Simple algebra manipulation?












3












$begingroup$


I am currently reading through this guide
http://www.math.ucsd.edu/~ebender/CombText/ch-10.pdf
on generating functions and I can't figure out how on page three he went from this formula to the next
$$frac{1}{(1-y) - xy} = frac{(frac{1}{1-y})}{1 - frac{xy}{1-y}}$$
It's got me stumped.






combinatorics algebra-precalculus







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Divide both the numerator and denominator by $1 - y$.
    $endgroup$
    – N. F. Taussig
    Jan 2 at 19:20






  • 1




    $begingroup$
    No idea why that didn't just click. Thanks.
    $endgroup$
    – fractallyconfused
    Jan 3 at 2:16
















3












$begingroup$


I am currently reading through this guide
http://www.math.ucsd.edu/~ebender/CombText/ch-10.pdf
on generating functions and I can't figure out how on page three he went from this formula to the next
$$frac{1}{(1-y) - xy} = frac{(frac{1}{1-y})}{1 - frac{xy}{1-y}}$$
It's got me stumped.






combinatorics algebra-precalculus







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Divide both the numerator and denominator by $1 - y$.
    $endgroup$
    – N. F. Taussig
    Jan 2 at 19:20






  • 1




    $begingroup$
    No idea why that didn't just click. Thanks.
    $endgroup$
    – fractallyconfused
    Jan 3 at 2:16














3












3








3


0



$begingroup$


I am currently reading through this guide
http://www.math.ucsd.edu/~ebender/CombText/ch-10.pdf
on generating functions and I can't figure out how on page three he went from this formula to the next
$$frac{1}{(1-y) - xy} = frac{(frac{1}{1-y})}{1 - frac{xy}{1-y}}$$
It's got me stumped.






combinatorics algebra-precalculus







share|cite|improve this question









$endgroup$




I am currently reading through this guide
http://www.math.ucsd.edu/~ebender/CombText/ch-10.pdf
on generating functions and I can't figure out how on page three he went from this formula to the next
$$frac{1}{(1-y) - xy} = frac{(frac{1}{1-y})}{1 - frac{xy}{1-y}}$$
It's got me stumped.






combinatorics algebra-precalculus




combinatorics algebra-precalculus






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 2 at 19:18









fractallyconfusedfractallyconfused

162




162








  • 2




    $begingroup$
    Divide both the numerator and denominator by $1 - y$.
    $endgroup$
    – N. F. Taussig
    Jan 2 at 19:20






  • 1




    $begingroup$
    No idea why that didn't just click. Thanks.
    $endgroup$
    – fractallyconfused
    Jan 3 at 2:16














  • 2




    $begingroup$
    Divide both the numerator and denominator by $1 - y$.
    $endgroup$
    – N. F. Taussig
    Jan 2 at 19:20






  • 1




    $begingroup$
    No idea why that didn't just click. Thanks.
    $endgroup$
    – fractallyconfused
    Jan 3 at 2:16








2




2




$begingroup$
Divide both the numerator and denominator by $1 - y$.
$endgroup$
– N. F. Taussig
Jan 2 at 19:20




$begingroup$
Divide both the numerator and denominator by $1 - y$.
$endgroup$
– N. F. Taussig
Jan 2 at 19:20




1




1




$begingroup$
No idea why that didn't just click. Thanks.
$endgroup$
– fractallyconfused
Jan 3 at 2:16




$begingroup$
No idea why that didn't just click. Thanks.
$endgroup$
– fractallyconfused
Jan 3 at 2:16










2 Answers
2






active

oldest

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2












$begingroup$


We can factor out $1-y$ and obtain
begin{align*}
frac{1}{(1-y)-xy}&=frac{1}{(1-y)left(1-frac{xy}{1-y}right)}\
&=frac{1}{1-y}cdotfrac{1}{1-frac{xy}{1-y}}
end{align*}







share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Multiplying numerator and denominator by $$1-yneq 0$$ we get
    $$frac{frac{1-y}{1-y}}{(1-frac{xy}{1-y})(1-y)}=frac{1}{1-y-xy}$$






    share|cite|improve this answer









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      2 Answers
      2






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      2 Answers
      2






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      2












      $begingroup$


      We can factor out $1-y$ and obtain
      begin{align*}
      frac{1}{(1-y)-xy}&=frac{1}{(1-y)left(1-frac{xy}{1-y}right)}\
      &=frac{1}{1-y}cdotfrac{1}{1-frac{xy}{1-y}}
      end{align*}







      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$


        We can factor out $1-y$ and obtain
        begin{align*}
        frac{1}{(1-y)-xy}&=frac{1}{(1-y)left(1-frac{xy}{1-y}right)}\
        &=frac{1}{1-y}cdotfrac{1}{1-frac{xy}{1-y}}
        end{align*}







        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$


          We can factor out $1-y$ and obtain
          begin{align*}
          frac{1}{(1-y)-xy}&=frac{1}{(1-y)left(1-frac{xy}{1-y}right)}\
          &=frac{1}{1-y}cdotfrac{1}{1-frac{xy}{1-y}}
          end{align*}







          share|cite|improve this answer









          $endgroup$




          We can factor out $1-y$ and obtain
          begin{align*}
          frac{1}{(1-y)-xy}&=frac{1}{(1-y)left(1-frac{xy}{1-y}right)}\
          &=frac{1}{1-y}cdotfrac{1}{1-frac{xy}{1-y}}
          end{align*}








          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 2 at 20:08









          Markus ScheuerMarkus Scheuer

          62.3k459149




          62.3k459149























              1












              $begingroup$

              Multiplying numerator and denominator by $$1-yneq 0$$ we get
              $$frac{frac{1-y}{1-y}}{(1-frac{xy}{1-y})(1-y)}=frac{1}{1-y-xy}$$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Multiplying numerator and denominator by $$1-yneq 0$$ we get
                $$frac{frac{1-y}{1-y}}{(1-frac{xy}{1-y})(1-y)}=frac{1}{1-y-xy}$$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Multiplying numerator and denominator by $$1-yneq 0$$ we get
                  $$frac{frac{1-y}{1-y}}{(1-frac{xy}{1-y})(1-y)}=frac{1}{1-y-xy}$$






                  share|cite|improve this answer









                  $endgroup$



                  Multiplying numerator and denominator by $$1-yneq 0$$ we get
                  $$frac{frac{1-y}{1-y}}{(1-frac{xy}{1-y})(1-y)}=frac{1}{1-y-xy}$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 2 at 19:22









                  Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                  77k42866




                  77k42866






























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