Please help with simplification of expression. [closed]












-1












$begingroup$


I have the following but cannot replicate the result. Could someone please help me?



begin{equation}
frac{x-1}{(x-2)(x-3)} = frac{1}{x-3} + frac{1}{(x-2)(x-3)}
end{equation}






calculus algebra-precalculus







share|cite|improve this question









$endgroup$



closed as off-topic by Saad, Adrian Keister, mrtaurho, Lee David Chung Lin, John Wayland Bales Jan 3 at 17:11


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Adrian Keister, mrtaurho, Lee David Chung Lin, John Wayland Bales

If this question can be reworded to fit the rules in the help center, please edit the question.


















    -1












    $begingroup$


    I have the following but cannot replicate the result. Could someone please help me?



    begin{equation}
    frac{x-1}{(x-2)(x-3)} = frac{1}{x-3} + frac{1}{(x-2)(x-3)}
    end{equation}






    calculus algebra-precalculus







    share|cite|improve this question









    $endgroup$



    closed as off-topic by Saad, Adrian Keister, mrtaurho, Lee David Chung Lin, John Wayland Bales Jan 3 at 17:11


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Adrian Keister, mrtaurho, Lee David Chung Lin, John Wayland Bales

    If this question can be reworded to fit the rules in the help center, please edit the question.
















      -1












      -1








      -1





      $begingroup$


      I have the following but cannot replicate the result. Could someone please help me?



      begin{equation}
      frac{x-1}{(x-2)(x-3)} = frac{1}{x-3} + frac{1}{(x-2)(x-3)}
      end{equation}






      calculus algebra-precalculus







      share|cite|improve this question









      $endgroup$




      I have the following but cannot replicate the result. Could someone please help me?



      begin{equation}
      frac{x-1}{(x-2)(x-3)} = frac{1}{x-3} + frac{1}{(x-2)(x-3)}
      end{equation}






      calculus algebra-precalculus




      calculus algebra-precalculus






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 3 at 10:36









      probablymeprobablyme

      134




      134




      closed as off-topic by Saad, Adrian Keister, mrtaurho, Lee David Chung Lin, John Wayland Bales Jan 3 at 17:11


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Adrian Keister, mrtaurho, Lee David Chung Lin, John Wayland Bales

      If this question can be reworded to fit the rules in the help center, please edit the question.




      closed as off-topic by Saad, Adrian Keister, mrtaurho, Lee David Chung Lin, John Wayland Bales Jan 3 at 17:11


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Adrian Keister, mrtaurho, Lee David Chung Lin, John Wayland Bales

      If this question can be reworded to fit the rules in the help center, please edit the question.






















          3 Answers
          3






          active

          oldest

          votes


















          5












          $begingroup$

          $$frac{x-1}{(x-2)(x-3)}=frac{(x-2)+1}{(x-2)(x-3)}=frac{1}{(x-3)}+frac{1}{(x-2)(x-3)}$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you so much!
            $endgroup$
            – probablyme
            Jan 3 at 10:53






          • 1




            $begingroup$
            lab bhattacharjee's is correct, this requires $x-2 neq 0$
            $endgroup$
            – kelalaka
            Jan 3 at 11:22



















          1












          $begingroup$

          $$dfrac{x-1}{(x-2)(x-3)}-dfrac1{(x-2)(x-3)}=dfrac{x-1-1}{(x-2)(x-3)}=dfrac1{x-3}$$ if $x-2ne0$






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            Using the method of partial fractions, the expression $ frac{x-1}{(x-2)(x-3)} $ can be decomposed (broken down) into the following equation:



            $$mathrm{(1)} qquad frac{x-1}{(x-2)(x-3)} = frac{A}{x-2} + frac{B}{x-3} $$



            where $ A $ and $ B $ are the constants to be solved, and $ x-2 $ and $ x-3 $ the factors in the denominators of the terms with $ A $ and $ B $ in the numerators, respectively.



