Finding solutions to $x + y = xy$ where $x > y$












0












$begingroup$


Finding solutions (no matter real or complex) for $x$ and $y$.



$$ x + y = xy$$



Where $x > y$.



Is this possible?



If I try to substitute 1 for $x$, it turns out:
begin{align}
(1) + y &= (1)y\
y + 1 &= y\
y - y = -1\
0 ≠ -1\
end{align}



EDIT: For complex solutions, disregard the condition $ x > y$






algebra-precalculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    There's an easy way to find such solutions. You got the only case where the solution is not defined. Anyway, what does it means $x>y$ for complex numbers??
    $endgroup$
    – Exodd
    Nov 30 '18 at 9:32






  • 1




    $begingroup$
    Take, fro example, any $y in (1,2)$ and define $x$ to be $frac y {y-1}$.
    $endgroup$
    – Kavi Rama Murthy
    Nov 30 '18 at 9:34






  • 3




    $begingroup$
    You say "no matter real or complex" but then you require $x>y$. That doesn't make sense for complex numbers.
    $endgroup$
    – Arthur
    Nov 30 '18 at 9:36










  • $begingroup$
    @Murthy It breaks the condition $x > y$...
    $endgroup$
    – MMJM
    Nov 30 '18 at 9:37






  • 1




    $begingroup$
    You may rewrite the constraints as $(x-1)(y-1)=1$ and $x-1>y-1$. This should be easy.
    $endgroup$
    – user1551
    Nov 30 '18 at 10:01
















0












$begingroup$


Finding solutions (no matter real or complex) for $x$ and $y$.



$$ x + y = xy$$



Where $x > y$.



Is this possible?



If I try to substitute 1 for $x$, it turns out:
begin{align}
(1) + y &= (1)y\
y + 1 &= y\
y - y = -1\
0 ≠ -1\
end{align}



EDIT: For complex solutions, disregard the condition $ x > y$






algebra-precalculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    There's an easy way to find such solutions. You got the only case where the solution is not defined. Anyway, what does it means $x>y$ for complex numbers??
    $endgroup$
    – Exodd
    Nov 30 '18 at 9:32






  • 1




    $begingroup$
    Take, fro example, any $y in (1,2)$ and define $x$ to be $frac y {y-1}$.
    $endgroup$
    – Kavi Rama Murthy
    Nov 30 '18 at 9:34






  • 3




    $begingroup$
    You say "no matter real or complex" but then you require $x>y$. That doesn't make sense for complex numbers.
    $endgroup$
    – Arthur
    Nov 30 '18 at 9:36










  • $begingroup$
    @Murthy It breaks the condition $x > y$...
    $endgroup$
    – MMJM
    Nov 30 '18 at 9:37






  • 1




    $begingroup$
    You may rewrite the constraints as $(x-1)(y-1)=1$ and $x-1>y-1$. This should be easy.
    $endgroup$
    – user1551
    Nov 30 '18 at 10:01














0












0








0





$begingroup$


Finding solutions (no matter real or complex) for $x$ and $y$.



$$ x + y = xy$$



Where $x > y$.



Is this possible?



If I try to substitute 1 for $x$, it turns out:
begin{align}
(1) + y &= (1)y\
y + 1 &= y\
y - y = -1\
0 ≠ -1\
end{align}



EDIT: For complex solutions, disregard the condition $ x > y$






algebra-precalculus







share|cite|improve this question











$endgroup$




Finding solutions (no matter real or complex) for $x$ and $y$.



$$ x + y = xy$$



Where $x > y$.



Is this possible?



If I try to substitute 1 for $x$, it turns out:
begin{align}
(1) + y &= (1)y\
y + 1 &= y\
y - y = -1\
0 ≠ -1\
end{align}



EDIT: For complex solutions, disregard the condition $ x > y$






algebra-precalculus




algebra-precalculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 30 '18 at 9:59









user1551

72k566126




72k566126










asked Nov 30 '18 at 9:30









MMJMMMJM

3081110




3081110












  • $begingroup$
    There's an easy way to find such solutions. You got the only case where the solution is not defined. Anyway, what does it means $x>y$ for complex numbers??
    $endgroup$
    – Exodd
    Nov 30 '18 at 9:32






  • 1




    $begingroup$
    Take, fro example, any $y in (1,2)$ and define $x$ to be $frac y {y-1}$.
    $endgroup$
    – Kavi Rama Murthy
    Nov 30 '18 at 9:34






  • 3




    $begingroup$
    You say "no matter real or complex" but then you require $x>y$. That doesn't make sense for complex numbers.
    $endgroup$
    – Arthur
    Nov 30 '18 at 9:36










  • $begingroup$
    @Murthy It breaks the condition $x > y$...
    $endgroup$
    – MMJM
    Nov 30 '18 at 9:37






