Simplify this complex fraction that I constructed (challenge question)

lookagain

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Spoiler update solution

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The least common denominator is 6xy.



12 + 1x  1y 1xy13 1x 1y + 1xy  =\displaystyle \dfrac{\dfrac{1}{2} \ + \ \dfrac{1}{x} \ - \ \dfrac{1}{y} \ - \dfrac{1}{xy}}{\dfrac{1}{3} \ -\dfrac{1}{x} \ - \dfrac{1}{y} \ + \ \dfrac{1}{xy}} \ \ =


(6xy1)12 + (6xy1)1x  (6xy1)1y (6xy1)1xy(6xy1)13 (6xy1)1x (6xy1)1y + (6xy1)1xy  =\displaystyle \dfrac{\bigg(\dfrac{6xy}{1}\bigg)\dfrac{1}{2} \ + \ \bigg(\dfrac{6xy}{1}\bigg)\dfrac{1}{x} \ - \ \bigg(\dfrac{6xy}{1}\bigg)\dfrac{1}{y} \ - \bigg(\dfrac{6xy}{1}\bigg)\dfrac{1}{xy}}{ \bigg( \dfrac{6xy}{1} \bigg) \dfrac{1}{3} \ - \bigg(\dfrac{6xy}{1}\bigg)\dfrac{1}{x} \ - \bigg(\dfrac{6xy}{1}\bigg)\dfrac{1}{y} \ + \ \bigg(\dfrac{6xy}{1}\bigg)\dfrac{1}{xy}} \ \ =


  3xy + 6y  6x 62xy  6y  6x + 6  \displaystyle \boxed{ \ \ \dfrac{3xy \ + \ 6y \ - \ 6x \ - 6}{2xy \ - \ 6y \ - \ 6x \ + \ 6} \ \ }
 
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