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    Algebra2guy's Avatar
    Algebra2guy Posts: 6, Reputation: 1
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    #1

    Oct 9, 2010, 08:38 AM
    Would like to solve via elimination
    w+2x+2y+z= -2
    w+3x-2y-z= -6
    -2w-x+3y+3z= 6
    w+4x+y-2z= -14
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #2

    Oct 9, 2010, 08:56 AM

    You might want to use the Gaussian elimination method to make it easier?

    If not, you can use the first and second equation to eliminate y and z directly.
    Then use the first and third equation to get another equation in w and x.

    Take those two equations and solve simultaneously.

    Now, you have the value of w and x, you can solve any to equations simultaneously.

    Post what you get! :)
    Algebra2guy's Avatar
    Algebra2guy Posts: 6, Reputation: 1
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    #3

    Oct 9, 2010, 12:44 PM
    Comment on Unknown008's post
    Still don't get it. :(
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #4

    Oct 10, 2010, 12:10 AM

    Hm...

    Take 1st and 2nd equations:

    w+2x+2y+z= -2
    w+3x-2y-z= -6

    Eliminate 2y and z by adding both equations:

    w+2x+2y+z + (w+3x-2y-z) = -2 + (-6)

    This becomes:

    2w + 5x + 0 + 0 = -8
    2w + 5x = -8... (A)

    I think I misread the second equation... but anyway, this only makes it a longer.
    Take 3rd and 4th (I pre-multiplied the 4th by 3 and the 3rd by 2)

    -4w - 2x + 6y + 6z = 12
    3w + 12x + 3y - 6z = -42

    Add;

    -w + 10x +9y = -30... (B)

    Take 2nd and 3rd (I pre-multiplied the 2nd by 3)

    3w + 9x - 6y - 3z = -18
    - 2w - x + 3y + 3z = 6

    Add;

    w + 8x - 3y + 0 = -12
    w + 8x - 3y = -12... (C)

    Multiply C by 3

    3w + 24x - 9y = -36... (D)

    Add D and B:

    2w + 34x = -66... (E)

    Now, take A and E together and solve simultaneously;
    2w + 5x = -8
    2w + 34x = -66

    Subtract;

    29x = -58
    x = -2

    Can you find the rest now? :)
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    Algebra2guy Posts: 6, Reputation: 1
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    #5

    Oct 10, 2010, 06:45 PM
    w=1
    x= -2
    y= -1
    z=3
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #6

    Oct 10, 2010, 10:23 PM

    Great! Yes, those are the values of w, x, y and z. Good job :)

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