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Integer Solutions to Linear Systems

  1. Mar 17, 2013 #1
    1. The problem statement, all variables and given/known data
    We were asked to solve it using Augmented matrix. I just need one more equation though.
    A jar of coins contains only dimes, nickels, and pennies. There are 98 coins in the jar, and the total value of the coins is $6.49.
    Set up the system of equations representing this system, in the variables d, n, and p.

    2. Relevant equations
    1 dime = 10 cents
    1 nickel = 5 cents
    1 penny = 1 cent

    3. The attempt at a solution
    I set up two equations:
    d+n+p = 98
    10d + 5n + p = 649

    but I can't find the third equation.

    p is a free variable.
    the 9 in 649 is the clue i think because only nickels and pennies can add up to 9
    and there are two possible ways:
    1 nickel + 4 pennies, OR
    9 pennies
    So d and n depend on p but I don't know how to express it in an equation.

    I tried 5n+p=9 but it didn't work.
    I tried to substitute 5n=9-p for n, but then it's not linear algebra and the answer was not correct.
     
  2. jcsd
  3. Mar 17, 2013 #2

    epenguin

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    Homework Helper
    Gold Member

    You could also have 14 or 19 or in general ? pennies. You could express that as a formula.
    Don't worry you don't always have to express everything as a formula.
    Although you can't eliminate from the equations in the ordinary way to reduce it to one unknown, you can reduce it to two unknowns - in different ways but try the simplest first. You can then try the same sort of logic - properties of whole numbers and divisibility etc.
    It is not guranteed that there is a unique solution.
     
  4. Mar 17, 2013 #3
    Thank you. I think I salved it!
    It should be two equations:
    x+y+z=98
    10x+5y+z=649

    From the Augmented Matrix I got:
    y=331/5 - 9/5(z)
    x=31.8+0.8z

    We just pretend that z is a constant ie. a free variable.
     
  5. Mar 18, 2013 #4

    SteamKing

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    If you set p = 4, 9, whatever, then you must adjust the two equations accordingly.
    For instance, ifyou assume p = 4, then the equations become:

    d + n = 94
    10d + 5n = 645

    you can then solve for d and n and get a unique solution for p = 4.
    If p is assumed to be 9, then there will be a different number of dimes and nickles.

    The solutions to d,n,p only have meaning if each is an integer. In order to satisfy the total amount of the change, p is restricted in which values it can assume, i.e., 4, 9, 14, etc. You can't have 0.8 of a dime.
     
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