            Rewriting equation $ (1) $ by multiplying the $ x-3 $ factor to $ A $ and the $ x-2 $ factor to $ B $:



            $$mathrm{(2)} qquad frac{x-1}{(x-2)(x-3)} = frac{A(x-3)}{(x-2)(x-3)} + frac{B(x-2)}{(x-2)(x-3)} $$



            The common denominator on both the left and right hand sides of equation $ (2) $ is consistent, and, focusing on the numerators of every term in the expression:



            $$mathrm{(3)} qquad x-1 = A(x-3) + B(x-2) $$



            When $ x = 3 $, the term with the coefficient $ A $ is wiped out, leaving with $ 2 = B cdot 1 $, or $$ B = 2 $$



            If $ x = 2 $, the term with the coefficient $ B $ vanishes; thus $ 1 = A cdot -1 $, and solving for $ A $, $$ A = -1 $$



            So the complete decomposition of $ frac{x-1}{(x-2)(x-3)} $ is as follows:



            $$ mathrm{(4)} qquad frac{x-1}{(x-2)(x-3)} = frac{-1}{x-2} + frac{2}{x-3} $$



            Now one of the right side terms in equation $ (4) $, $ frac{2}{x-3} $, can be rewritten as $$ frac{2}{x-3} = frac{1}{x-3} + frac{1}{x-3} $$ and hence



            $$ mathrm{(5)} qquad frac{x-1}{(x-2)(x-3)} = frac{-1}{x-2} + frac{1}{x-3} + frac{1}{x-3} $$



            Grouping the first two right side terms together and multiplying each term by its respective factors $ x-3 $ and $ x-2 $,



            $$ mathrm{(6)} qquad frac{x-1}{(x-2)(x-3)} = frac{-1(x-3) + 1(x-2)}{(x-2)(x-3)} + frac{1}{x-3} $$



            Distributing one of the numerators on the right side, $ -1(x-3) + 1(x-2) $, and simplifying:



            $$ qquad frac{x-1}{(x-2)(x-3)} = frac{1}{(x-2)(x-3)} + frac{1}{x-3} $$ as expected.






            share|cite|improve this answer









            $endgroup$




















              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              5












              $begingroup$

              $$frac{x-1}{(x-2)(x-3)}=frac{(x-2)+1}{(x-2)(x-3)}=frac{1}{(x-3)}+frac{1}{(x-2)(x-3)}$$






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                Thank you so much!
                $endgroup$
                – probablyme
                Jan 3 at 10:53






              • 1




                $begingroup$
                lab bhattacharjee's is correct, this requires $x-2 neq 0$
                $endgroup$
                – kelalaka
                Jan 3 at 11:22
















              5












              $begingroup$

              $$frac{x-1}{(x-2)(x-3)}=frac{(x-2)+1}{(x-2)(x-3)}=frac{1}{(x-3)}+frac{1}{(x-2)(x-3)}$$






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                Thank you so much!
                $endgroup$
                – probablyme
                Jan 3 at 10:53






              • 1




                $begingroup$
                lab bhattacharjee's is correct, this requires $x-2 neq 0$
                $endgroup$
                – kelalaka
                Jan 3 at 11:22














              5












              5








              5





              $begingroup$

              $$frac{x-1}{(x-2)(x-3)}=frac{(x-2)+1}{(x-2)(x-3)}=frac{1}{(x-3)}+frac{1}{(x-2)(x-3)}$$






              share|cite|improve this answer











              $endgroup$



              $$frac{x-1}{(x-2)(x-3)}=frac{(x-2)+1}{(x-2)(x-3)}=frac{1}{(x-3)}+frac{1}{(x-2)(x-3)}$$







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Jan 3 at 10:40

























              answered Jan 3 at 10:38









              weilam06weilam06

              16011




              16011












              • $begingroup$
                Thank you so much!
                $endgroup$
                – probablyme
                Jan 3 at 10:53






              • 1




                $begingroup$
                lab bhattacharjee's is correct, this requires $x-2 neq 0$
                $endgroup$
                – kelalaka
                Jan 3 at 11:22


















              • $begingroup$
                Thank you so much!
                $endgroup$
                – probablyme
                Jan 3 at 10:53






              • 1




                $begingroup$
                lab bhattacharjee's is correct, this requires $x-2 neq 0$
                $endgroup$
                – kelalaka
                Jan 3 at 11:22
















              $begingroup$
              Thank you so much!
              $endgroup$
              – probablyme
              Jan 3 at 10:53




              $begingroup$
              Thank you so much!
              $endgroup$
              – probablyme
              Jan 3 at 10:53




              1




              1




              $begingroup$
              lab bhattacharjee's is correct, this requires $x-2 neq 0$
              $endgroup$
              – kelalaka
              Jan 3 at 11:22




              $begingroup$
              lab bhattacharjee's is correct, this requires $x-2 neq 0$
              $endgroup$
              – kelalaka
              Jan 3 at 11:22











