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    x+yz=2
    y+xz=4
    z+xy=3
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    (Original post by kamallikenan)
    x+yz=2
    y+xz=4
    z+xy=3
    If you gave a bit of background to where this came from you might get a better response.
    If you are setting a puzzle for interested readers then say so. There is a place for that kind of thing but that place is not in this thread.

    Now to the solution. Effectively this is a quintic and most people would solve a quintic numerically. There may be an analytic method in special cases but I think there is no general analytic solution for quintics.

    It is a quintic because you need to get an equation in one variable and since all three equations involve all three variables, each substitution gives you an increase in powers of the unknowns involved.

    My single unchecked calculation gives me y=1.75 plus four complex solutions which I did on Wolfram alpha solving

    (y-4)(1-y^2)^2+3(2-3y)(1-y^2)-(2-3y)^2=0
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    Make x the subject in the 1st equation.

    Then sub in for x in the other 2 equations.

    Try that and see where it goes.
 
 
 
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Updated: January 31, 2017
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