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Math Help - Help solving systems with 3 variables

  1. #1
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    Exclamation Help solving systems with 3 variables

    Ok so i keep having trouble with these problems, can anyone help? You have to use elimination ...but it uses 3 variables... x,y,z

    3x + y + z = 2
    4x - 2y + 3z = -4
    2x + 2y + 2z = 8
    Last edited by I hate Math; October 26th 2008 at 07:13 PM.
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  2. #2
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    Combine the second two equations to eliminate the Y's. Then solve one of the variables in the new equation (say X) in terms of the other (Z). Then multiply and combine two of the original equations to get the X's gone. Then solve for Y in terms of Z. Then substitute the values in for X and Y (the ones in terms of Z) so that the entire equation is all Z's. Then solve and substitute the Z value into the two-variable equations to solve for X and Y. Then check to make sure that it works.

    (If it doesn't work, try getting rid of the same variable both times you combine equations. Then multiply and combine the two new equations to eliminate one variable to find the value of the other. Then substitute the value again, and check.)
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  3. #3
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    Quote Originally Posted by I hate Math View Post
    Ok so i keep having trouble with these problems, can anyone help? You have to use elimination ...but it uses 3 variables... x,y,z

    3x + y + z = 2
    4x - 2y + 3z = -4
    2x + 2y + 2z = 8
    (1) 3x + y + z = 2
    (2) 4x - 2y + 3z = -4
    (3) 2x + 2y + 2z = 8

    Multiply equation (1) by -2 and add it to equation (3).

    (1) -6x - 2y - 2z = -4
    (3) 2x + 2y + 2z = 8
    -----------------------
    (4) -4x = 4

    x = -1



    Add equations (2) and (3)

    (2) 4x - 2y + 3z = -4
    (3) 2x + 2y + 2z = 8
    ----------------------
    (5) 6x + 5z = 4

    Substitute x = -1 into (5)

    (5) 6x + 5z = 4
    6(-1) + 5z = 4
    -6 + 5z = 4

    z = 2

    Substitute x = -1 and z = 2 into equation (1)

    (1) 3x + y + z = 2
    3(-1) + y + (2) = 2
    -3 + y + 2 = 2
    y = 2 + 3 - 2

    y = 3


    Solution (-1, 3, 2)
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