  • 1




    $begingroup$
    You may rewrite the constraints as $(x-1)(y-1)=1$ and $x-1>y-1$. This should be easy.
    $endgroup$
    – user1551
    Nov 30 '18 at 10:01


















  • $begingroup$
    There's an easy way to find such solutions. You got the only case where the solution is not defined. Anyway, what does it means $x>y$ for complex numbers??
    $endgroup$
    – Exodd
    Nov 30 '18 at 9:32






  • 1




    $begingroup$
    Take, fro example, any $y in (1,2)$ and define $x$ to be $frac y {y-1}$.
    $endgroup$
    – Kavi Rama Murthy
    Nov 30 '18 at 9:34






  • 3




    $begingroup$
    You say "no matter real or complex" but then you require $x>y$. That doesn't make sense for complex numbers.
    $endgroup$
    – Arthur
    Nov 30 '18 at 9:36










  • $begingroup$
    @Murthy It breaks the condition $x > y$...
    $endgroup$
    – MMJM
    Nov 30 '18 at 9:37






  • 1




    $begingroup$
    You may rewrite the constraints as $(x-1)(y-1)=1$ and $x-1>y-1$. This should be easy.
    $endgroup$
    – user1551
    Nov 30 '18 at 10:01
















$begingroup$
There's an easy way to find such solutions. You got the only case where the solution is not defined. Anyway, what does it means $x>y$ for complex numbers??
$endgroup$
– Exodd
Nov 30 '18 at 9:32




$begingroup$
There's an easy way to find such solutions. You got the only case where the solution is not defined. Anyway, what does it means $x>y$ for complex numbers??
$endgroup$
– Exodd
Nov 30 '18 at 9:32




1




1




$begingroup$
Take, fro example, any $y in (1,2)$ and define $x$ to be $frac y {y-1}$.
$endgroup$
– Kavi Rama Murthy
Nov 30 '18 at 9:34




$begingroup$
Take, fro example, any $y in (1,2)$ and define $x$ to be $frac y {y-1}$.
$endgroup$
– Kavi Rama Murthy
Nov 30 '18 at 9:34




3




3




$begingroup$
You say "no matter real or complex" but then you require $x>y$. That doesn't make sense for complex numbers.
$endgroup$
– Arthur
Nov 30 '18 at 9:36




$begingroup$
You say "no matter real or complex" but then you require $x>y$. That doesn't make sense for complex numbers.
$endgroup$
– Arthur
Nov 30 '18 at 9:36












$begingroup$
@Murthy It breaks the condition $x > y$...
$endgroup$
– MMJM
Nov 30 '18 at 9:37




$begingroup$
@Murthy It breaks the condition $x > y$...
$endgroup$
– MMJM
Nov 30 '18 at 9:37




1




1




$begingroup$
You may rewrite the constraints as $(x-1)(y-1)=1$ and $x-1>y-1$. This should be easy.
$endgroup$
– user1551
Nov 30 '18 at 10:01




$begingroup$
You may rewrite the constraints as $(x-1)(y-1)=1$ and $x-1>y-1$. This should be easy.
$endgroup$
– user1551
Nov 30 '18 at 10:01










3 Answers
3






active

oldest

votes


















2












$begingroup$

Consider
$$
A^2 - u A + u = 0,
$$
where $u = x+y =xy$, then $x,y$ are the roots of the equation above about $A$. $x >y$ implies $x, y in mathbb R$ [for complex numbers, we could define $<$, but none of them can guarantee that each pair $(x,y)$ is comparable ]. Then the discriminant
$$
u^2 - 4u > 0 iff u < 0 vee u >4.
$$
So to get one solution, pick some $u > 4$, say $u = 8$, then solve the quadratic, we have
$$
x = 4+2sqrt 2, y = 4-2sqrt 2.
$$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Solving for y:
    $$y=dfrac{x}{x-1}$$
    You want $xgt y$ so.
    $$xgtdfrac{x}{x-1}$$
    and you get:
    $xgt 2$






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      Here is a proof that gives the exhaustive set of solutions.



      Equation $$x+y=xy text{with constraint} x>y, tag{1}$$



      with the following change of variables (see the remark by @user1551):



      $$x=u+1 text{and} y=v+1 tag{2}$$



      is equivalent to the simpler problem:



      $$uv=1 text{with constraint} u>v tag{3}$$



      (the constraint on $u$ and $v$ is due to $x>y iff x-1>y-1$).



      Thus it suffices to consider




      • either any $0<v<1$ ; the ordered pair $(u=frac{1}{v},v)$ is a solution.


      • or any $v<-1$ ; the ordered pair $(u=frac{1}{v},v)$ is a solution.


