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3 variables 3 equations (2 linear 1 quadratic)

  1. Jan 20, 2013 #1
    1. The problem statement, all variables and given/known data
    Find three numbers x,y,z satisfying the following equations:
    9x-6y-10z=1
    -6x+4y+7z=0
    x2+y2+z2=9


    2. Relevant equations



    3. The attempt at a solution
    The book suggests that the best approach is to express 2 unknowns in terms of the third unknown from the first 2 linear equations and, by substituting these expressions in the last equation, obtain a quadratic equation for the third unknown.
    So i try this:
    4(9x-6y-10z=1)= 36x-24y-40z=4
    6(-6x+4y+7z=0)= -36x+24y+42z=0
    (36x-24y-40z=4) + (-36x+24y+42z=0)= 2z=4
    z=2
    I figure I've gone wrong somewhere because this is not the strategy outlined in the book (it's also not the correct answer for z according to the book)
     
  2. jcsd
  3. Jan 20, 2013 #2

    Ray Vickson

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    If you have stated the problem correctly, you are right: z = 2, no matter what the book says. In fact, there are two solutions; they both have z = 2 but have different x,y values.
     
  4. Jan 20, 2013 #3
    Thanks for your reply. On further inspection I realise that the book does in fact say that z is equal to 2. However this still leaves me a bit confused about the purpose of the quadratic equation since it seems it is superfluous to the requirements for providing a solution.

    The problem is question 1A, page 44 (in the pdf digital copy. Or 22 in the hard copy)
    of this book:
    http://www.scribd.com/doc/39412845/Mathematics-and-Plausible-Reasoning-Vol1-Induction-and-Analogy-in-Mathematics-Polya-1954 [Broken]
     
    Last edited by a moderator: May 6, 2017
  5. Jan 20, 2013 #4

    Ray Vickson

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    If you know z = 2 you can put this into the first and third equations (for example), to get two equations in the two unknowns x and y. One of the equations is linear, so that helps.
     
    Last edited by a moderator: May 6, 2017
  6. Jan 20, 2013 #5
    Thanks for the help I'll try that.
     
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