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Math Help - System of Equations - Help

  1. #1
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    System of Equations - Help

    Solve the following system of equations using the elimination method

    3x-9y=10 (1)
    4x+y=7 (2)

    I have completed this question, but I would like someone to look over it to make sure that I did the right thing, because it does not look like the right answer to me. Any help or comments would be appreciated.
    Thanks!

    My Answer:
    3x-9y=10 (1)
    4x+y=7 (2)
    By elimination:
    Take equation (2) and multiply by 9
    Thus, (2) x 9 gives you 36x-9y = 63 (3)
    Add equation 3 to 1
    Which gives you 39x = 73 ... divide both sides by 39 and x= 73/39 or 1.87

    Substitute x (1.87) into Equation #1 to find y:
    3(73/39)-9y=10
    ..This is where I get confused.
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  2. #2
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    He4llo, cnmath16!

    Solve the following system of equations using the elimination method

    . . \begin{array}{cccc}3x-9y&=&10 & [1] \\<br />
4x+y&=&7 & [2] \end{array}


    My Answer:

    \begin{array}{cccccc}\text{Multiply [2] by 9:} & 36x + 9y &=& 63 \\<br />
\text{Add [1]:} & 3x - 9y &=& 10 \end{array}


    Which gives you: . 39x \:= \:73 \quad\Rightarrow\quad x \:=\:\tfrac{73}{39} . . . . Good!

    Substitute into [2]: . 4\left(\tfrac{73}{39}\right) + y \;=\;7 \quad\Rightarrow\quad y \:=\:7 - \tfrac{292}{39} \quad\Rightarrow\quad y \:=\:\text{-}\tfrac{19}{39}


    Therefore: . \begin{Bmatrix}\; x &=&\dfrac{73}{39}\; \\ \\[-3mm] \; y &=& \text{-}\dfrac{19}{39}\; \end{Bmatrix}

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