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Math Help - Equation

  1. #1
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    Equation

    Can someone help me with this one?

    <br />
\frac {5}{x+2} - \frac{1}{x-4} -\frac{5}{2x+4} = 0<br />
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  2. #2
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    \frac {5}{x+2} - \frac{1}{x-4} -\frac{5}{2x+4} = 0

    \frac {5(x-4)-1(x+2)}{(x+2)(x-4)} - \frac{5}{2x+4} = 0

    \frac {5x-20-x-2}{(x+2)(x-4)}-\frac{5}{2x+4} = 0

    \frac {4x-22}{(x+2)(x-4)}-\frac{5}{2x+4} = 0

    \frac {(4x-22)(2x+4)-5(x+2)(x-4)}{(x+2)(x-4)(2x+4)}= 0 (cross multiply)

    (4x-22)(2x+4)-5(x+2)(x-4) = 0
    8x^2+16x-44x-88-5(x^2-4x+2x-8) = 0
    8x^2+16x-44x-88-5x^2+10x+40 = 0
    3x^2-18x-48 = 0
    (x-8)(3x+6) = 0
    x=8 or x = -2

    Hope it helps.

    Quote Originally Posted by hkerbest View Post
    Can someone help me with this one?

    <br />
\frac {5}{x+2} - \frac{1}{x-4} -\frac{5}{2x+4} = 0<br />
    Last edited by tester85; January 26th 2009 at 06:03 AM.
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  3. #3
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    Quote Originally Posted by hkerbest View Post
    Can someone help me with this one?

    <br />
\frac {5}{x+2} - \frac{1}{x-4} -\frac{5}{2x+4} = 0<br />
    first get the lcd that is 2(x+2)(x-2)

     \frac{5*2(x-2)-1(2)-5(x-2)}{2(x+2)(x-2)}

     \frac{10x-20- 2 -5x+10}{2(x+2)(x-2)}

    \frac{5x-8}{2(x-4)}

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  4. #4
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    Quote Originally Posted by princess_21 View Post
    first get the lcd that is 2(x+2)(x-2)

     \frac{5*2(x-2)-1(2)-5(x-2)}{2(x+2)(x-2)}

     \frac{10x-20- 2 -5x+10}{2(x+2)(x-2)}

    \frac{5x-8}{2(x-4)}

    Sorry i don't think this is right. As the answer you got is x = 8/5, when you resubstitute it back to the solution. You will not get back zero. I will format my previous solution using latex shortly.
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  5. #5
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    Talking

    right, oops my mistake..
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