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Math Help - help me solve this system of linear equation.

  1. #1
    Newbie JoanneMac's Avatar
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    Exclamation help me solve this system of linear equation.

    in 3 variables, please. I'm tearing my hair out!!

    3x^2+4y+z=7
    2y+z=3
    -5x^2+3y+8z=-31

    Sorry to be such a pest, but math isn't my best subject.
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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by JoanneMac View Post
    in 3 variables, please. I'm tearing my hair out!!

    3x^2+4y+z=7
    2y+z=3
    -5x^2+3y+8z=-31

    Sorry to be such a pest, but math isn't my best subject.
    Hello Joanne,

    Tearing your hair out will not help.

    (1) 3x^2+4y+z=7

    (2) 2y+z=3

    (3) -5x^2+3y+8z=-31

    Use (1) and (3) to eliminate x

    15x^2+20y+5z=35

    -15x^2+9y+24z=-93
    ----------------------------------
    (4) 29y+29z=-58

    Solve (2) for z and substitute in (4)

    (2) z=3-2y

    (4) 29y+29(3-2y)=-58

    29y+87-58y=-58

    -29y=-145

    \boxed{y=5}

    Substitute y=5 into (2) to find z

    z=3-2(5)

    \boxed{z=-7}

    Substitute y=5 and z=-7 into (1)

    (1) 3x^2+4(5)+(-7)=7

    3x^2+20=14

    3x^2=-6

    x^2=-2

    \boxed{x=\pm i\sqrt{2}}
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  3. #3
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    Hello, JoanneMac!

    I must assume that complex roots are expected.


    \begin{array}{cccc}3x^2 + 4y + z &=& 7 & {\color{blue}[1]} \\ \qquad\quad<br />
2y + z &=& 3 & {\color{blue}[2]} \\ \text{-}5x^2 + 3y + 8z &=& \text{-}31 & {\color{blue}[3]} \end{array}

    From [2], we have: . z \:=\:3 - 2y

    \begin{array}{ccccccccc}\text{Substitute into {\color{blue}[1]}:} & 3x^2 + 4y + (3-2y) &=& 7 & \Rightarrow & 3x^2 + 2y &=& 4 & {\color{blue}[4]} \\<br />
\text{Substitute into {\color{blue}[3]}:} & \text{-}5x^2 + 3y + 8(3-2y) &=& \text{-}31 & \Rightarrow & \text{-}5x^2 - 13y &=& \text{-}55 & {\color{blue}[5]} \end{array}

    \begin{array}{cccccccc}\text{Multiply {\color{blue}[4]} by 13:} & 39x^2 + 26y &=& 52 \\ \text{Multiply {\color{blue}[5]} by 2:} & \text{-}10x^2 - 26y &=& \text{-}110 \end{array}

    Add: . 29x^2 \:=\:-58 \quad\Rightarrow\quad x^2 \:=\:-2\quad\Rightarrow\quad\boxed{ x \:=\:\pm i\sqrt{2}}


    Substitute into [4]: . 3(\pm i\sqrt{2})^2 + 2y \:=\:4 \quad\Rightarrow\quad -6 +2y \:=\:4 \quad\Rightarrow\quad 2y \:=\:10  \quad\Rightarrow\quad\boxed{ y \:=\:5}


    Substitute into [2]: . 2(5) + z \:=\:3 \quad\Rightarrow\quad\boxed{ z \:=\:-7}

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