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Math Help - use cramers rule to solve each linear system help please

  1. #1
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    Unhappy use cramers rule to solve each linear system help please

    use cramers rule to solve each linear system

    could somone show me how to do these 5 problems or if u could just show me a few it would be great i got a test tommorow and im clueless so fast help would be appreciated

    x-y=7
    4x+5y=46





    -10x+3y=18
    3x-4y=-24





    9x+2y=12
    5x-3y=19




    2x+y=24
    -x+6y=14






    4x-y=1
    7x+8y=-47
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  2. #2
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    Quote Originally Posted by shadower4 View Post
    use cramers rule to solve each linear system


    x-y=7
    4x+5y=46


    x\displaystyle  = \left|<br />
  \begin{array}{ c c }<br />
     7 & -1 \\<br />
     46 & 5<br />
  \end{array} \right|<br />
\div \left|<br />
  \begin{array}{ c c }<br />
     1 & -1 \\<br />
     4 & 5<br />
  \end{array} \right|


    y\displaystyle  = \left|<br />
  \begin{array}{ c c }<br />
     1 & 7 \\<br />
     4 & 46<br />
  \end{array} \right|<br />
\div \left|<br />
  \begin{array}{ c c }<br />
     1 & -1 \\<br />
     4 & 5<br />
  \end{array} \right|

    Can you see the pattern?

    If not look here

    Cramer's rule - Wikipedia, the free encyclopedia
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  3. #3
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    Lexington, MA (USA)
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    Hello, shadower4!

    I'll do the first one . . .


    \text{Use Cramer's Rule to solve each linear system.}

    . . \begin{array}{ccc}x-y &=& 7 \\ 4x+5y&=&46 \end{array}

    . . . . . . . . . . . . .
    x-coefficients . y-coefficients
    . . . . . . . . . . . . . . . . . . . \downarrow \quad\, \downarrow
    We have these numbers: . \begin{array}{|cc|c|}<br />
1 & \text{-}1 & 7 \\ 4 & 5 & 46 \end{array}
    . . . . . . . . . . . . . . . . . . . . . . . . . \uparrow
    . . . . . . . . . . . . . . . . . . . . . . .
    constants

    [1] Find \,D, the determinant of the coefficients:

    . . \begin{array}{|cc|} 1 & \text{-}1 \\ 4 & 5\end{array} \;=\;(1)(5) - (\text{-}1)(4) \:=\:5 + 4 \;=\;9


    [2] Find \,D_x, the determinant with the \,x-coefficients replaced by the constants.

    . . D_x \;=\;\begin{vmatrix}7 & \text{-}1 \\ 46 & 5 \end{vmatrix} \;=\;(7)(5) - (\text{-}1)(46) \;=\;35 + 46 \;=\;81

    . . \text{Divide by }D\!:\;\;x \;=\;\dfrac{81}{9} \;=\;9


    [3] Find \,D_y, the determinant with the \,y-coefficients replaced by the constants.

    . . D_x \;=\;\begin{vmatrix}1 & 7 \\ 4 & 46 \end{vmatrix} \;=\;(1)(46) - (7)(4) \;=\;46 - 28 \;=\;18

    . . \text{Divide by }D\!:\;\;y \;=\;\dfrac{18}{9} \;=\;2


    Therefore: . \begin{Bmatrix}x &=& 9 \\ y &=& 2\end{Bmatrix}

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