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A problem in algebric geometry

  1. Jun 28, 2009 #1
    1. The problem statement, all variables and given/known data

    Hello,
    I am trying to solve a problem in algebraic geometry where three unknown variables x,y and z needs to be determined to further progress in this problem.
    The equations involving the variables x,y and z are 10y+28z-20=A;
    18z-30x+60=A;-84x-18y+204=A.
    Where A=3(x^2+y^2+z^2)=3(5x+5y+11z-20)
    also 3x+y+z=8.


    Could you suggest an effective method to find the possible values of x,y and z.

    If you have any ideas, anything will be appreciated. Thanks

    2. Relevant equations

    3x+y+z=8.

    3. The attempt at a solution
     
  2. jcsd
  3. Jun 28, 2009 #2

    Dick

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    Treat A as another variable. Use the linear equations to eliminate as many variables as you can from the set {x,y,z,A}. Substitute the results into the quadratic.
     
  4. Jun 29, 2009 #3
    Thanks for the hint Dick. Could you give me some elaboration on this method. Its difficult to eliminate any variable from the set {x,y,z,A}. I always get into equation of one variable in terms of other two.
    Or it will be helpful to give me some e-links where I can find similar problems (Topics) discussed.
     
  5. Jun 29, 2009 #4

    Dick

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    I don't see what the problem is. Start with 3x+y+z=8, so z=8-3x-y. Put that into all the other linear equations. Boom. z eliminated and one less equation, at least. Your linear equations aren't independent, so you'll find some of them turning into the same thing. There's also matrix type methods.
    http://en.wikipedia.org/wiki/System_of_linear_equations
     
  6. Jun 30, 2009 #5
    As I tried to solve by using
    10y+28z-20=18z-30x+60
    18z-30x+60=-84x-18y+204
    10y+28z-20=-84x-18y+204

    All of the equation turn into 3x+y+z=8 since

    30x+10y+10z=80 /: 10
    54x+18y+18z=144 /: 18
    84x+28y+28z=224 /: 28

    or the same equation 3x+y+z=8

    Now the only remaining equations are
    3(x^2+y^2+z^2)=3(5x+5y+11z-20)

    and 3x+y+z=8
     
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