Algebra word problem...?

Joe has a collection of nickels and dimes that is worth $8.55. If the number of dimes was tripled and the number of nickels was decreased by 2, the value of the coins would be $10.45. How many nickels and dimes does he have?

Update:

any thoughts at all??

Comments

  • x = Number of Nickels, y - Number of Dimes

    Nickels are worth .05 and dimes are worth .10

    $0.05x + $0.10y = $8.55 (I hate working with decimals like this so I multiply by 100)

    5x + 10y = 855

    5(x - 2y) = 855

    x -2y = 171

    [x = 2y + 171]

    $0.05(x - 2) + 3($0.10y) = $10.45

    5(x - 2) + 30y = 1045

    5x - 10 + 30y = 1045

    5x - 30y = 1055

    5(x + 6y) = 1055

    x + 6y = 211

    [x = 211 - 6y]

    211 - 6y = 171 - 2y

    211 - 4y = 171

    40 = 4y

    [y = 10] *

    x = 171 - 2y

    x = 171 - 2(10)

    x = 171 - 20

    [x = 151] *

    $0.05x + $0.10y = $8.55

    $0.05(151) + $ 0.10(10) = $8.55

    $7.55 + $1.00 = $8.55

    $8.55 = $8.55

    check

    $0.05(x - 2) + 3($0.10y) = $10.45

    $0.05(151 - 2) + 3($0.10[10]) = $10.45

    $0.05(149) + 3($1.00) = $10.45

    $7.45 + $3.00 = $10.45

    $10.45 = $10.45

    Check

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