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Math Help - On the tip of my fingers but can't get it...

  1. #1
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    On the tip of my fingers but can't get it...

    I feel really stupid posting this question as I know Im on the verge of finding a way to solve it but I just can't think of it.

    Here it is...

    Solve for A, B, C, and D in the following equations:

    A+C=0
    B-D=0
    4A+3D+2C=7
    A+4B+6C=-6

    Thanks again for any and all help
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  2. #2
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    Quote Originally Posted by forkball42 View Post
    I feel really stupid posting this question as I know Im on the verge of finding a way to solve it but I just can't think of it.

    Here it is...

    Solve for A, B, C, and D in the following equations:

    A+C=0
    B-D=0
    4A+3D+2C=7
    A+4B+6C=-6

    Thanks again for any and all help
    Solve the second equation for D:
    D = B

    Then
    A+C=0
    4A+3B+2C=7
    A+4B+6C=-6

    Solve the first equation for C:
    C = -A

    So
    2A+3B=7
    -5A+4B=-6

    Solve the first equation for B:
    B = -\frac{2}{3}A + \frac{7}{3}

    Thus
    -\frac{23}{3}A + \frac{28}{3}=-6

    or
    A = 2

    I leave the rest to you.

    -Dan
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  3. #3
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    Hello, forkball42!

    A slightly different approach . . .


    Solve for A, B, C, and D:

    . . \begin{array}{cccc}A+C &=& 0 & [1]\\<br />
B-D & = & 0 & [2]\\4A+2C+3D& =&7 & [3]\\ A+4B+6C&=&\text{-}6 & [4]\end{array}

    \begin{array}{ccccc}<br />
\text{From [1], we have:} & C & = & \text{-}A& [5] \\<br />
\text{From [2], we have:} & D & = & B & [6]\end{array}


    Substitute [5] and [6] into [3] and [4]:

    . . \begin{array}{cccccccc}4A + 2(\text{-}A) + 3(B ) & = & 7 & \Rightarrow & 2A + 3B & = & 7& [7] \\<br /> <br />
A + 4B + 6(\text{-}A) & = & \text{-}6 & \Rightarrow & \text{-}5A + 4B  & = & \text{-}6& [8]\end{array}


    \begin{array}{cccc}\text{Multiply [7] by 4:} & 8A + 12B & = & 28 \\<br />
\text{Multiply [8] by -3:} & 15A - 12B & = & 18\end{array}

    Add: . 23A \:=\:46\quad\Rightarrow\quad\boxed{ A \,=\,2}

    . . Substitute into [5]: . \boxed{C \,=\,\text{-}2}

    Substitute into [7]: . 2(2) + 3B \:=\:7\quad\Rightarrow\quad\boxed{ B \,= \,1}

    . . Substitute into [6]: . \boxed{D \,= \,1}

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