algebra word problem- help!!?

A local comedy club put on a play. Adult tickets cost $6.50 each while children's tickets sold for $3.75. There were 290 people who attended the performances. The total receipts from the sale of tickets amounted to $1656.75 how many adults and children attended?

Comments

  • 6.50A + 3.75C = 1656

    A + C = 290

    A = 290 -C

    6.50(290 - C) + 3.75C = 1656.75

    1885 - 6.50C + 3.75C = 1656.75

    -2.75C = -228.25

    C = 83

    A = 290 - 83 = 207

  • Adults = a , and children = c.

    a + c = 290, so a = 290 -c

    6.50a + 3.75 c = 1656.75

    6.50(290 - c ) +3.75 c = 1656.75 ( substitute a in the second equation by 290-c from the first equation).

    1885 - 6.5c + 3.75c =1656.75

    -2.75 c =1656.75 - 1885

    c =83 If 83 children, then 207 adults ( 83*3.75 ) + (207*6.50)=1656.75

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