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Math Help - 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; November 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.

    \frac{x - 2}{3}  +  \frac{y - 1}{4}  =  <br />
\frac{13}{12} (eq.1)

    \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.

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

    \frac{2 - x}{2} + \frac{3 + y}{3} = \frac{11}{6} (eq.2)
    I'll do \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: x - 2 + \frac{3(y - 1)}{4} = \frac{3(13)}{12}

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

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

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

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

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

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

    Solve: 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: \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 12 to get:

    <br />
4x-8+3y-3=13<br />

    rearrange to:

    <br />
4x + 3y =24<br />

    Multiply the second equation through by 6 to get:

    <br />
6-3x+6+2y=11<br />

    rearrange:

    <br />
-3x+2y=-1<br />

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

    ... <br />
4x + 3y =24<br />
    <br />
-3x+2y=-1<br />


    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:
    \begin{array}{l}<br />
 4x + 3y = 24 \\ <br />
  - 3x + 2y =  - 1 \\ <br />
 \end{array}<br />

    Solve first equation for x:
    \begin{array}{l}<br />
 4x + 3y = 24 \\ <br />
 4x = 24 - 3y \\ <br />
 x = \frac{{24 - 3y}}{4} \\ <br />
 \end{array}<br />

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

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


    So solutions are x=3,y=4
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