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












0














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
























  • 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??
    – Exodd
    Nov 30 at 9:32






  • 1




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






  • 3




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










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






  • 1




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
















0














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
























  • 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??
    – Exodd
    Nov 30 at 9:32






  • 1




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






  • 3




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










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






  • 1




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














0












0








0







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















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 at 9:59









user1551

71.2k566125




71.2k566125










asked Nov 30 at 9:30









MMJM

3081110




3081110












  • 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??
    – Exodd
    Nov 30 at 9:32






  • 1




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






  • 3




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










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






  • 1




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


















  • 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??
    – Exodd
    Nov 30 at 9:32






  • 1




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






  • 3




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










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






  • 1




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
















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??
– Exodd
Nov 30 at 9:32




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??
– Exodd
Nov 30 at 9:32




1




1




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




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




3




3




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




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












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




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




1




1




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




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










3 Answers
3






active

oldest

votes


















2














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





























    1














    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





























      1














      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























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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        2














        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


























          2














          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
























            2












            2








            2






            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












            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 at 9:41









            xbh

            5,6551522




            5,6551522























                1














                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


























                  1














                  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
























                    1












                    1








                    1






                    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












                    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 at 9:48









                    Riccardo.Alestra

                    5,92412053




                    5,92412053























                        1














                        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




























                          1














                          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


























                            1












                            1








                            1






                            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














                            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 at 12:29

























                            answered Nov 30 at 11:15









                            Jean Marie

                            28.7k41949




                            28.7k41949






























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