              1












              $begingroup$

              $$dfrac{x-1}{(x-2)(x-3)}-dfrac1{(x-2)(x-3)}=dfrac{x-1-1}{(x-2)(x-3)}=dfrac1{x-3}$$ if $x-2ne0$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                $$dfrac{x-1}{(x-2)(x-3)}-dfrac1{(x-2)(x-3)}=dfrac{x-1-1}{(x-2)(x-3)}=dfrac1{x-3}$$ if $x-2ne0$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  $$dfrac{x-1}{(x-2)(x-3)}-dfrac1{(x-2)(x-3)}=dfrac{x-1-1}{(x-2)(x-3)}=dfrac1{x-3}$$ if $x-2ne0$






                  share|cite|improve this answer









                  $endgroup$



                  $$dfrac{x-1}{(x-2)(x-3)}-dfrac1{(x-2)(x-3)}=dfrac{x-1-1}{(x-2)(x-3)}=dfrac1{x-3}$$ if $x-2ne0$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 3 at 10:41









                  lab bhattacharjeelab bhattacharjee

                  224k15156274




                  224k15156274























                      0












                      $begingroup$

                      Using the method of partial fractions, the expression $ frac{x-1}{(x-2)(x-3)} $ can be decomposed (broken down) into the following equation:



                      $$mathrm{(1)} qquad frac{x-1}{(x-2)(x-3)} = frac{A}{x-2} + frac{B}{x-3} $$



                      where $ A $ and $ B $ are the constants to be solved, and $ x-2 $ and $ x-3 $ the factors in the denominators of the terms with $ A $ and $ B $ in the numerators, respectively.



                      Rewriting equation $ (1) $ by multiplying the $ x-3 $ factor to $ A $ and the $ x-2 $ factor to $ B $:



                      $$mathrm{(2)} qquad frac{x-1}{(x-2)(x-3)} = frac{A(x-3)}{(x-2)(x-3)} + frac{B(x-2)}{(x-2)(x-3)} $$



                      The common denominator on both the left and right hand sides of equation $ (2) $ is consistent, and, focusing on the numerators of every term in the expression:



                      $$mathrm{(3)} qquad x-1 = A(x-3) + B(x-2) $$



                      When $ x = 3 $, the term with the coefficient $ A $ is wiped out, leaving with $ 2 = B cdot 1 $, or $$ B = 2 $$



                      If $ x = 2 $, the term with the coefficient $ B $ vanishes; thus $ 1 = A cdot -1 $, and solving for $ A $, $$ A = -1 $$



                      So the complete decomposition of $ frac{x-1}{(x-2)(x-3)} $ is as follows:



                      $$ mathrm{(4)} qquad frac{x-1}{(x-2)(x-3)} = frac{-1}{x-2} + frac{2}{x-3} $$



                      Now one of the right side terms in equation $ (4) $, $ frac{2}{x-3} $, can be rewritten as $$ frac{2}{x-3} = frac{1}{x-3} + frac{1}{x-3} $$ and hence



                      $$ mathrm{(5)} qquad frac{x-1}{(x-2)(x-3)} = frac{-1}{x-2} + frac{1}{x-3} + frac{1}{x-3} $$



                      Grouping the first two right side terms together and multiplying each term by its respective factors $ x-3 $ and $ x-2 $,



                      $$ mathrm{(6)} qquad frac{x-1}{(x-2)(x-3)} = frac{-1(x-3) + 1(x-2)}{(x-2)(x-3)} + frac{1}{x-3} $$



                      Distributing one of the numerators on the right side, $ -1(x-3) + 1(x-2) $, and simplifying:



                      $$ qquad frac{x-1}{(x-2)(x-3)} = frac{1}{(x-2)(x-3)} + frac{1}{x-3} $$ as expected.






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        Using the method of partial fractions, the expression $ frac{x-1}{(x-2)(x-3)} $ can be decomposed (broken down) into the following equation:



                        $$mathrm{(1)} qquad frac{x-1}{(x-2)(x-3)} = frac{A}{x-2} + frac{B}{x-3} $$



                        where $ A $ and $ B $ are the constants to be solved, and $ x-2 $ and $ x-3 $ the factors in the denominators of the terms with $ A $ and $ B $ in the numerators, respectively.



