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Math Help - Simplify fractions

  1. #1
    Newbie TriForce's Avatar
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    Simplify fractions

    This problem is driving me nuts... it should be rather easy but...

    \dfrac{\dfrac{1}{a}-\dfrac{1}{a+b}}{\dfrac{1}{b}-\dfrac{1}{a+b}} =

    \dfrac{\dfrac{\dfrac{b}{a}}{a+b}}{\dfrac{\dfrac{a}  {b}}{a+b}}} =

    \dfrac{\dfrac{b}{a}}{a+b}*\dfrac{a+b}{\dfrac{a}{b}  }=

    \dfrac{\dfrac{b}{a}*(a+b)}{(a+b)*\dfrac{a}{b}}=

    Think the above is correct, but not the following:

    \dfrac{\dfrac{ab+{b}^{2}}{a}}{\dfrac{ab+{a}^{2}}{a  }}=

    \frac{ab+{b}^{2}}{a}*\frac{a}{ab+{a}^{2}}=

    \frac{{a}^{2}b+a{b}^{2}}{{a}^{2}b+{a}^{3}}=

    \dfrac{ab+{b}^{2}}{ab+{a}^{2}} which is not correct the answer should be \dfrac{{b}^{2}}{{a}^{2}}
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  2. #2
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    earboth's Avatar
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    Quote Originally Posted by TriForce View Post
    This problem is driving me nuts... it should be rather easy but...

    \dfrac{\dfrac{1}{a}-\dfrac{1}{a+b}}{\dfrac{1}{b}-\dfrac{1}{a+b}} =

    \dfrac{\dfrac{\dfrac{b}{a}}{a+b}}{\dfrac{\dfrac{a}  {b}}{a+b}}} =

    \dfrac{\dfrac{b}{a}}{a+b}*\dfrac{a+b}{\dfrac{a}{b}  }=

    \dfrac{\dfrac{b}{a}*(a+b)}{(a+b)*\dfrac{a}{b}}=

    Think the above is correct, but not the following:

    \dfrac{\dfrac{ab+{b}^{2}}{a}}{\dfrac{ab+{a}^{2}}{a  }}=

    \frac{ab+{b}^{2}}{a}*\frac{a}{ab+{a}^{2}}=

    \frac{{a}^{2}b+a{b}^{2}}{{a}^{2}b+{a}^{3}}=

    \dfrac{ab+{b}^{2}}{ab+{a}^{2}} which is not correct the answer should be \dfrac{{b}^{2}}{{a}^{2}}
    I don't understand what you've done. Here is the method I would have used:

    \dfrac{\dfrac{1}{a}-\dfrac{1}{a+b}}{\dfrac{1}{b}-\dfrac{1}{a+b}} = \dfrac{\frac{a+b-a}{a(a+b)}}{\frac{a+b-b}{b(a+b)}}=\dfrac{b}{a(a+b)} \cdot \dfrac{b(a+b)}{a}

    Cancel the term in parantheses and you're done.
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  3. #3
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    Hello, TriForce!

    \dfrac{\dfrac{1}{a}-\dfrac{1}{a+b}}{\dfrac{1}{b}-\dfrac{1}{a+b}} =

    Multiply numerator and denominator by the LCD, . ab(a+b)



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  4. #4
    Newbie TriForce's Avatar
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    Quote Originally Posted by earboth View Post
    I don't understand what you've done. Here is the method I would have used:

    \dfrac{\dfrac{1}{a}-\dfrac{1}{a+b}}{\dfrac{1}{b}-\dfrac{1}{a+b}} = \dfrac{\frac{a+b-a}{a(a+b)}}{\frac{a+b-b}{b(a+b)}}=\dfrac{b}{a(a+b)} \cdot \dfrac{b(a+b)}{a}

    Cancel the term in parantheses and you're done.
    Ah yes, that works too

    Btw, in my solution should have done:

    \dfrac{\dfrac{1}{a}-\dfrac{1}{a+b}}{\dfrac{1}{b}-\dfrac{1}{a+b}} =

    \dfrac{\dfrac{\dfrac{b}{a}}{a+b}}{\dfrac{\dfrac{a}  {b}}{a+b}}} =

    \dfrac{\dfrac{b}{a}}{a+b}*\dfrac{a+b}{\dfrac{a}{b}  }=

    This part different

    \dfrac{\dfrac{b}{a}}{1}*\dfrac{1}{\dfrac{a}{b}}=

    \dfrac{\dfrac{b}{a}}{\dfrac{a}{b}}=

    \dfrac{b}{a}*{\dfrac{b}{a}=

    \dfrac{{b}^{2}}{{a}^{2}}
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