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Algebra

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Nadia has 32 coins made up of nickles, dimes, and quarters. The sum of the number of nickels and the number of quarters is three times the number of dimes. If the total value of the coins is $4.60, how many of each kind does she have?
I can't figure out how to make this word problem into 3 equations.

  • Algebra -

    x nickels
    y dimes and
    z quarters:

    equation base on number of coins
    x + y + z = 32 ------ #1

    equation based on value of coins

    5x + 10y + 25z = 460, which reduces to
    x + 2y + 5z = 92 ----- #2

    "The sum of the number of nickels and the number of quarters is three times the number of dimes"
    ----> x + z = 3y

    x -3y + z = 0 ----- #3

    There are your 3 equations,
    if you do #1 - #3 , you get
    4y = 32
    y = 8

    so #1 becomes
    x + 8 + z = 32
    x + z = 24

    #2 becomes
    x + 16 + 5z = 92
    x + 5z = 76

    subtract those two ...
    4z = 52
    z = 13

    back into the 1st
    x + 8 + 13 = 32
    x = 11

    looks like 11 nickels, 8 dimes and 13 quarters

    Check:
    number of coins = 11+8+13 = 32
    value of coins = 5(11) + 10(8) + 25(13) = 460
    is 11+13 equal to 3 times 8 ??
    yes!
    All is good

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