Problem in solving inequality.












0












$begingroup$


In my book I found this following inequality :
$frac{2x}{x-1}>x$



In my book, they solved this in a very hard way which I don't understand properly. But I tried to do this in another way. I divided each side by $x$ and then solved. But I didn't get the full answer, only half part of it. So why can't I divided both side by $x$ and then solve? Is there any other not so difficult way to solve this?






inequality







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    In my book I found this following inequality :
    $frac{2x}{x-1}>x$



    In my book, they solved this in a very hard way which I don't understand properly. But I tried to do this in another way. I divided each side by $x$ and then solved. But I didn't get the full answer, only half part of it. So why can't I divided both side by $x$ and then solve? Is there any other not so difficult way to solve this?






    inequality







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      In my book I found this following inequality :
      $frac{2x}{x-1}>x$



      In my book, they solved this in a very hard way which I don't understand properly. But I tried to do this in another way. I divided each side by $x$ and then solved. But I didn't get the full answer, only half part of it. So why can't I divided both side by $x$ and then solve? Is there any other not so difficult way to solve this?






      inequality







      share|cite|improve this question









      $endgroup$




      In my book I found this following inequality :
      $frac{2x}{x-1}>x$



      In my book, they solved this in a very hard way which I don't understand properly. But I tried to do this in another way. I divided each side by $x$ and then solved. But I didn't get the full answer, only half part of it. So why can't I divided both side by $x$ and then solve? Is there any other not so difficult way to solve this?






      inequality




      inequality






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 2 at 10:38









      Asif IqubalAsif Iqubal

      1397




      1397






















          4 Answers
          4






          active

          oldest

          votes


















          3












          $begingroup$

          You can not divide by $x$ because $x$ can be positive or $x$ can be negative.



          It's $$frac{2x}{x-1}-x>0$$ or
          $$frac{x(3-x)}{x-1}>0,$$ which gives the answer:
          $$(-infty,0)cup(1,3).$$
          I used the intervals method.



          We need to draw the $x$ axis and to put there points $0$, $1$ and $3$.



          Now, on $(3,+infty)$ the expression $frac{x(3-x)}{x-1}$ is negative and if we pass a point $3$ then the sign changes.



          Thus, the sign on $(1,3)$ is $+$.



          If we pass a point $1$ then the sign changes again.



          Thus, the sign on $(0,1)$ is $-$.



          And if we pass a point $0$ then the sign still changes.



          Thus, the sign on $(-infty,0)$ is $+$.



          Now, we can write the answer.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Now there is my problem How am I going to deal with three factors? Won't it be very lengthy?
            $endgroup$
            – Asif Iqubal
            Jan 2 at 10:47










          • $begingroup$
            @Asif Iqubal It's the intervals method.
            $endgroup$
            – Michael Rozenberg
            Jan 2 at 10:49










          • $begingroup$
            Can you be kind to elaborate please?
            $endgroup$
            – Asif Iqubal
            Jan 2 at 10:50










          • $begingroup$
            @Asif Iqubal I added something. See now.
            $endgroup$
            – Michael Rozenberg
            Jan 2 at 11:00



















          2












          $begingroup$

          You cannot cancel $x$ because $x$ is unknown, in particular, you do not know the sign of $x$. If $x$ is negative then the inequality becomes reversed. The standard way to solve the inequality of this type is to move all terms to the same side, add together, factorize numerator and denominator and study the sign.



          In this example:
          $$
          frac{2x}{x-1}-x>0
          $$
          $$
          frac{3x-x^2}{x-1}>0
          $$
          $$
          frac{x(3-x)}{x-1}>0
          $$
          A possible change of the sign is at the zeros of the numerator and the denominator, i.e. at $0$, $1$ and $3$. When the whole expression becomes strictly positive?






          share|cite|improve this answer











          $endgroup$





















            1












            $begingroup$

            Dividing by $x$ is a good idea, but one has to realize that there is a problem if $x=0$.



            Actually, if $x<0$ the inequality direction will change ! Same thing if you multiply by $x-1$ : the inequality direction will change if $x-1<0$.



            Once you take care about these points, you should find easily the full solution.






            share|cite|improve this answer









            $endgroup$





















              1












              $begingroup$

              We require $frac{2x}{x-1}-x>0$ or $frac{x(x-3)}{x-1}<0$, which is true for $xin(-infty,0)cup(1,3)$. It can be solved using the wavy-curve method.






              share|cite|improve this answer









              $endgroup$













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                4 Answers
                4






                active

                oldest

                votes








                4 Answers
                4






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                3












                $begingroup$

                You can not divide by $x$ because $x$ can be positive or $x$ can be negative.



