Math help?

Solve the problem by setting up and solving a system of three linear equations in three variables.

A box contains $7.00 in nickels, dimes, and quarters. There are 41 coins in all, and the sum of the numbers of nickels and dimes is 3 less than the number of quarters. How many coins of each kind are there?

nickels?

dimes ?

quarters?

2 Answers

Relevance
  • Mike G
    Lv 7
    6 months ago
    Favorite Answer

    N+D+Q = 41

    N+D = Q-3

    Q-3+Q = 41

    Q = 22

    N+D = 19

    5N+10D+25Q = 700

    5N+10(19-N) = 150

    -5N = -40

    N = 8

    Nickels = 8

    Dimes = 11

    Quarters = 22

    • Login to reply the answers
  • 6 months ago

    n + d + q = 41

    n + d + 3 = q

    5n + 10d + 25q = 700

    n + d = q – 3

    n + d + q = 41

    (q – 3) + q = 41

    2q = 41 + 3 = 44

    q = 22

    5n + 10d + 25q = 700

    5n + 10d + 25•22 = 700

    5n + 10d + 550 = 700

    5n + 10d = 150

    n + 2d = 30

    n + d = q – 3

    n + d = 22 – 3

    n + d = 19

    n + 2d = 30

    n + 2d = 30

    –n – d = –19

    d = 11

    n = 19 – d = 8

    so we have

    q = 22

    d = 11

    n = 8

    check

    n + d + q = 41

    8 + 11 + 22 = 41 ok

    n + d + 3 = q

    8 + 11 + 3 = 22 ok

    5n + 10d + 25q = 700

    40 + 110 + 550 = 700 ok

    • Login to reply the answers
Still have questions? Get your answers by asking now.