                        Rewriting equation $ (1) $ by multiplying the $ x-3 $ factor to $ A $ and the $ x-2 $ factor to $ B $:



                        $$mathrm{(2)} qquad frac{x-1}{(x-2)(x-3)} = frac{A(x-3)}{(x-2)(x-3)} + frac{B(x-2)}{(x-2)(x-3)} $$



                        The common denominator on both the left and right hand sides of equation $ (2) $ is consistent, and, focusing on the numerators of every term in the expression:



                        $$mathrm{(3)} qquad x-1 = A(x-3) + B(x-2) $$



                        When $ x = 3 $, the term with the coefficient $ A $ is wiped out, leaving with $ 2 = B cdot 1 $, or $$ B = 2 $$



                        If $ x = 2 $, the term with the coefficient $ B $ vanishes; thus $ 1 = A cdot -1 $, and solving for $ A $, $$ A = -1 $$



                        So the complete decomposition of $ frac{x-1}{(x-2)(x-3)} $ is as follows:



                        $$ mathrm{(4)} qquad frac{x-1}{(x-2)(x-3)} = frac{-1}{x-2} + frac{2}{x-3} $$



                        Now one of the right side terms in equation $ (4) $, $ frac{2}{x-3} $, can be rewritten as $$ frac{2}{x-3} = frac{1}{x-3} + frac{1}{x-3} $$ and hence



                        $$ mathrm{(5)} qquad frac{x-1}{(x-2)(x-3)} = frac{-1}{x-2} + frac{1}{x-3} + frac{1}{x-3} $$



                        Grouping the first two right side terms together and multiplying each term by its respective factors $ x-3 $ and $ x-2 $,



                        $$ mathrm{(6)} qquad frac{x-1}{(x-2)(x-3)} = frac{-1(x-3) + 1(x-2)}{(x-2)(x-3)} + frac{1}{x-3} $$



                        Distributing one of the numerators on the right side, $ -1(x-3) + 1(x-2) $, and simplifying:



                        $$ qquad frac{x-1}{(x-2)(x-3)} = frac{1}{(x-2)(x-3)} + frac{1}{x-3} $$ as expected.






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          Using the method of partial fractions, the expression $ frac{x-1}{(x-2)(x-3)} $ can be decomposed (broken down) into the following equation:



                          $$mathrm{(1)} qquad frac{x-1}{(x-2)(x-3)} = frac{A}{x-2} + frac{B}{x-3} $$



                          where $ A $ and $ B $ are the constants to be solved, and $ x-2 $ and $ x-3 $ the factors in the denominators of the terms with $ A $ and $ B $ in the numerators, respectively.



                          Rewriting equation $ (1) $ by multiplying the $ x-3 $ factor to $ A $ and the $ x-2 $ factor to $ B $:



                          $$mathrm{(2)} qquad frac{x-1}{(x-2)(x-3)} = frac{A(x-3)}{(x-2)(x-3)} + frac{B(x-2)}{(x-2)(x-3)} $$



                          The common denominator on both the left and right hand sides of equation $ (2) $ is consistent, and, focusing on the numerators of every term in the expression:



                          $$mathrm{(3)} qquad x-1 = A(x-3) + B(x-2) $$



                          When $ x = 3 $, the term with the coefficient $ A $ is wiped out, leaving with $ 2 = B cdot 1 $, or $$ B = 2 $$



                          If $ x = 2 $, the term with the coefficient $ B $ vanishes; thus $ 1 = A cdot -1 $, and solving for $ A $, $$ A = -1 $$



                          So the complete decomposition of $ frac{x-1}{(x-2)(x-3)} $ is as follows:



                          $$ mathrm{(4)} qquad frac{x-1}{(x-2)(x-3)} = frac{-1}{x-2} + frac{2}{x-3} $$



                          Now one of the right side terms in equation $ (4) $, $ frac{2}{x-3} $, can be rewritten as $$ frac{2}{x-3} = frac{1}{x-3} + frac{1}{x-3} $$ and hence



                          $$ mathrm{(5)} qquad frac{x-1}{(x-2)(x-3)} = frac{-1}{x-2} + frac{1}{x-3} + frac{1}{x-3} $$



                          Grouping the first two right side terms together and multiplying each term by its respective factors $ x-3 $ and $ x-2 $,



                          $$ mathrm{(6)} qquad frac{x-1}{(x-2)(x-3)} = frac{-1(x-3) + 1(x-2)}{(x-2)(x-3)} + frac{1}{x-3} $$



                          Distributing one of the numerators on the right side, $ -1(x-3) + 1(x-2) $, and simplifying:



                          $$ qquad frac{x-1}{(x-2)(x-3)} = frac{1}{(x-2)(x-3)} + frac{1}{x-3} $$ as expected.






                          share|cite|improve this answer









                          $endgroup$



                          Using the method of partial fractions, the expression $ frac{x-1}{(x-2)(x-3)} $ can be decomposed (broken down) into the following equation:



                          $$mathrm{(1)} qquad frac{x-1}{(x-2)(x-3)} = frac{A}{x-2} + frac{B}{x-3} $$



                          where $ A $ and $ B $ are the constants to be solved, and $ x-2 $ and $ x-3 $ the factors in the denominators of the terms with $ A $ and $ B $ in the numerators, respectively.



