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Math Help - Gaus Elimination Help

  1. #1
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    Question Gaus Elimination Help

    Ok im totally stumped because the variables got switched up on me... here is the problem

    A furniture copmany makes loungers, chairs, and footstools made out of wood, fabric and stuffing. The number of units of each of these materials needed for each of the products is given in the table below. How many of each product can be made if tehre are 54 units of wood, 63 units of fabric, and 43 units of stuffing availabe?

    __________Wood__Fabric__Stuffing
    Lounger_____1_____2_______2
    Chair_______2_____2_______1
    Footstool___3_____1_______1
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  2. #2
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    Hello, Kdocimo90!

    A furniture company makes loungers, chairs, and footstools made out of wood,
    fabric and stuffing. The number of units of each of these materials needed
    for each of the products is given in the table below.

    How many of each product can be made if there are 54 units of wood,
    63 units of fabric, and 43 units of stuffing availabe?

    \begin{array}{|c||c|c|c|}<br />
& \text{Wood} & \text{Fabric} & \text{Stuffing} \\ \hline \hline<br />
\text{Lounger }(x) & 1 & 2 & 2 \\ \hline<br />
\text{Chair }(y) & 2 & 2 & 1 \\ \hline<br />
\text{Stool }(z) & 3 & 1 & 1 \\ \hline \hline<br />
\text{Total} & 54 & 63 & 43 \\ \hline \end{array}
    The equations are: . \begin{array}{ccc}x + 2y + 3z &=& 54 \\ 2x + 2y + z &=& 63 \\ 2x + y + z &=& 43 \end{array}


    We have: . \begin{array}{|ccc|c|}<br />
1 & 2 & 3 & 54 \\ 2 & 2 & 1 & 63 \\ 2 & 1 & 1 & 43 \end{array}


    \begin{array}{c}\\ R_2-2R_1 \\ R_3-R_2\end{array} \begin{array}{|ccc|c|}<br />
1 & 2 & 3 & 54 \\ 0 & \text{-}2 & \text{-}5 & \text{-}45 \\ 0 & \text{-}1 & 0 & \text{-}20 \end{array}


    \begin{array}{c}R_1+R_2 \\ \text{-}1\cdot R_2 \\ \text{-}1\cdot R_3 \end{array} \begin{array}{|ccc|c|} 1& 0 & \text{-}2 & 9 \\ 0 & 2 & 5 & 45 \\ 0 & 1 & 0 & 20 \end{array}


    \begin{array}{c} \\ R_2-R_3 \\ \\ \end{array} \begin{array}{|ccc|c|}<br />
1 & 0 & -2 & 9 \\ 0 & 1 & 5 & 25 \\ 0 & 1& 0 & 20 \end{array}


    \begin{array}{c}\\ \\ R_3-R_2 \end{array} \begin{array}{|ccc|c|}<br />
1 & 0 & \text{-}2 & 9 \\ 0 & 1 & 5 & 25 \\ 0 & 0 & \text{-}5 & \text{-}5 \end{array}


    \begin{array}{c} \\ \\ \text{-}\frac{1}{5}\cdot R_3\end{array} \begin{array}{|ccc|c|}<br />
1 & 0 & \text{-}2 & 9 \\ 0 & 1 & 5 & 25 \\ 0 & 0 & 1 & 1 \end{array}


    \begin{array}{c}R_1+2R_3 \\ R_2 - 5R_3 \\ \\ \end{array} \begin{array}{|ccc|c|} 1&0 & 0 & 11 \\ 0 & 1 & 0 & 20 \\ 0 & 0 & 1 & 1 \end{array}


    Therefore: . x = 11,\;y = 20,\;z = 1

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