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Math Help - Algebra.. with 8 variables.

  1. #1
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    Algebra.. with 8 variables.

    Ok i have these 4 equations:
    a^2 + bc + w^2 + xy = (a+w)^2 + (b+x)(c+y)
    ca + cd + yw + yz = (c+y)(a+w) + (c+y)(d+z)
    ba + bd + xw + xz = (b+x)(a+w) + (b+x)(d+z)
    cb + d^2 + yx +z^2 = (c+y)(b+x) + (d+z)^2
    From those i have derived:
    0 = 2aw + by + xc (1)
    0 = 2dz + by + xc (2)
    0 = c(w+z) + y(a+d) (3)
    0 = b(w+z) + x(a+d) (4)
    I'm trying to find what values (integers only) satisfy each of these equations..


    so i do (1) - (2)
    0 = 2aw - 2dz
    2dz = 2aw
    dz = aw
    (3)/(4)
    0 = (c+y) / (b+x)
    Thus
    c+y = 0
    b+x /= 0
    dz = aw
    Deriving some values:
    a = 3, b = 2, c = 4, d = 6
    w = 4, x = 1, y = -4, z = 2
    None of these satisfy the original equations.
    So trying to gather some more rules:

    (3)+(4)
    0 = c(w+z) + y(a+d) + b(w+z) + x(a+d)
    0 = (c+B)(w+z) + (y+x)(a+d)
    (c+B)(w+z) = -(y+x)(a+d)
    (c+B)(w+z)/(a+d) = -(y+x)
    So give random values to a and d
    a = 2, d = 4
    Derive w and z from this (the equation aw = dz)
    w = 6 z = 3


    Sub in:
    ((c+B)(6+3))/(2+4) = -(y+x)
    -(y+x) must be an integer so:


    Give values to c and b so that:
    9(c+B)/(6) = int
    b = 1 c = 3


    Thus
    -(y+x) = 6
    c + y = 0 so:
    1 + y = 0, so y = -1
    Sub in:
    -(-1+x) = 6
    -1 + x = -6
    x = -5
    So using the values
    a = 2, b = 3, c = 1, d = 4, w = 6, x = -5, y = -1, z = 3
    and subbing back into the original 4 equations.....
    IS WRONG!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!
    so can anyone see where my error/s are?
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  2. #2
    Bar0n janvdl's Avatar
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    Quote Originally Posted by thaiboy View Post
    Ok i have these 4 equations:
    a^2 + bc + w^2 + xy = (a+w)^2 + (b+x)(c+y)
    ca + cd + yw + yz = (c+y)(a+w) + (c+y)(d+z)
    ba + bd + xw + xz = (b+x)(a+w) + (b+x)(d+z)
    cb + d^2 + yx +z^2 = (c+y)(b+x) + (d+z)^2
    From those i have derived:
    0 = 2aw + by + xc (1)
    0 = 2dz + by + xc (2)
    0 = c(w+z) + y(a+d) (3)
    0 = b(w+z) + x(a+d) (4)
    I'm trying to find what values (integers only) satisfy each of these equations..


    so i do (1) - (2)
    0 = 2aw - 2dz
    2dz = 2aw
    dz = aw
    (3)/(4)
    0 = (c+y) / (b+x)
    Thus
    c+y = 0
    b+x /= 0
    dz = aw
    Deriving some values:
    a = 3, b = 2, c = 4, d = 6
    w = 4, x = 1, y = -4, z = 2
    None of these satisfy the original equations.
    So trying to gather some more rules:

    (3)+(4)
    0 = c(w+z) + y(a+d) + b(w+z) + x(a+d)
    0 = (c+B)(w+z) + (y+x)(a+d)
    (c+B)(w+z) = -(y+x)(a+d)
    (c+B)(w+z)/(a+d) = -(y+x)
    So give random values to a and d
    a = 2, d = 4
    Derive w and z from this (the equation aw = dz)
    w = 6 z = 3


    Sub in:
    ((c+B)(6+3))/(2+4) = -(y+x)
    -(y+x) must be an integer so:


    Give values to c and b so that:
    9(c+B)/(6) = int
    b = 1 c = 3


    Thus
    -(y+x) = 6
    c + y = 0 so:
    1 + y = 0, so y = -1
    Sub in:
    -(-1+x) = 6
    -1 + x = -6
    x = -5
    So using the values
    a = 2, b = 3, c = 1, d = 4, w = 6, x = -5, y = -1, z = 3
    and subbing back into the original 4 equations.....
    IS WRONG!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!
    so can anyone see where my error/s are?
    You have already posted this here.

    Please do not double post. It is against the rules.
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  3. #3
    Newbie
    Joined
    Jun 2008
    Posts
    3
    sorry, i posted it in the wrong forum.
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