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Math Help - Three-Variable Linear Systems

  1. #1
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    Three-Variable Linear Systems

    I've tried solving this problem for a couple hours and I am stumped
    a little help would be much appreciated

    X + Y + Z = 18
    X - Y - Z = 12
    3X - 2Y + 4Z = 4

    Thanks in Advanced
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  2. #2
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    You need to solve


    \left[ \begin{array}{c}<br />
x\\<br />
y\\<br />
z\end{array}\right]=\left[ \begin{array}{ccc}<br />
1 & 1 & 1 \\<br />
1 & -1 & -1 \\<br />
3 & -2 & 4 \end{array} \right]^{-1} \left[ \begin{array}{c}<br />
18\\<br />
12\\<br />
4\end{array}\right]

    You you know how to find this inverse?
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  3. #3
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    i dont know how to find the inverse
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  4. #4
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    You can use this Matrix calculator

    Assuming you know simple matrix operations like mulitplication and finding determinants,

    To solve the system consider cramer's Method.

    Cramer's rule - Wikipedia, the free encyclopedia

    If you don't know these elementary ideas I suggest this problem should be solved using subtitution and elimination.
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  5. #5
    Super Member Quacky's Avatar
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    (1) X + Y + Z = 18
    (2) X - Y - Z = 12
    (3) 3X - 2Y + 4Z = 4

    Here's my simplistic and basic method

    Rearranging (1) to isolate X
    (4) X=18-Z-Y
    Substituting into (2)
    18-Z-Y-Y-Z=12
    18-2Z-2Y=12
    Rearranging this gives:
    2Y+2Z=6
    Taking out (and then dividing through by) a common factor of 2 gives
    Y+Z=3
    Substituting this into equation 1 gives
    X+3=18
    X=15

    Substituting that into all equations leaves:
    Y+Z=3
    2Y-4Z=41
    Which can then be solved easily simultaneously to give
    Y=3-Z
    2(3-Z)-4Z=41
    6-6Z=41
    -6Z=35
    Z=-5.833
    Then substituting this back into the above equations will give
    Y=8.8333
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  6. #6
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    Hello, jasimmons86!

    What methods are you allowed to use?


    \begin{array}{cccc}x + y + z &=& 18 & [1] \\<br />
x - y - z &=& 12 & [2] \\<br />
3x - 2y + 4z &=& 4 & [3] \end{array}

    Add [1] and [2]: / 2x \:=\:30 \quad\Rightarrow\quad \boxed{x \;=\;15}


    \begin{array}{ccccccc}\text{Substitute into [2]:} & 15 - y - z \:=\:12 & \Rightarrow &  y + z \:=\:3 & [3] \\<br />
\text{Substitute into [3]:} & 45 - 2y + 4z \:=\:4 & \Rightarrow & 2y - 4z \:=\:41 & [4] \end{array}

    . . \begin{array}{ccccc}4 \times [3]: & 4y + 4z \:=\:12 \\<br />
\text{Add [4]:} & 2y - 4z \:=\:41 \\ \hline \\[-3mm]<br />
\text{We have: }& 6y\qquad \:=\:53 \end{array}

    Hence: . \boxed{y \:=\:\frac{53}{6}}


    Substitute into [3]: . \frac{53}{6}+ z \:=\:3 \quad\Rightarrow\quad\boxed{ z \:=\:-\frac{35}{6}}

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