                It's $$frac{2x}{x-1}-x>0$$ or
                $$frac{x(3-x)}{x-1}>0,$$ which gives the answer:
                $$(-infty,0)cup(1,3).$$
                I used the intervals method.



                We need to draw the $x$ axis and to put there points $0$, $1$ and $3$.



                Now, on $(3,+infty)$ the expression $frac{x(3-x)}{x-1}$ is negative and if we pass a point $3$ then the sign changes.



                Thus, the sign on $(1,3)$ is $+$.



                If we pass a point $1$ then the sign changes again.



                Thus, the sign on $(0,1)$ is $-$.



                And if we pass a point $0$ then the sign still changes.



                Thus, the sign on $(-infty,0)$ is $+$.



                Now, we can write the answer.






                share|cite|improve this answer











                $endgroup$













                • $begingroup$
                  Now there is my problem How am I going to deal with three factors? Won't it be very lengthy?
                  $endgroup$
                  – Asif Iqubal
                  Jan 2 at 10:47










                • $begingroup$
                  @Asif Iqubal It's the intervals method.
                  $endgroup$
                  – Michael Rozenberg
                  Jan 2 at 10:49










                • $begingroup$
                  Can you be kind to elaborate please?
                  $endgroup$
                  – Asif Iqubal
                  Jan 2 at 10:50










                • $begingroup$
                  @Asif Iqubal I added something. See now.
                  $endgroup$
                  – Michael Rozenberg
                  Jan 2 at 11:00
















                3












                $begingroup$

                You can not divide by $x$ because $x$ can be positive or $x$ can be negative.



                It's $$frac{2x}{x-1}-x>0$$ or
                $$frac{x(3-x)}{x-1}>0,$$ which gives the answer:
                $$(-infty,0)cup(1,3).$$
                I used the intervals method.



                We need to draw the $x$ axis and to put there points $0$, $1$ and $3$.



                Now, on $(3,+infty)$ the expression $frac{x(3-x)}{x-1}$ is negative and if we pass a point $3$ then the sign changes.



                Thus, the sign on $(1,3)$ is $+$.



                If we pass a point $1$ then the sign changes again.



                Thus, the sign on $(0,1)$ is $-$.



                And if we pass a point $0$ then the sign still changes.



                Thus, the sign on $(-infty,0)$ is $+$.



                Now, we can write the answer.






                share|cite|improve this answer











                $endgroup$













                • $begingroup$
                  Now there is my problem How am I going to deal with three factors? Won't it be very lengthy?
                  $endgroup$
                  – Asif Iqubal
                  Jan 2 at 10:47










                • $begingroup$
                  @Asif Iqubal It's the intervals method.
                  $endgroup$
                  – Michael Rozenberg
                  Jan 2 at 10:49










                • $begingroup$
                  Can you be kind to elaborate please?
                  $endgroup$
                  – Asif Iqubal
                  Jan 2 at 10:50










                • $begingroup$
                  @Asif Iqubal I added something. See now.
                  $endgroup$
                  – Michael Rozenberg
                  Jan 2 at 11:00














                3












                3








                3





                $begingroup$

                You can not divide by $x$ because $x$ can be positive or $x$ can be negative.



                It's $$frac{2x}{x-1}-x>0$$ or
                $$frac{x(3-x)}{x-1}>0,$$ which gives the answer:
                $$(-infty,0)cup(1,3).$$
                I used the intervals method.



                We need to draw the $x$ axis and to put there points $0$, $1$ and $3$.



                Now, on $(3,+infty)$ the expression $frac{x(3-x)}{x-1}$ is negative and if we pass a point $3$ then the sign changes.



                Thus, the sign on $(1,3)$ is $+$.



                If we pass a point $1$ then the sign changes again.



                Thus, the sign on $(0,1)$ is $-$.



                And if we pass a point $0$ then the sign still changes.



                Thus, the sign on $(-infty,0)$ is $+$.



                Now, we can write the answer.






                share|cite|improve this answer











                $endgroup$



                You can not divide by $x$ because $x$ can be positive or $x$ can be negative.



                It's $$frac{2x}{x-1}-x>0$$ or
                $$frac{x(3-x)}{x-1}>0,$$ which gives the answer:
                $$(-infty,0)cup(1,3).$$
                I used the intervals method.



