,100 Multiply both sides by 2

    9 ( s + c ) + 2 * 7 s + 2 * 5 c = 2 * 6,100

    9 s + 9 c + 14 s + 10 c = 12,200

    23 s + 19 c = 12,200

    Now you must solve 2 equations with 2 unknow:

    c + s = 600

    23 s + 19 c = 12,200

    Try that.

    The solutions are : c = 400 , s = 200

    400 childrens

    200 students

    Proof:

    a = ( s + c ) / 2

    a = ( 400 + 200 ) / 2 = 600 / 2 = 300

    300 adults

    $9 * a + $ 7 * s + $5 * c = $6,100

    $9 * 300 + $ 7 * 200 + $5 * 400 = $6,100

    $2,700 + $1,400 + $2,000 = $6,100

    $6,100 = $6,100

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  3. A theater is presenting a program on drinking and driving for students and their parents or other responsible adults. The proceeds will be donated to a local alcohol information center. Admission is $ 10.00 for adults and $ 5.00 for students.​ However, this situation has two​ constraints: The theater can hold no more than 210 people and for every two​ adults, there must be at least one student. How many adults and students should attend to raise the maximum amount of​ money?
    To raise the maximum amount of​ money,
    nothing adults and
    nothing students should attend.

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