      • there are no other solutions, because other choices for $v$ give $u<v$.



      Then switch back to variables $x,y$ using (2).



      One can see a way to interpret what we do in terms of a supplementary image : (quadric) surface $z=xy$ intersected by plane with equation $z=1$ along the hyperbola whose horizontal projection has equation $y=frac{1}{x}$.



      enter image description here



      Remark : you cannot have complex solutions because $x>y$ doesn't make sense in $mathbb{C}$.






      share|cite|improve this answer











      $endgroup$













        Your Answer





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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        2












        $begingroup$

        Consider
        $$
        A^2 - u A + u = 0,
        $$
        where $u = x+y =xy$, then $x,y$ are the roots of the equation above about $A$. $x >y$ implies $x, y in mathbb R$ [for complex numbers, we could define $<$, but none of them can guarantee that each pair $(x,y)$ is comparable ]. Then the discriminant
        $$
        u^2 - 4u > 0 iff u < 0 vee u >4.
        $$
        So to get one solution, pick some $u > 4$, say $u = 8$, then solve the quadratic, we have
        $$
        x = 4+2sqrt 2, y = 4-2sqrt 2.
        $$






        share|cite|improve this answer









        $endgroup$


















          2












          $begingroup$

          Consider
          $$
          A^2 - u A + u = 0,
          $$
          where $u = x+y =xy$, then $x,y$ are the roots of the equation above about $A$. $x >y$ implies $x, y in mathbb R$ [for complex numbers, we could define $<$, but none of them can guarantee that each pair $(x,y)$ is comparable ]. Then the discriminant
          $$
          u^2 - 4u > 0 iff u < 0 vee u >4.
          $$
          So to get one solution, pick some $u > 4$, say $u = 8$, then solve the quadratic, we have
          $$
          x = 4+2sqrt 2, y = 4-2sqrt 2.
          $$






          share|cite|improve this answer









          $endgroup$
















            2












            2








            2





            $begingroup$

            Consider
            $$
            A^2 - u A + u = 0,
            $$
            where $u = x+y =xy$, then $x,y$ are the roots of the equation above about $A$. $x >y$ implies $x, y in mathbb R$ [for complex numbers, we could define $<$, but none of them can guarantee that each pair $(x,y)$ is comparable ]. Then the discriminant
            $$
            u^2 - 4u > 0 iff u < 0 vee u >4.
            $$
            So to get one solution, pick some $u > 4$, say $u = 8$, then solve the quadratic, we have
            $$
            x = 4+2sqrt 2, y = 4-2sqrt 2.
            $$






            share|cite|improve this answer









            $endgroup$



            Consider
            $$
            A^2 - u A + u = 0,
            $$
            where $u = x+y =xy$, then $x,y$ are the roots of the equation above about $A$. $x >y$ implies $x, y in mathbb R$ [for complex numbers, we could define $<$, but none of them can guarantee that each pair $(x,y)$ is comparable ]. Then the discriminant
            $$
            u^2 - 4u > 0 iff u < 0 vee u >4.
            $$
            So to get one solution, pick some $u > 4$, say $u = 8$, then solve the quadratic, we have
            $$
            x = 4+2sqrt 2, y = 4-2sqrt 2.
            $$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 30 '18 at 9:41









            xbhxbh

            5,8681522




            5,8681522























                1












                $begingroup$

                Solving for y:
                $$y=dfrac{x}{x-1}$$
                You want $xgt y$ so.
                $$xgtdfrac{x}{x-1}$$
                and you get:
                $xgt 2$






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  Solving for y:
                  $$y=dfrac{x}{x-1}$$
                  You want $xgt y$ so.
                  $$xgtdfrac{x}{x-1}$$
                  and you get:
                  $xgt 2$






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    Solving for y:
                    $$y=dfrac{x}{x-1}$$
                    You want $xgt y$ so.
                    $$xgtdfrac{x}{x-1}$$
                    and you get:
                    $xgt 2$






                    share|cite|improve this answer









                    $endgroup$



                    Solving for y:
                    $$y=dfrac{x}{x-1}$$
                    You want $xgt y$ so.
                    $$xgtdfrac{x}{x-1}$$
                    and you get:
                    $xgt 2$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 30 '18 at 9:48









                    Riccardo.AlestraRiccardo.Alestra

                    5,95212153




                    5,95212153























                        1












                        $begingroup$

                        Here is a proof that gives the exhaustive set of solutions.



                        Equation $$x+y=xy text{with constraint} x>y, tag{1}$$



                        with the following change of variables (see the remark by @user1551):



                        $$x=u+1 text{and} y=v+1 tag{2}$$



                        is equivalent to the simpler problem:



                        $$uv=1 text{with constraint} u>v tag{3}$$



                        (the constraint on $u$ and $v$ is due to $x>y iff x-1>y-1$).