                          Rewriting equation $ (1) $ by multiplying the $ x-3 $ factor to $ A $ and the $ x-2 $ factor to $ B $:



                          $$mathrm{(2)} qquad frac{x-1}{(x-2)(x-3)} = frac{A(x-3)}{(x-2)(x-3)} + frac{B(x-2)}{(x-2)(x-3)} $$



                          The common denominator on both the left and right hand sides of equation $ (2) $ is consistent, and, focusing on the numerators of every term in the expression:



                          $$mathrm{(3)} qquad x-1 = A(x-3) + B(x-2) $$



                          When $ x = 3 $, the term with the coefficient $ A $ is wiped out, leaving with $ 2 = B cdot 1 $, or $$ B = 2 $$



                          If $ x = 2 $, the term with the coefficient $ B $ vanishes; thus $ 1 = A cdot -1 $, and solving for $ A $, $$ A = -1 $$



                          So the complete decomposition of $ frac{x-1}{(x-2)(x-3)} $ is as follows:



                          $$ mathrm{(4)} qquad frac{x-1}{(x-2)(x-3)} = frac{-1}{x-2} + frac{2}{x-3} $$



                          Now one of the right side terms in equation $ (4) $, $ frac{2}{x-3} $, can be rewritten as $$ frac{2}{x-3} = frac{1}{x-3} + frac{1}{x-3} $$ and hence



                          $$ mathrm{(5)} qquad frac{x-1}{(x-2)(x-3)} = frac{-1}{x-2} + frac{1}{x-3} + frac{1}{x-3} $$



                          Grouping the first two right side terms together and multiplying each term by its respective factors $ x-3 $ and $ x-2 $,



                          $$ mathrm{(6)} qquad frac{x-1}{(x-2)(x-3)} = frac{-1(x-3) + 1(x-2)}{(x-2)(x-3)} + frac{1}{x-3} $$



                          Distributing one of the numerators on the right side, $ -1(x-3) + 1(x-2) $, and simplifying:



                          $$ qquad frac{x-1}{(x-2)(x-3)} = frac{1}{(x-2)(x-3)} + frac{1}{x-3} $$ as expected.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 3 at 16:17









                          Marvin CohenMarvin Cohen

                          43115




                          43115















                              Popular posts from this blog

                              MongoDB - Not Authorized To Execute Command

                              Image
                              .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I have successfully enabled authorization on MongoDB and I have created an account on the admin database and then I created an account for my database called test. The following connection string to connect to my test database works successfully: mongo --host 192.168.17.52 --port 27017 -u user1 -p password --authenticationDatabase test Only problem I have now is, I cannot execute commands such as: show dbs. I get the following error when I try to do so: "errmsg" : "not authorized on admin to execute command { listDatabases: 1.0, lsid: { id: UUID("a1d5bc0d-bc58-485e-b232-270758a89455") }, $db: "admin...
                              Read more

                              How to fix TextFormField cause rebuild widget in Flutter

                              Image
                              .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I've got the same issue at #9280. I tried all solutions from that topic with/without GlobalKey for Scaffold and Form but the problem wasn't fixed. I tried to build a authentication form with login and signup. But the keyboard doesn't show when I tap the TextFormField at the first time. Then I try to input some letter from physical keyboard, the widget is rebuilt. Like this: login screen Nothing to be log when I run flutter run --verbose, so I logged manually. Every tapping on field, the widget called dispose -> init -> build. [√] Flutter (Channel beta, v1.0.0, on Microsoft Windows [Version 10.0.17763.195], ...
                              Read more

                              in spring boot 2.1 many test slices are not allowed anymore due to multiple @BootstrapWith

                              Image
                              1 I tried to upgrade a yummy sandwich made of two test slices (@JsonTest and @JdbcTest in my case, crunchy test code in between) adding spring boot 2.1 flavour to it. But it seems it was not much of a success. I cannot annotate my tests with many @...Test since they are now each bringing their own XxxTestContextBootstrapper. It used to work when they all used same SpringBootTestContextBootstrapper. @RunWith(SpringRunner.class) @JdbcTest @JsonTest public class Test { @Test public void test() { System.out.printn("Hello, World !"); } } The error I get from BootstrapUtils is illegalStateException : Configuration error: found multiple declarations of @BootstrapWith for test class I understand I might be doing something wrong here but is there an easy way I could load both Json and Jdbc contex...
                              Read more