How to separate an equation into partial fractions?












0












$begingroup$


I am looking at a math question that has simplified this:



enter image description here



into this:



enter image description here



Can somebody explain the process for how this simplification was made? i.e. how does the denominator get broken down to those two terms?






partial-fractions







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Numerator $1=y+1-y$
    $endgroup$
    – lab bhattacharjee
    Dec 27 '18 at 4:13






  • 1




    $begingroup$
    This can be done with partial fraction decomposition. en.wikipedia.org/wiki/Partial_fraction_decomposition
    $endgroup$
    – CyclotomicField
    Dec 27 '18 at 4:14
















0












$begingroup$


I am looking at a math question that has simplified this:



enter image description here



into this:



enter image description here



Can somebody explain the process for how this simplification was made? i.e. how does the denominator get broken down to those two terms?






partial-fractions







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Numerator $1=y+1-y$
    $endgroup$
    – lab bhattacharjee
    Dec 27 '18 at 4:13






  • 1




    $begingroup$
    This can be done with partial fraction decomposition. en.wikipedia.org/wiki/Partial_fraction_decomposition
    $endgroup$
    – CyclotomicField
    Dec 27 '18 at 4:14














0












0








0





$begingroup$


I am looking at a math question that has simplified this:



enter image description here



into this:



enter image description here



Can somebody explain the process for how this simplification was made? i.e. how does the denominator get broken down to those two terms?






partial-fractions







share|cite|improve this question









$endgroup$




I am looking at a math question that has simplified this:



enter image description here



into this:



enter image description here



Can somebody explain the process for how this simplification was made? i.e. how does the denominator get broken down to those two terms?






partial-fractions




partial-fractions






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 27 '18 at 4:11









StartecStartec

1236




1236








  • 1




    $begingroup$
    Numerator $1=y+1-y$
    $endgroup$
    – lab bhattacharjee
    Dec 27 '18 at 4:13






  • 1




    $begingroup$
    This can be done with partial fraction decomposition. en.wikipedia.org/wiki/Partial_fraction_decomposition
    $endgroup$
    – CyclotomicField
    Dec 27 '18 at 4:14














  • 1




    $begingroup$
    Numerator $1=y+1-y$
    $endgroup$
    – lab bhattacharjee
    Dec 27 '18 at 4:13






  • 1




    $begingroup$
    This can be done with partial fraction decomposition. en.wikipedia.org/wiki/Partial_fraction_decomposition
    $endgroup$
    – CyclotomicField
    Dec 27 '18 at 4:14








1




1




$begingroup$
Numerator $1=y+1-y$
$endgroup$
– lab bhattacharjee
Dec 27 '18 at 4:13




$begingroup$
Numerator $1=y+1-y$
$endgroup$
– lab bhattacharjee
Dec 27 '18 at 4:13




1




1




$begingroup$
This can be done with partial fraction decomposition. en.wikipedia.org/wiki/Partial_fraction_decomposition
$endgroup$
– CyclotomicField
Dec 27 '18 at 4:14




$begingroup$
This can be done with partial fraction decomposition. en.wikipedia.org/wiki/Partial_fraction_decomposition
$endgroup$
– CyclotomicField
Dec 27 '18 at 4:14










2 Answers
2






active

oldest

votes


















2












$begingroup$

Here's a quick and easy way for simple or basic fractions$$frac 1{y(1-y)}=frac {color{blue}{1-y}+y}{y(color{blue}{1-y})}=frac {1-y}{y(1-y)}+frac y{y(1-y)}=frac 1y+frac 1{1-y}color{brown}{=frac 1y-frac 1{y-1}}$$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    We write $$ frac {1}{y(1-y)}=frac {A}{y}+frac {B}{1-y}$$



    Upon taking common denominator , we get $$frac {A(1-y)+By}{Y(1-y)}=frac {1}{y(1-y)}$$