                We need to draw the $x$ axis and to put there points $0$, $1$ and $3$.



                Now, on $(3,+infty)$ the expression $frac{x(3-x)}{x-1}$ is negative and if we pass a point $3$ then the sign changes.



                Thus, the sign on $(1,3)$ is $+$.



                If we pass a point $1$ then the sign changes again.



                Thus, the sign on $(0,1)$ is $-$.



                And if we pass a point $0$ then the sign still changes.



                Thus, the sign on $(-infty,0)$ is $+$.



                Now, we can write the answer.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 2 at 11:16

























                answered Jan 2 at 10:45









                Michael RozenbergMichael Rozenberg

                106k1893198




                106k1893198












                • $begingroup$
                  Now there is my problem How am I going to deal with three factors? Won't it be very lengthy?
                  $endgroup$
                  – Asif Iqubal
                  Jan 2 at 10:47










                • $begingroup$
                  @Asif Iqubal It's the intervals method.
                  $endgroup$
                  – Michael Rozenberg
                  Jan 2 at 10:49










                • $begingroup$
                  Can you be kind to elaborate please?
                  $endgroup$
                  – Asif Iqubal
                  Jan 2 at 10:50










                • $begingroup$
                  @Asif Iqubal I added something. See now.
                  $endgroup$
                  – Michael Rozenberg
                  Jan 2 at 11:00


















                • $begingroup$
                  Now there is my problem How am I going to deal with three factors? Won't it be very lengthy?
                  $endgroup$
                  – Asif Iqubal
                  Jan 2 at 10:47










                • $begingroup$
                  @Asif Iqubal It's the intervals method.
                  $endgroup$
                  – Michael Rozenberg
                  Jan 2 at 10:49










                • $begingroup$
                  Can you be kind to elaborate please?
                  $endgroup$
                  – Asif Iqubal
                  Jan 2 at 10:50










                • $begingroup$
                  @Asif Iqubal I added something. See now.
                  $endgroup$
                  – Michael Rozenberg
                  Jan 2 at 11:00
















                $begingroup$
                Now there is my problem How am I going to deal with three factors? Won't it be very lengthy?
                $endgroup$
                – Asif Iqubal
                Jan 2 at 10:47




                $begingroup$
                Now there is my problem How am I going to deal with three factors? Won't it be very lengthy?
                $endgroup$
                – Asif Iqubal
                Jan 2 at 10:47












                $begingroup$
                @Asif Iqubal It's the intervals method.
                $endgroup$
                – Michael Rozenberg
                Jan 2 at 10:49




                $begingroup$
                @Asif Iqubal It's the intervals method.
                $endgroup$
                – Michael Rozenberg
                Jan 2 at 10:49












                $begingroup$
                Can you be kind to elaborate please?
                $endgroup$
                – Asif Iqubal
                Jan 2 at 10:50




                $begingroup$
                Can you be kind to elaborate please?
                $endgroup$
                – Asif Iqubal
                Jan 2 at 10:50












                $begingroup$
                @Asif Iqubal I added something. See now.
                $endgroup$
                – Michael Rozenberg
                Jan 2 at 11:00




                $begingroup$
                @Asif Iqubal I added something. See now.
                $endgroup$
                – Michael Rozenberg
                Jan 2 at 11:00











                2












                $begingroup$

                You cannot cancel $x$ because $x$ is unknown, in particular, you do not know the sign of $x$. If $x$ is negative then the inequality becomes reversed. The standard way to solve the inequality of this type is to move all terms to the same side, add together, factorize numerator and denominator and study the sign.



                In this example:
                $$
                frac{2x}{x-1}-x>0
                $$
                $$
                frac{3x-x^2}{x-1}>0
                $$
                $$
                frac{x(3-x)}{x-1}>0
                $$
                A possible change of the sign is at the zeros of the numerator and the denominator, i.e. at $0$, $1$ and $3$. When the whole expression becomes strictly positive?






                share|cite|improve this answer











                $endgroup$


















                  2












                  $begingroup$

                  You cannot cancel $x$ because $x$ is unknown, in particular, you do not know the sign of $x$. If $x$ is negative then the inequality becomes reversed. The standard way to solve the inequality of this type is to move all terms to the same side, add together, factorize numerator and denominator and study the sign.



