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Math Help - Algebraic fractions

  1. #1
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    Algebraic fractions

    \frac{x+3}{x}-\frac{2}{x-2}=0

    Will somebody please explain how I can get a minus sign between the first fraction and the second so that I can put the minus sign between the division lines of the first and second fractions please?

    Thanks

    David

    Thank you Quacky for that useful information

    Now for the next part.

    \(x-2)(x+3)\div(x)(x-2)} -\(2x)\div(x^2 -2)=0

    It does not seem to be formatting as previously, I need more help and experience as I must be doing something incorrect?

    I will have to come back to this later?
    Last edited by David Green; January 15th 2012 at 03:17 PM. Reason: still learning latex
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  2. #2
    Super Member Quacky's Avatar
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    Re: Algebraic fractions

    You have written [tex]\frac{x+3}{x}[/tex][tex]\frac{2}{x-2}[/tex]

    You shouldn't tag each individual fraction. Tag the whole equation, as below, and then you can simply slot in the minus sign:

    [TEX]\frac{x+3}{x}-\frac{2}{x-2}=0[/TEX] gives:

    \frac{x+3}{x}-\frac{2}{x-2}=0
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  3. #3
    Super Member Quacky's Avatar
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    Re: Algebraic fractions

    Quote Originally Posted by David Green View Post
    Now for the next part.

    \(x-2)(x+3)\div(x)(x-2)} -\(2x)\div(x^2 -2)=0

    It does not seem to be formatting as previously, I need more help and experience as I must be doing something incorrect?

    I will have to come back to this later?
    The code:

    [TEX]\frac{(x+2)(x+3)}{(x)(x-2)}-\frac{2x}{x^2-2}=0[/TEX] gives:

    \frac{(x+2)(x+3)}{(x)(x-2)}-\frac{2x}{x^2-2}=0
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  4. #4
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    Re: Algebraic fractions

    \frac{(x-2)(x+3)}{(x)(x-2)}-\frac{(2x)}{(x)(x-2)}=0

    \frac{(x^2+3x-2x-6)}{(x^2-2x)}-\frac{(2x)}{(x^2-2x)} =

    \(x^2+x-6)}-{(2x)}=x+6

    Somewhere here I am making a mistake in the maths but at present I can't see it, and although I am brand new to latex, it's not the latex that is incorrect it is my maths

    The answer does not seem correct to me?

    Any help greatly appreciated to point out my mistake

    David
    Last edited by David Green; January 16th 2012 at 10:05 AM. Reason: Made mistake in last line
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  5. #5
    Super Member Quacky's Avatar
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    Re: Algebraic fractions

    Quote Originally Posted by David Green View Post
    \frac{(x-2)(x+3)}{(x)(x-2)}-\frac{(2x)}{(x)(x-2)}=0

    \frac{(x^2+3x-2x-6)}{(x^2-2x)}-\frac{(2x)}{(x^2-2x)} =

    \(x^2+x-6)}-{(2x)}=x+6

    Somewhere here I am making a mistake in the maths but at present I can't see it, and although I am brand new to latex, it's not the latex that is incorrect it is my maths

    The answer does not seem correct to me?

    Any help greatly appreciated to point out my mistake

    David
    Your working out is good, your notation is a problem.

    \frac{(x^2+3x-2x-6)}{(x^2-2x)}-\frac{(2x)}{(x^2-2x)} =0

    \frac{(x^2+x-6)}{(x^2-2x)}-\frac{(2x)}{(x^2-2x)} =0

    \frac{(x^2+x-6-2x)}{(x^2-2x)}=0

    x^2-x-6=0

    This is a factorable quadratic that you can now solve using your favourite method.
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