    Therefore we want to have,$$ A(1-y)+B(y)=1$$



    Let $y=1$ and we get $B=1$ and let $y=0$ to get $A=1$



    Thus $$ frac {1}{y(1-y)}=frac {1}{y}+frac {1}{1-y}=frac {1}{y}-frac {1} {y-1} $$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Can you explain the part "Let y=1 and we get B=1 and let y=0 to get A=1"
      $endgroup$
      – Startec
      Dec 27 '18 at 7:48










    • $begingroup$
      We like to have the identity for every value of $y$, $ A(1−y)+B(y)=1$. If you plug $y=1$ in the identity you get $A(1-1)+B(1)=1$ which is the same as $B=1$. Similarly with $y=0$
      $endgroup$
      – Mohammad Riazi-Kermani
      Dec 27 '18 at 9:50











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Here's a quick and easy way for simple or basic fractions$$frac 1{y(1-y)}=frac {color{blue}{1-y}+y}{y(color{blue}{1-y})}=frac {1-y}{y(1-y)}+frac y{y(1-y)}=frac 1y+frac 1{1-y}color{brown}{=frac 1y-frac 1{y-1}}$$






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Here's a quick and easy way for simple or basic fractions$$frac 1{y(1-y)}=frac {color{blue}{1-y}+y}{y(color{blue}{1-y})}=frac {1-y}{y(1-y)}+frac y{y(1-y)}=frac 1y+frac 1{1-y}color{brown}{=frac 1y-frac 1{y-1}}$$






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Here's a quick and easy way for simple or basic fractions$$frac 1{y(1-y)}=frac {color{blue}{1-y}+y}{y(color{blue}{1-y})}=frac {1-y}{y(1-y)}+frac y{y(1-y)}=frac 1y+frac 1{1-y}color{brown}{=frac 1y-frac 1{y-1}}$$






        share|cite|improve this answer









        $endgroup$



        Here's a quick and easy way for simple or basic fractions$$frac 1{y(1-y)}=frac {color{blue}{1-y}+y}{y(color{blue}{1-y})}=frac {1-y}{y(1-y)}+frac y{y(1-y)}=frac 1y+frac 1{1-y}color{brown}{=frac 1y-frac 1{y-1}}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 27 '18 at 4:15









        Frank W.Frank W.

        3,7031321




        3,7031321























            1












            $begingroup$

            We write $$ frac {1}{y(1-y)}=frac {A}{y}+frac {B}{1-y}$$



            Upon taking common denominator , we get $$frac {A(1-y)+By}{Y(1-y)}=frac {1}{y(1-y)}$$



            Therefore we want to have,$$ A(1-y)+B(y)=1$$



            Let $y=1$ and we get $B=1$ and let $y=0$ to get $A=1$



            Thus $$ frac {1}{y(1-y)}=frac {1}{y}+frac {1}{1-y}=frac {1}{y}-frac {1} {y-1} $$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Can you explain the part "Let y=1 and we get B=1 and let y=0 to get A=1"
              $endgroup$
              – Startec
              Dec 27 '18 at 7:48










            • $begingroup$
              We like to have the identity for every value of $y$, $ A(1−y)+B(y)=1$. If you plug $y=1$ in the identity you get $A(1-1)+B(1)=1$ which is the same as $B=1$. Similarly with $y=0$
              $endgroup$
              – Mohammad Riazi-Kermani
              Dec 27 '18 at 9:50
















            1












            $begingroup$

            We write $$ frac {1}{y(1-y)}=frac {A}{y}+frac {B}{1-y}$$



            Upon taking common denominator , we get $$frac {A(1-y)+By}{Y(1-y)}=frac {1}{y(1-y)}$$



            Therefore we want to have,$$ A(1-y)+B(y)=1$$



            Let $y=1$ and we get $B=1$ and let $y=0$ to get $A=1$



            Thus $$ frac {1}{y(1-y)}=frac {1}{y}+frac {1}{1-y}=frac {1}{y}-frac {1} {y-1} $$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Can you explain the part "Let y=1 and we get B=1 and let y=0 to get A=1"
              $endgroup$
              – Startec
              Dec 27 '18 at 7:48










