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Infinite number of solutions

  1. Dec 31, 2012 #1
    1. The problem statement, all variables and given/known data

    I worked out until the last part of the question and 3 equations with 3 unknowns got reduced to this:

    x - 2y + 3z = 1
    x + 3z = 3



    3. The attempt at a solution


    y = 1,
    x = 3 -3z

    Letting x = λ where λ is any real number,

    (x,y,z) = (3,1,0) + λ(-3,0,1)


    It wouldn't make a difference if i let z be λ instead right?
     
  2. jcsd
  3. Dec 31, 2012 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    The solution you give is the same as [itex]x= 3- 3\lambda[/itex], [itex]y= 1[/itex], and [itex]z= \lambda[/itex]. Since [itex]z= \lambda[/itex], it doesn't matter which you use. You can write the solution as [itex](x, y, z)= (3-\lambda, 1, \lambda)[/itex] or as [itex](x, y, z)= (3-z, 1, z)[/itex].
     
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