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Solving a Linear equation with 3 unknown variables

  1. Nov 10, 2013 #1
    3x + 4y + 2y = 1

    The solutions for x, y, and z is { ( (1/3-4l-2m) | 3l | 2m ) }, where y = l and z = m.

    I've tried this method, presupposing y = l and m = z, then it came to

    I. l = (1 - 3x -2m) / 4

    II. m = (1 - 3x -4l) / 2

    If I try to put in either (I) or (II) to x, it would come to the first equation.

    I don't have any idea what should I do next.

    Thanks.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 10, 2013 #2

    Ray Vickson

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    Homework Helper

    You cannot do more, in general. The equation defines a 2-dimensional plane in three dimensions, so will have infinitely many point in it; that is, you have a region defined by two "free" variables. Assuming your equation was supposed to be 3x + 4y + 2z = 1 (you wrote 3x + 4y + 2y = 1 !)
    we can just say that
    [tex] z = \frac{1}{3}\, (1-4y - 2z) [/tex]
    I'm not sure whether or not you wrote this, because your use of parentheses was sloppy and so it is impossible to tell what gets divided by 3; and besides that you used some other weird notation whose meaning is unclear.
     
  4. Nov 11, 2013 #3

    NascentOxygen

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    Ray, is your equation supposed to be
    [tex] x = \frac{1}{3}\, (1-4y - 2z) [/tex]
     
  5. Nov 11, 2013 #4

    Ray Vickson

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    Yes. (He says, as he cringes and slaps forehead.)
     
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