            • $begingroup$
              We like to have the identity for every value of $y$, $ A(1−y)+B(y)=1$. If you plug $y=1$ in the identity you get $A(1-1)+B(1)=1$ which is the same as $B=1$. Similarly with $y=0$
              $endgroup$
              – Mohammad Riazi-Kermani
              Dec 27 '18 at 9:50














            1












            1








            1





            $begingroup$

            We write $$ frac {1}{y(1-y)}=frac {A}{y}+frac {B}{1-y}$$



            Upon taking common denominator , we get $$frac {A(1-y)+By}{Y(1-y)}=frac {1}{y(1-y)}$$



            Therefore we want to have,$$ A(1-y)+B(y)=1$$



            Let $y=1$ and we get $B=1$ and let $y=0$ to get $A=1$



            Thus $$ frac {1}{y(1-y)}=frac {1}{y}+frac {1}{1-y}=frac {1}{y}-frac {1} {y-1} $$






            share|cite|improve this answer









            $endgroup$



            We write $$ frac {1}{y(1-y)}=frac {A}{y}+frac {B}{1-y}$$



            Upon taking common denominator , we get $$frac {A(1-y)+By}{Y(1-y)}=frac {1}{y(1-y)}$$



            Therefore we want to have,$$ A(1-y)+B(y)=1$$



            Let $y=1$ and we get $B=1$ and let $y=0$ to get $A=1$



            Thus $$ frac {1}{y(1-y)}=frac {1}{y}+frac {1}{1-y}=frac {1}{y}-frac {1} {y-1} $$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 27 '18 at 4:27









            Mohammad Riazi-KermaniMohammad Riazi-Kermani

            41.6k42061




            41.6k42061












            • $begingroup$
              Can you explain the part "Let y=1 and we get B=1 and let y=0 to get A=1"
              $endgroup$
              – Startec
              Dec 27 '18 at 7:48










            • $begingroup$
              We like to have the identity for every value of $y$, $ A(1−y)+B(y)=1$. If you plug $y=1$ in the identity you get $A(1-1)+B(1)=1$ which is the same as $B=1$. Similarly with $y=0$
              $endgroup$
              – Mohammad Riazi-Kermani
              Dec 27 '18 at 9:50


















            • $begingroup$
              Can you explain the part "Let y=1 and we get B=1 and let y=0 to get A=1"
              $endgroup$
              – Startec
              Dec 27 '18 at 7:48










            • $begingroup$
              We like to have the identity for every value of $y$, $ A(1−y)+B(y)=1$. If you plug $y=1$ in the identity you get $A(1-1)+B(1)=1$ which is the same as $B=1$. Similarly with $y=0$
              $endgroup$
              – Mohammad Riazi-Kermani
              Dec 27 '18 at 9:50
















            $begingroup$
            Can you explain the part "Let y=1 and we get B=1 and let y=0 to get A=1"
            $endgroup$
            – Startec
            Dec 27 '18 at 7:48




            $begingroup$
            Can you explain the part "Let y=1 and we get B=1 and let y=0 to get A=1"
            $endgroup$
            – Startec
            Dec 27 '18 at 7:48












            $begingroup$
            We like to have the identity for every value of $y$, $ A(1−y)+B(y)=1$. If you plug $y=1$ in the identity you get $A(1-1)+B(1)=1$ which is the same as $B=1$. Similarly with $y=0$
            $endgroup$
            – Mohammad Riazi-Kermani
            Dec 27 '18 at 9:50




            $begingroup$
            We like to have the identity for every value of $y$, $ A(1−y)+B(y)=1$. If you plug $y=1$ in the identity you get $A(1-1)+B(1)=1$ which is the same as $B=1$. Similarly with $y=0$
            $endgroup$
            – Mohammad Riazi-Kermani
            Dec 27 '18 at 9:50


















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