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Math Help - Linear Equation by Elimination

  1. #1
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    Linear Equation by Elimination

    Solve by elimination.
    x+2/6 - 3(y+2)/2 = 1

    x-2/2 + y-1/3 = 0


    For the second one, this is what i did:

    multiply by -3 (x-2/2 + y-1/3 = 0)

    y = 3x - 4

    For the first one:

    multiply by 2 (x+2/6 - 3(y+2)/2 = 1)

    1/3x + 2/3 -3y -6 = 2

    Am i doing this right so far? i'm really confused at this one please help
    Last edited by Vince604; December 2nd 2012 at 05:31 PM.
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  2. #2
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    Re: Linear Equation by Elimination

    Do you mean:
     x+2/6 - 3(y+2)/2 = 1

     x-2/2 + y-1/3 = 0
    (this is how you wrote it)
    or

     \frac{x+2}{6} - \frac{3(y+2)}{2} = 1
     \frac{x-2}{2} + \frac{y-1}{3} = 0
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  3. #3
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    Re: Linear Equation by Elimination

    Some brackets used where they're needed would be nice...
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    Re: Linear Equation by Elimination

    Quote Originally Posted by Scopur View Post


     \frac{x+2}{6} - \frac{3(y+2)}{2} = 1
     \frac{x-2}{2} + \frac{y-1}{3} = 0
    yea thanks, this. how would you simplify it, so that its easier to find x,y?
    Last edited by Vince604; December 2nd 2012 at 07:17 PM.
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    Re: Linear Equation by Elimination

    Anyone?
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  6. #6
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    Re: Linear Equation by Elimination

    Take the 2nd equation and isolate for  y .

     \frac{y-1}{3} = - \frac{x-2}{2}

     y-1 = - \frac{3}{2}(x-2)

     y-1 = -\frac{3}{2}x + 3

     y = -\frac{3}{2}x + 4

    Now take this and plug it into your first equation for y.
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