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Math Help - Solving systems of equations by determinants, elimination or substitution

  1. #1
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    Solving systems of equations by determinants, elimination or substitution

    Solve the given sytem of equations

    9x+12y+2z=23
    21x+2y-5z=-25
    -18x+6y-4z=-11

    Solve the given system of equations by use of determinants

    26L-52M-52N=39
    45L+96M+40N=-80
    55L+62M-11N=-48

    Solve the given system of equation by determinants

    6541x+4397y=-7732
    3309x-8755y=7622

    Solve the given system of equations by either by either substitution or elimination

    0.66x+0.66y=-0.77
    0.33x+1.32y=1.43
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  2. #2
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    Bump

    Anyone know the answers to these?
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by OzzMan View Post
    Bump

    Anyone know the answers to these?
    Please don't "bump." It's rude and see rule #10 here.

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by OzzMan View Post
    Solve the given sytem of equations

    9x+12y+2z=23
    21x+2y-5z=-25
    -18x+6y-4z=-11
    Since you have to do a number of these by determinants I'll do this one that way. Specifically, I'm using Cramer's rule.
    \left ( \begin{matrix} 9 & 12 & 2 \\ 21 & 2 & -5 \\ -18 & 6 & -4 \end{matrix} \right ) \left ( \begin{matrix} x \\ y \\ z \end{matrix} \right ) = \left ( \begin{matrix} 23 \\ -25 \\ -11 \end{matrix} \right )

    First we need the determinant of the coefficient matrix. This is 2610.

    Now to solve the system. By Cramer's rule we take the vector on the RHS and replace the columns of the coefficient matrix one by one:
    x = \frac{1}{2610} \left | \begin{matrix} 23 & 12 & 2 \\ -25 & 2 & -5 \\ -11 & 6 & -4 \end{matrix} \right | = \frac{-290}{2610} = -\frac{1}{9}

    y = \frac{1}{2610} \left | \begin{matrix} 9 & 23 & 2 \\ 21 & -25 & -5 \\ -18 & -11 & -4 \end{matrix} \right | = \frac{3045}{2610} = \frac{7}{6}

    y = \frac{1}{2610} \left | \begin{matrix} 9 & 12 & 23 \\ 21 & 2 & -25 \\ -18 & 6 & -11 \end{matrix} \right | = \frac{13050}{2610} = 5

    -Dan
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  5. #5
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    Quote Originally Posted by OzzMan View Post
    ...

    Solve the given system of equations by either by either substitution or elimination

    0.66x+0.66y=-0.77
    0.33x+1.32y=1.43
    Hello,

    \left|~\begin{array}{lcr}0.66x+0.66y&=&-0.77 \\0.33x+1.32y&=&1.43\end{array}\right. multiply the 2nd equation by (-2):

    \left|~\begin{array}{lcr}0.66x+0.66y&=&-0.77 \\-0.66x-2.64y&=&-2.86\end{array}\right. add columnwise(?):

    -1.98y = -3.63~\iff~y = \frac{11}{6} plug in this value into the first (or second) equation and solve for x. I've got: \boxed{x = -3}
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    Is there a simple way to evaluate this system of equations without a calculator?

    26L-52M-52N=39
    45L+96M+40N=-80
    55L+62M-11N=-48
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by OzzMan View Post
    Is there a simple way to evaluate this system of equations without a calculator?

    26L-52M-52N=39
    45L+96M+40N=-80
    55L+62M-11N=-48
    Simple? No. Without a calculator? Yes. But it's the exact same method and it will take you a while. I'd use Cramer's rule for this as that has less computation than most other methods I know. (Gaussian elimination would be a good alternate as it is easy to check your work.) I would think of this as an excellent way to exercise your algebra skills, but if you are being graded on it, use the calculator!

    -Dan
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