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Systems of Linear Eqs.

  1. May 23, 2013 #1
    1. The problem statement, all variables and given/known data
    A boy finds £2.10 in 20p, 10p and 2p coins. If there are 17 coins in all how many of each can he have?


    2. Relevant equations
    Row ops.



    3. The attempt at a solution
    I'm trying to come up with a few equations with which I can create a matrix.
    x1 = amount of 20p.
    x2 = amount of 10p.
    x3 = amount of 2p.
    So, 0.2(x1) + 0.1(x2) + 0.02(x3) = 2.10 and
    x1 + x2 + x3 = 17.
    Then, place the above in an augmented matrix and solve. Does this sound about right?
     
  2. jcsd
  3. May 23, 2013 #2

    danago

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    Gold Member

    Have you tried solving it in the way that you mentioned and checking if the result makes sense? Just try it and then interpret your results. If they make sense, then great! If not, try to figure out what went wrong. It looks like you've already made a good attempt, so you may as well follow through with your idea to see what happens :smile: In my opinion it's one of the best ways to learn!
     
  4. May 23, 2013 #3
    Well, having worked through you come out with a matrix that has a parameter (which is pretty clear). The answer at the back of the book is saying that there are only 2 possible values for each x which sort of contradicts the fact that there are parameters in the solution.
     
  5. May 23, 2013 #4

    danago

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    Gold Member

    Remember the context of the question though--if the x's represent the number of coins, they can only have non-negative integer values, which restricts the solution set. Of course there will be more solutions mathematically speaking, but not in the context of the question.
     
  6. May 23, 2013 #5

    Borek

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    Staff: Mentor

    No, it doesn't. Number of coins can be only a natural number, which puts additional constraint on the system.

    This is actually the same problem we face when trying to balance chemical reactions by the algebraic method - almost always there is not enough equations, but we know that the coefficients have to be positive and non zero integers, and additionally we want them to be smallest possible.

    Edit: danago was faster.
     
  7. May 23, 2013 #6
    Brilliant! Got it now. Thanks for the help!
     
  8. May 23, 2013 #7

    Ray Vickson

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    Science Advisor
    Homework Helper

    It is better to write the equations as x1+10x2+20x3=210 and x1+x2+x3=17, so you can deal with exact fractions instead of decimal numbers.

    You can (for example) solve for x1 and x2 in terms of x3. Then you can evaluate the solution for x3 = 0, 2, 3, ... and see if you ever get non-negative integer numbers for x1 and x2. There will only be a few possibilities, because if x3 is too large one of x1 or x2 will become < 0.
     
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