                        Thus it suffices to consider




                        • either any $0<v<1$ ; the ordered pair $(u=frac{1}{v},v)$ is a solution.


                        • or any $v<-1$ ; the ordered pair $(u=frac{1}{v},v)$ is a solution.


                        • there are no other solutions, because other choices for $v$ give $u<v$.



                        Then switch back to variables $x,y$ using (2).



                        One can see a way to interpret what we do in terms of a supplementary image : (quadric) surface $z=xy$ intersected by plane with equation $z=1$ along the hyperbola whose horizontal projection has equation $y=frac{1}{x}$.



                        enter image description here



                        Remark : you cannot have complex solutions because $x>y$ doesn't make sense in $mathbb{C}$.






                        share|cite|improve this answer











                        $endgroup$


















                          1












                          $begingroup$

                          Here is a proof that gives the exhaustive set of solutions.



                          Equation $$x+y=xy text{with constraint} x>y, tag{1}$$



                          with the following change of variables (see the remark by @user1551):



                          $$x=u+1 text{and} y=v+1 tag{2}$$



                          is equivalent to the simpler problem:



                          $$uv=1 text{with constraint} u>v tag{3}$$



                          (the constraint on $u$ and $v$ is due to $x>y iff x-1>y-1$).



                          Thus it suffices to consider




                          • either any $0<v<1$ ; the ordered pair $(u=frac{1}{v},v)$ is a solution.


                          • or any $v<-1$ ; the ordered pair $(u=frac{1}{v},v)$ is a solution.


                          • there are no other solutions, because other choices for $v$ give $u<v$.



                          Then switch back to variables $x,y$ using (2).



                          One can see a way to interpret what we do in terms of a supplementary image : (quadric) surface $z=xy$ intersected by plane with equation $z=1$ along the hyperbola whose horizontal projection has equation $y=frac{1}{x}$.



                          enter image description here



                          Remark : you cannot have complex solutions because $x>y$ doesn't make sense in $mathbb{C}$.






                          share|cite|improve this answer











                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            Here is a proof that gives the exhaustive set of solutions.



                            Equation $$x+y=xy text{with constraint} x>y, tag{1}$$



                            with the following change of variables (see the remark by @user1551):



                            $$x=u+1 text{and} y=v+1 tag{2}$$



                            is equivalent to the simpler problem:



                            $$uv=1 text{with constraint} u>v tag{3}$$



                            (the constraint on $u$ and $v$ is due to $x>y iff x-1>y-1$).



                            Thus it suffices to consider




                            • either any $0<v<1$ ; the ordered pair $(u=frac{1}{v},v)$ is a solution.


                            • or any $v<-1$ ; the ordered pair $(u=frac{1}{v},v)$ is a solution.


                            • there are no other solutions, because other choices for $v$ give $u<v$.



                            Then switch back to variables $x,y$ using (2).



                            One can see a way to interpret what we do in terms of a supplementary image : (quadric) surface $z=xy$ intersected by plane with equation $z=1$ along the hyperbola whose horizontal projection has equation $y=frac{1}{x}$.



                            enter image description here



                            Remark : you cannot have complex solutions because $x>y$ doesn't make sense in $mathbb{C}$.






                            share|cite|improve this answer











                            $endgroup$



                            Here is a proof that gives the exhaustive set of solutions.



                            Equation $$x+y=xy text{with constraint} x>y, tag{1}$$



                            with the following change of variables (see the remark by @user1551):



                            $$x=u+1 text{and} y=v+1 tag{2}$$



                            is equivalent to the simpler problem:



                            $$uv=1 text{with constraint} u>v tag{3}$$



                            (the constraint on $u$ and $v$ is due to $x>y iff x-1>y-1$).



                            Thus it suffices to consider




                            • either any $0<v<1$ ; the ordered pair $(u=frac{1}{v},v)$ is a solution.


                            • or any $v<-1$ ; the ordered pair $(u=frac{1}{v},v)$ is a solution.


                            • there are no other solutions, because other choices for $v$ give $u<v$.



                            Then switch back to variables $x,y$ using (2).



                            One can see a way to interpret what we do in terms of a supplementary image : (quadric) surface $z=xy$ intersected by plane with equation $z=1$ along the hyperbola whose horizontal projection has equation $y=frac{1}{x}$.



                            enter image description here



                            Remark : you cannot have complex solutions because $x>y$ doesn't make sense in $mathbb{C}$.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Nov 30 '18 at 12:29

























                            answered Nov 30 '18 at 11:15









                            Jean MarieJean Marie

                            28.8k41949




                            28.8k41949






























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                                Verónica Boquete

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                                Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl ...
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                                Ida-Boy-Ed-Garten

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                                Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich ...
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