Ann had three less than twice as many nickels as dimes. If the total value of his coins was $1.45 how many....?

Ann had three less than twice as many nickels as dimes. If the total value of her coins was $1.45 how many of each coin did she have?

Please try to show or tell your work.

5 Answers

Relevance
  • 1 decade ago
    Favorite Answer

    Let d = dimes

    Let 2d-3 = nickles

    10d + 5(2d-3) = 145

    10d + 10d-15 = 145

    20d-15 = 145

    20d = 160

    d = 8

    2d-3 = 13

    8 dimes and 13 nickles

  • 4 years ago

    the sides of this concern are: sixteen dimes and a million nickel = 17 money 0 dimes and 33 nickels = 33 money the respond would desire to be someplace in between yet on the threshold of the 1st shrink. 15 dimes and 3 nickels = 18 money 21-18=3 So, 12 dimes and 9 nickels = 21 money is the respond.

  • 1 decade ago

    145 = 5n + 10d

    where n = nickels and d = dimes

    also,

    2d - 3 = n

    Using substitution,

    145 = 5(2d-3) +10d

    distribute the 5

    145 = 10d -15 + 10d

    combine like terms and add 15

    160 = 20d

    divide by 20

    d= 8

    plug that into one of the formulas

    n = 2 * 8 -3

    n=13

    There are 13 nickels and 8 dimes.

  • 1 decade ago

    8 dimes 13 nickels

  • How do you think about the answers? You can sign in to vote the answer.
  • 1 decade ago

    i think its like 13

    nickles and 8 dimes

    IDK im pretty sure im wrong but this is a guess

Still have questions? Get your answers by asking now.