                  In this example:
                  $$
                  frac{2x}{x-1}-x>0
                  $$
                  $$
                  frac{3x-x^2}{x-1}>0
                  $$
                  $$
                  frac{x(3-x)}{x-1}>0
                  $$
                  A possible change of the sign is at the zeros of the numerator and the denominator, i.e. at $0$, $1$ and $3$. When the whole expression becomes strictly positive?






                  share|cite|improve this answer











                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    You cannot cancel $x$ because $x$ is unknown, in particular, you do not know the sign of $x$. If $x$ is negative then the inequality becomes reversed. The standard way to solve the inequality of this type is to move all terms to the same side, add together, factorize numerator and denominator and study the sign.



                    In this example:
                    $$
                    frac{2x}{x-1}-x>0
                    $$
                    $$
                    frac{3x-x^2}{x-1}>0
                    $$
                    $$
                    frac{x(3-x)}{x-1}>0
                    $$
                    A possible change of the sign is at the zeros of the numerator and the denominator, i.e. at $0$, $1$ and $3$. When the whole expression becomes strictly positive?






                    share|cite|improve this answer











                    $endgroup$



                    You cannot cancel $x$ because $x$ is unknown, in particular, you do not know the sign of $x$. If $x$ is negative then the inequality becomes reversed. The standard way to solve the inequality of this type is to move all terms to the same side, add together, factorize numerator and denominator and study the sign.



                    In this example:
                    $$
                    frac{2x}{x-1}-x>0
                    $$
                    $$
                    frac{3x-x^2}{x-1}>0
                    $$
                    $$
                    frac{x(3-x)}{x-1}>0
                    $$
                    A possible change of the sign is at the zeros of the numerator and the denominator, i.e. at $0$, $1$ and $3$. When the whole expression becomes strictly positive?







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Jan 2 at 10:50

























                    answered Jan 2 at 10:43









                    A.Γ.A.Γ.

                    22.8k32656




                    22.8k32656























                        1












                        $begingroup$

                        Dividing by $x$ is a good idea, but one has to realize that there is a problem if $x=0$.



                        Actually, if $x<0$ the inequality direction will change ! Same thing if you multiply by $x-1$ : the inequality direction will change if $x-1<0$.



                        Once you take care about these points, you should find easily the full solution.






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          Dividing by $x$ is a good idea, but one has to realize that there is a problem if $x=0$.



                          Actually, if $x<0$ the inequality direction will change ! Same thing if you multiply by $x-1$ : the inequality direction will change if $x-1<0$.



                          Once you take care about these points, you should find easily the full solution.






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            Dividing by $x$ is a good idea, but one has to realize that there is a problem if $x=0$.



                            Actually, if $x<0$ the inequality direction will change ! Same thing if you multiply by $x-1$ : the inequality direction will change if $x-1<0$.



                            Once you take care about these points, you should find easily the full solution.






                            share|cite|improve this answer









                            $endgroup$



                            Dividing by $x$ is a good idea, but one has to realize that there is a problem if $x=0$.



                            Actually, if $x<0$ the inequality direction will change ! Same thing if you multiply by $x-1$ : the inequality direction will change if $x-1<0$.



                            Once you take care about these points, you should find easily the full solution.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 2 at 10:43









                            SergeDSergeD

                            214




                            214























                                1












                                $begingroup$

                                We require $frac{2x}{x-1}-x>0$ or $frac{x(x-3)}{x-1}<0$, which is true for $xin(-infty,0)cup(1,3)$. It can be solved using the wavy-curve method.






                                share|cite|improve this answer









                                $endgroup$


















                                  1












                                  $begingroup$

                                  We require $frac{2x}{x-1}-x>0$ or $frac{x(x-3)}{x-1}<0$, which is true for $xin(-infty,0)cup(1,3)$. It can be solved using the wavy-curve method.






                                  share|cite|improve this answer









                                  $endgroup$
















                                    1












                                    1








                                    1





                                    $begingroup$

                                    We require $frac{2x}{x-1}-x>0$ or $frac{x(x-3)}{x-1}<0$, which is true for $xin(-infty,0)cup(1,3)$. It can be solved using the wavy-curve method.






                                    share|cite|improve this answer









                                    $endgroup$



                                    We require $frac{2x}{x-1}-x>0$ or $frac{x(x-3)}{x-1}<0$, which is true for $xin(-infty,0)cup(1,3)$. It can be solved using the wavy-curve method.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Jan 2 at 10:48









                                    Shubham JohriShubham Johri

                                    5,204718




                                    5,204718






























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