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Math Help - [SOLVED] Help with solve equation

  1. #1
    Newbie onymousillusion's Avatar
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    Exclamation [SOLVED] Help with solve equation

    2x/x^2-4 = 4/x^2-4 - 3/x+2

    The back of the book said that it's no solution but I get an answer of x = -2/5
    I plug it back it make sure it doesn't make the denominator go to 0 and it doesn't so I'm not doing something right.... any help would be appreciated
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  2. #2
    Super Member craig's Avatar
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    Quote Originally Posted by onymousillusion View Post
    2x/x^2-4 = 4/x^2-4 - 3/x+2

    The back of the book said that it's no solution but I get an answer of x = -2/5
    I plug it back it make sure it doesn't make the denominator go to 0 and it doesn't so I'm not doing something right.... any help would be appreciated
    Could you write out your equation using the fraction latex tag, it's unclear what exactly is a fraction and what isn't.

    For example, using \frac{a}{b} to get \frac{a}{b}
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  3. #3
    Newbie onymousillusion's Avatar
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    Thanks for the tip didn't no you could do that...
    well here is all the step i did.
    <br />
\frac{2x}{x^2-4} = \frac{4}{x^2-4} - \frac{3}{x+2}<br />

    <br />
\frac{2x}{(x-2)(x+2)} = \frac{4}{(x-2)(x+2)} - \frac{3}{x+2}<br />

    <br />
\frac{2x(x-2)(x+2)}{(x-2)(x+2)} = \frac{4(x-2)(x+2)}{(x-2)(x+2)} = \frac{3(x+2)(x-2)}{x+2}<br />

    than I canceled and got

    <br />
2x = 4 - 3x - 6<br />

    and you can see how I got x = -2/5
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  4. #4
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by onymousillusion View Post
    2x/x^2-4 = 4/x^2-4 - 3/x+2

    The back of the book said that it's no solution but I get an answer of x = -2/5
    I plug it back it make sure it doesn't make the denominator go to 0 and it doesn't so I'm not doing something right.... any help would be appreciated
    I assume your question is

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

    or, \frac{2x}{x^{2}-4} - \frac{4}{x^{2}-4} = - \frac{3}{x+2}

    or, \frac{2x-4}{x^{2}-4} = - \frac{3}{x+2}

    or, \frac{2(x-2)}{(x+2)(x-2)} = - \frac{3}{x+2}

    or, \frac{2}{x+2} =  - \frac{3}{x+2}

    or,  2 =  - \frac{3(x+2)}{x+2}

    or,  2 =  -3

    which is why there is no solution
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  5. #5
    Super Member craig's Avatar
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    Quote Originally Posted by onymousillusion View Post
    Thanks for the tip didn't no you could do that...
    well here is all the step i did.
    <br />
\frac{2x}{x^2-4} = \frac{4}{x^2-4} - \frac{3}{x+2}<br />

    <br />
\frac{2x}{(x-2)(x+2)} = \frac{4}{(x-2)(x+2)} - \frac{3}{x+2}<br />

    <br />
\frac{2x(x-2)(x+2)}{(x-2)(x+2)} = \frac{4(x-2)(x+2)}{(x-2)(x+2)} = \frac{3(x+2)(x-2)}{x+2}<br />

    than I canceled and got

    <br />
2x = 4 - 3x - 6<br />

    and you can see how I got x = -2/5
    From what I can see your equation should be 2x = 4 - 3x -(-6), which will lead you to x=2
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  6. #6
    Newbie onymousillusion's Avatar
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    I see what I did wrong thanks again everyone....
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  7. #7
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by craig View Post
    From what I can see your equation should be 2x = 4 - 3x -(-6), which will lead you to x=2
    I think the answers x=-2/5 and x=2 are both incorrect!

    Does x=2 satisfy the given equation? you get 0 = -3/4

    there is no solution to this equation
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