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Thread: Help....I need this for monday

  1. #1
    NettyJay
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    Help....I need this for monday

    Hi, I am really stuck with this equations. I am so lost and dont know what to do. Please help.

    x - 2 + y - 1 = 13 (eq.1)
    3 4 12

    2 - x + 3 + y = 11 (eq.2)
    2 3 6
    Last edited by NettyJay; Nov 17th 2006 at 03:48 AM. Reason: it didnt come out the way i typed it
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  2. #2
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by NettyJay View Post
    Hi, I am really stuck with this equations. I am so lost and dont know what to do. Please help.

    x - 2 + y - 1 = 13 (eq.1)
    3 4 12

    2 - x + 3 + y = 11 (eq.2)
    2 3 6
    Did you mean:
    Quote Originally Posted by NettyJay View Post
    Hi, I am really stuck with this equations. I am so lost and dont know what to do. Please help.

    $\displaystyle \frac{x - 2}{3} + \frac{y - 1}{4} =
    \frac{13}{12}$ (eq.1)

    $\displaystyle \frac{2 - x}{2} + \frac{3 + y}{3} = \frac{11}{6}$ (eq.2)
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  3. #3
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by NettyJay View Post
    Hi, I am really stuck with this equations. I am so lost and dont know what to do. Please help.

    Hi, I am really stuck with this equations. I am so lost and dont know what to do. Please help.

    $\displaystyle \frac{x - 2}{3} + \frac{y - 1}{4} = \frac{13}{12} $ (eq.1)

    $\displaystyle \frac{2 - x}{2} + \frac{3 + y}{3} = \frac{11}{6} $(eq.2)
    I'll do $\displaystyle \frac{x - 2}{3} + \frac{y - 1}{4} = \frac{13}{12} $ you try the other one.

    First thing to do is get rid of denominators, so multiply everything by 3 to get: $\displaystyle x - 2 + \frac{3(y - 1)}{4} = \frac{3(13)}{12} $

    Now multiply by 4: $\displaystyle 4x - 4(2) + 3(y - 1) = \frac{4(39)}{12} $

    Simplify: $\displaystyle 4x -8 + 3y - 3 = \frac{156}{12} $

    Thus: $\displaystyle 4x+ 3y - 11 = \frac{156}{12} $

    Add 11 to both sides: $\displaystyle 4x+ 3y = \frac{156}{12}+11 $

    Subtract 3y from both sides: $\displaystyle 4x = \frac{156}{12}+\frac{12(11)}{12}-3y$

    Then divide by 4: $\displaystyle x = \frac{156}{4(12)}+\frac{12(11)}{4(12)}-\frac{3y}{4}$

    Solve: $\displaystyle x = \frac{156}{48}+\frac{132}{48}-\frac{12(3y)}{12(4)}=\frac{156+138}{48}-\frac{36y}{48}=\frac{294-36y}{48}$

    That can be simplified to: $\displaystyle \boxed{x=\frac{49-6y}{8}}$
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  4. #4
    MHF Contributor Quick's Avatar
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    Was this supposed to be a system of equations, meaning was I sposed to find the value of x and y to fit both of them?
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by NettyJay View Post
    Hi, I am really stuck with this equations. I am so lost and dont know what to do. Please help.

    x - 2 + y - 1 = 13 (eq.1)
    ...3....... 4..... 12

    2 - x + 3 + y = 11 (eq.2)
    ..2........ 3....... 6
    Multiply the first equation through by $\displaystyle 12$ to get:

    $\displaystyle
    4x-8+3y-3=13
    $

    rearrange to:

    $\displaystyle
    4x + 3y =24
    $

    Multiply the second equation through by $\displaystyle 6$ to get:

    $\displaystyle
    6-3x+6+2y=11
    $

    rearrange:

    $\displaystyle
    -3x+2y=-1
    $

    So you have a bog standard pair of simultaneous linear equations:

    ...$\displaystyle
    4x + 3y =24
    $
    $\displaystyle
    -3x+2y=-1
    $


    RonL
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  6. #6
    Senior Member OReilly's Avatar
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    Just to extend CaptainBlack's post.

    You've got system of equations to solve:
    $\displaystyle \begin{array}{l}
    4x + 3y = 24 \\
    - 3x + 2y = - 1 \\
    \end{array}
    $

    Solve first equation for $\displaystyle x$:
    $\displaystyle \begin{array}{l}
    4x + 3y = 24 \\
    4x = 24 - 3y \\
    x = \frac{{24 - 3y}}{4} \\
    \end{array}
    $

    Substitute $\displaystyle x$ into second equation:
    $\displaystyle \begin{array}{l}
    - 3 \cdot (\frac{{24 - 3y}}{4}{\rm{)}} + 2y{\rm{ }} = - 1 \\
    - \frac{{72 - 9y}}{4} + 2y = - 1 \\
    \frac{{ - 72 + 9y + 8y}}{4} = - 1 \\
    \frac{{17y - 72}}{4} = - 1 \\
    17y - 72 = - 4 \\
    17y = 68 \\
    y = 4 \\
    \end{array}
    $

    Now we can find $\displaystyle x$ from first equation by substituting $\displaystyle y$:
    $\displaystyle \begin{array}{l}
    4x + 3 \cdot 4 = 24 \\
    4x + 12 = 24 \\
    4x = 12 \\
    x = 3 \\
    \end{array}
    $


    So solutions are $\displaystyle x=3,y=4$
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