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Math Help - Some quick help on solving rational equations

  1. #1
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    Some quick help on solving rational equations

    How can I solve for X with a question with a form like this :

    (2/x+1)+5 = 1/x

    and also

    x - 5/x = 4

    I seem to struggle when they single something out with a denominator of 1 for example the +5 in (2/x+1)+5 and x in x - 5/x

    First time posting btw
    Hopefully, I'll receive some replies.
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  2. #2
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    Re: Some quick help on solving rational equations

    Quote Originally Posted by Dante View Post
    How can I solve for X with a question with a form like this :

    (2/x+1)+5 = 1/x

    and also

    x - 5/x = 4

    I seem to struggle when they single something out with a denominator of 1 for example the +5 in (2/x+1)+5 and x in x - 5/x

    First time posting btw
    Hopefully, I'll receive some replies.

    \frac{2}{x+1} + 5 = \frac{1}{x}

    common denominator ... common denominator ... common denominator

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

    the numerators form the equation ...

    2x + 5x(x+1) = x+1

    solve the resulting quadratic equation keeping in mind that x cannot equal 0 or -1
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  3. #3
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    Re: Some quick help on solving rational equations

    Quote Originally Posted by skeeter View Post
    \frac{2}{x+1} + 5 = \frac{1}{x}

    common denominator ... common denominator ... common denominator

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

    the numerators form the equation ...

    2x + 5x(x+1) = x+1

    solve the resulting quadratic equation keeping in mind that x cannot equal 0 or -1
    Oh thank you! Your reply made it a lot clearer, just one question though ... How did the right side turn into x+1 shouldn't it cancel out and turn into x ?

    Edit : Oh never mind, I see what happened now. Thanks !
    Last edited by Dante; October 16th 2011 at 04:42 PM.
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