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Math Help - Simultaneous equations

  1. #1
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    Simultaneous equations

    Solve for x,y,and z. Q.1 4x - y + 3z = 9 3x + y - z = 5 2x + 3y - 2z = 1 Q.2 2a + 4b + 5c = 2 5a + 3b - 2c = 13 3a - 2b - 3c = 0
    Last edited by derekv; August 1st 2007 at 02:04 AM.
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  2. #2
    Senior Member tukeywilliams's Avatar
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    I would write these as matrices:

     \begin{bmatrix} 4 & -1 & -3 \\ 3 & 1 & -1 \\ 2 & 3 & -2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 9 \\ 5 \\ 1 \end{bmatrix} and then row reduce to find  x,y,z .

     \begin{bmatrix} 2 & 4 & 5 \\ 5 & 3 & -2 \\ 3 & -2 & 3 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 2 \\ 13\\ 0 \end{bmatrix}
    Last edited by tukeywilliams; August 1st 2007 at 02:09 AM.
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  3. #3
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    Quote Originally Posted by tukeywilliams View Post
    I would write these as matrices:

     \begin{bmatrix} 4 & -1 & -3 \\ 3 & 1 & -1 \\ 2 & 3 & -2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 9 \\ 5 \\ 1 \end{bmatrix} and then row reduce to find  x,y,x .

     \begin{bmatrix} 2 & 4 & 5 \\ 5 & 3 & -2 \\ 3 & -2 & 3 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 2 \\ 13\\ 0 \end{bmatrix}
    Could you explain what you mean please? I don't understand.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by derekv View Post
    Solve for x,y,and z. Q.1 4x - y + 3z = 9 3x + y - z = 5 2x + 3y - 2z = 1
    If you don't know how to do row reduction, how about substitution?
    4x - y + 3z = 9
    3x + y - z = 5
    2x + 3y - 2z = 1

    Take the second equation and solve it for z:
    z = 3x + y - 5

    Now insert this value of z into the other two equations:
    4x - y + 3(3x + y - 5) = 9 \implies 13x + 2y = 24
    2x + 3y - 2(3x + y - 5) = 1 \implies -4x + y = -9

    So we need to solve
    13x + 2y = 24
    -4x + y = -9

    Now solve the bottom equation for y:
    y = 4x - 9

    and insert this value into the top equation:
    13x + 2(4x - 9) = 24 \implies 21x = 42

    So we have
    21x = 42

    or
    x = 2.

    Thus
    y = 4(2) - 9 = -1

    and
    z = 3(2) + (-1) - 5 = 0

    Thus the solution to the system is (x, y, z) = (2, -1, 0).

    -Dan
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  5. #5
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    Hello, derekv!

    I'll solve #1 by Elimination . . . it takes a while, though.


    1) Solve for x, y and z.

    . . \begin{array}{cccc}4x - y + 3z & = & 9 & {\color{blue}[1]} \\  3x + y - z & = & 5 & {\color{blue}[2]} \\  2x + 3y - 2z & = & 1 & {\color{blue}[3]}\end{array}
    Add {\color{blue}[1]} and {\color{blue}[2]}: . 7x + 2z \:=\:14\;\;{\color{blue}[4]}


    \begin{array}{cccc}\text{Multiply {\color{blue}[2]} by 3:} & 9x+ 3y - 3z & =& 15 \\<br />
\text{Subtract {\color{blue}[3]}:} & 2x + 3y - 2z & = & 1\end{array}
    . . and we have: . 7x - z \:=\:14\;\;{\color{blue}[5]}


    \begin{array}{cccc}\text{We have {\color{blue}[4]}:} & 7x + 2z& = & 14 \\<br />
\text{Subtract {\color{blue}[5]}:} & 7x - z & = & 14 \end{array}
    . . and we have: . 3z \,=\,0\quad\Rightarrow\quad\boxed{z\,=\,0}

    Substitute into {\color{blue}[5]}: . 7x - 0 \:=\:14\quad\Rightarrow\quad\boxed{x \,=\,2}

    Substitute into {\color{blue}[2]}: . 3(2) + y - 0 \:=\:5\quad\Rightarrow\quad\boxed{y \,=\,\text{-}1}


    Answer: . (x,\,y,\,z) \:=\:(2,\,\text{-}1,\,0)

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  6. #6
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    You make it so much easier to understand, thank you both.
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