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Math Help - system of Equations (specific type)

  1. #1
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    system of Equations (specific type)

    can we solve this system

    a+11b+7c=12,
    8a+3b+3c=9,
    a^2 +a+bc=12!

    please help because i need the solution of these types of equations in my research and i stuck here
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  2. #2
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    Re: system of Equations (specific type)

    Quote Originally Posted by awaisysf View Post
    can we solve this system

    a+11b+7c=12,
    8a+3b+3c=9,
    a^2 +a+bc=12!

    please help because i need the solution of these types of equations in my research and i stuck here
    1. Solve the first two equations for a and b. I've got

    a = \frac{3(21-4c)}{85}~\vee~b=\frac{87-53c}{85}

    2. Plug in these terms into the 3rd equation. You'll get a quadratic equation in c. Solve for c. You'll get 2 complex solutions.

    3. Plug in the values of c into the equations of a and b to determine their values.
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  3. #3
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    Re: system of Equations (specific type)

    but i want to calculate in a finite field like F13
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    Re: system of Equations (specific type)

    Quote Originally Posted by awaisysf View Post
    a^2 +a+bc=12!
    12! = 479,001,600 ; typo?
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    Re: system of Equations (specific type)

    Quote Originally Posted by awaisysf View Post
    but i want to calculate in a finite field like F13
    Well, if a^2 + a + bc = 12, then:
    c = [4863 +- 85SQRT(-183543)] / 8722 (as earboth was trying to tell you...)

    But if: Well, if a^2 + a + bc = -12, then:
    c = [4863 +- 85SQRT(235113)] / 8722

    CHECK your equations !!
    Last edited by Wilmer; July 22nd 2012 at 08:15 PM.
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    Re: system of Equations (specific type)

    ok. the equations are a+11b+7c=12, 8a+3b+3c=9, a^2 +a+bc=12. and i need solution of this system in finite field F13.
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  7. #7
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    Re: system of Equations (specific type)

    from the first equation we have:

    a = 2b + 6c + 12 (mod 13) (*)

    therefore:

    8(2b + 6c + 12) + 3b + 3c = 9 (mod 13)
    3b + 9c + 5 + 3b + 3c = 9 (mod 13)
    6b + 12c = 4 (mod 13)
    6b = c + 4 (mod 13)
    11(6b) = 11(c + 4) (mod 13)
    b = 11c + 5 (mod 13)

    substituting this back in (*), we get:

    a = 2(11c + 5) + 6c + 12 = 9c + 10 + 6c + 12 = 2c + 9 (mod 13)

    this agrees with earboth's answer, if you reduce mod 13:

    a = 3(21 - 4c)/85 = 3(8 - 4c)/7 = 6(8 - 4c) = 6(8 + 9c) = 2c + 9

    b = (87 - 53c)/85 = (9 - c)/7 = 2(9 - c) = 2(9 + 12c) = 11c + 5

    finally, we turn to the last equation:

    a2 + a + bc = 12

    (2c + 9)2 + 2c + 9 + (11c + 5)c = 12 (mod 13)
    4c2 + 10c + 3 + 2c + 9 + 11c2 + 5c = 12 (mod 13)
    2c2 + 4c + 12 = 12 (mod 13)
    2c2 + 4c = 0 (mod 13)
    c(2c + 4) = 0 (mod 13)

    whereby we see that the two solutions are c = 0, and c = 11. since a quadratic may introduce extraneous solutions, we must check each one separately in the original equations.

    c = 0 leads to: a = 9, and b = 5.

    9 + 11(5) = 9 + 3 = 12 (mod 13), so this is consistent with the first equation.
    8(9) + 3(5) = 7 + 2 = 9 (mod 13), so this is consistent with the second equation.
    (9)2 + 9 = 3 + 9 = 12 (mod 13), so this is consistent with the third equation. thus (a,b,c) = (9,5,0) is a solution.

    c = 11 leads to: a = 5, and b = 9.

    5 + 11(9) + 7(11) = 5 + 8 + 12 = 12 (mod 13), so this checks out.
    8(5) + 3(9) + 3(11) = 1 + 1 + 7 = 9 (mod 13), so this checks out as well.
    (5)2 + 5 + (9)(11) = 12 + 5 + 8 = 12 (mod 13), so this also checks out. thus (a,b,c) = (5,9,10) is also a solution.

    note: check my arithmetic, as these computations have been somewhat involved, and i may have made errors.
    Thanks from awaisysf
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    Re: system of Equations (specific type)

    Quote Originally Posted by Deveno View Post
    a = 2b + 6c + 12 (mod 13)
    This is new to me; equation a + 11b + 7c = 12 is changed to above...

    I can see that 11b - 13b = -2b and 7c - 13c = -6c :
    a - 2b - 6c = 12
    a = 2b + 6c + 12

    BUT why?
    And why not 1a - 13a = -12a? Or 12 - 13 = -1?

    Know a site where I can read up on this?
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    Re: system of Equations (specific type)

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