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Thread: 3x3 arithmetic equations

  1. #1
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    3x3 arithmetic equations

    Been playing with 3x3 equations with coefficients in an arithmetic series..
    E.g

    x+2y+3z=4
    5x+6y+7z=8
    9x+10y+11z= 12

    Or even this

    x+2y+3z=4
    9x+11y+13z=15
    4x+1y+-2z= -5

    ( where each separate line is an arithmetic series).

    It appears that these types of equations ( been playing with wolfram) appear have no solutions, always??

    Can anyone see an easy way why?
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  2. #2
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    Re: 3x3 arithmetic equations

    Quote Originally Posted by rodders View Post
    ( where each separate line is an arithmetic series).
    That makes no sense...

    Clarify your problem.
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  3. #3
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    Re: 3x3 arithmetic equations

    Quote Originally Posted by DenisB View Post
    That makes no sense...

    Clarify your problem.
    Sorry, I mean the coeff are in an arithmetic series.
    So in the first set of equations the coeff are 1,2,3,4 ( first equation), second equation (5,6,7,8), third equation (9,10,11,12).

    In the second set of equations the coeff are 1,2,3,4 ( first eqn), second equation are (9,11,13,15), and third equation (4,1,-2,-5)

    So these sequences are arithmetic and I am interested in the equations formed by using these sequences?

    Does that help?
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  4. #4
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    Re: 3x3 arithmetic equations

    Not really.
    You're trying to solve a system of 3 equations.

    Can you solve this one:
    x + y + z = 9
    x + y - z = 1
    2x - y + 3z = 13
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  5. #5
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    Re: 3x3 arithmetic equations

    The first system

    x+2y+3z=4
    5x+6y+7z=8
    9x+10y+11z= 12

    has a solution $x=0, y=-1, z=2$

    actually it has infinitely many solutions
    Last edited by Idea; Feb 10th 2018 at 05:55 AM.
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  6. #6
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    Re: 3x3 arithmetic equations

    Quote Originally Posted by rodders View Post
    x+2y+3z=4
    5x+6y+7z=8
    9x+10y+11z= 12
    Nothing wrong with coefficients being arithmetic sequences,
    but you can't simply assign at random any results, like your 4,8,12 above.

    With following realistic values, you'll get a unique solution:
    1x+ 2y + 3z = 14
    5x +6y + 7z = 38
    9x + 10y + 11z = 62

    OK??
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  7. #7
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    Re: 3x3 arithmetic equations

    Quote Originally Posted by Idea View Post
    The first system
    x+2y+3z=4
    5x+6y+7z=8
    9x+10y+11z= 12
    has a solution $x=0, y=-1, z=2$
    actually it has infinitely many solutions
    Sure does! Another is a=1, b=-3, c=3
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  8. #8
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    Re: 3x3 arithmetic equations

    Thanks for all this. I think I have worked out what I was doing wrong!
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