Finding solutions to $x + y = xy$ where $x > y$
$begingroup$
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
edited Nov 30 '18 at 9:59
user1551
72k566126
asked Nov 30 '18 at 9:30
MMJMMMJM
3081110
$endgroup$
add a comment |
$begingroup$
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
edited Nov 30 '18 at 9:59
user1551
72k566126
asked Nov 30 '18 at 9:30
MMJMMMJM
3081110
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
edited Nov 30 '18 at 9:59
user1551
72k566126
asked Nov 30 '18 at 9:30
MMJMMMJM
3081110
$endgroup$
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
edited Nov 30 '18 at 9:59
user1551
72k566126
asked Nov 30 '18 at 9:30
MMJMMMJM
3081110
edited Nov 30 '18 at 9:59
user1551
72k566126
asked Nov 30 '18 at 9:30
MMJMMMJM
3081110
edited Nov 30 '18 at 9:59
user1551
72k566126
edited Nov 30 '18 at 9:59
user1551
72k566126
edited Nov 30 '18 at 9:59
user1551
72k566126
72k566126
asked Nov 30 '18 at 9:30
MMJMMMJM
3081110
asked Nov 30 '18 at 9:30
MMJMMMJM
3081110
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
add a comment |
$begingroup$
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
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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
$$
x = 4+2sqrt 2, y = 4-2sqrt 2.
$$
answered Nov 30 '18 at 9:41
xbhxbh
5,8681522
$endgroup$
add a comment |
$begingroup$
Solving for y:
$$y=dfrac{x}{x-1}$$
You want $xgt y$ so.
$$xgtdfrac{x}{x-1}$$
and you get:
$xgt 2$
answered Nov 30 '18 at 9:48
Riccardo.AlestraRiccardo.Alestra
5,95212153
$endgroup$
add a comment |
$begingroup$
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 $y=frac{1}{x}$.
Remark : you cannot have complex solutions because $x>y$ doesn't make sense in $mathbb{C}$.
answered Nov 30 '18 at 11:15
Jean MarieJean Marie
28.8k41949
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
$$
x = 4+2sqrt 2, y = 4-2sqrt 2.
$$
answered Nov 30 '18 at 9:41
xbhxbh
5,8681522
$endgroup$
add a comment |
$begingroup$
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
$$
x = 4+2sqrt 2, y = 4-2sqrt 2.
$$
answered Nov 30 '18 at 9:41
xbhxbh
5,8681522
$endgroup$
add a comment |
$begingroup$
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
$$
x = 4+2sqrt 2, y = 4-2sqrt 2.
$$
answered Nov 30 '18 at 9:41
xbhxbh
5,8681522
$endgroup$
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
$$
x = 4+2sqrt 2, y = 4-2sqrt 2.
$$
answered Nov 30 '18 at 9:41
xbhxbh
5,8681522
answered Nov 30 '18 at 9:41
xbhxbh
5,8681522
answered Nov 30 '18 at 9:41
xbhxbh
5,8681522
answered Nov 30 '18 at 9:41
xbhxbh
5,8681522
5,8681522
add a comment |
add a comment |
$begingroup$
Solving for y:
$$y=dfrac{x}{x-1}$$
You want $xgt y$ so.
$$xgtdfrac{x}{x-1}$$
and you get:
$xgt 2$
answered Nov 30 '18 at 9:48
Riccardo.AlestraRiccardo.Alestra
5,95212153
$endgroup$
add a comment |
$begingroup$
Solving for y:
$$y=dfrac{x}{x-1}$$
You want $xgt y$ so.
$$xgtdfrac{x}{x-1}$$
and you get:
$xgt 2$
answered Nov 30 '18 at 9:48
Riccardo.AlestraRiccardo.Alestra
5,95212153
$endgroup$
add a comment |
$begingroup$
Solving for y:
$$y=dfrac{x}{x-1}$$
You want $xgt y$ so.
$$xgtdfrac{x}{x-1}$$
and you get:
$xgt 2$
answered Nov 30 '18 at 9:48
Riccardo.AlestraRiccardo.Alestra
5,95212153
$endgroup$
Solving for y:
$$y=dfrac{x}{x-1}$$
You want $xgt y$ so.
$$xgtdfrac{x}{x-1}$$
and you get:
$xgt 2$
answered Nov 30 '18 at 9:48
Riccardo.AlestraRiccardo.Alestra
5,95212153
answered Nov 30 '18 at 9:48
Riccardo.AlestraRiccardo.Alestra
5,95212153
answered Nov 30 '18 at 9:48
Riccardo.AlestraRiccardo.Alestra
5,95212153
answered Nov 30 '18 at 9:48
Riccardo.AlestraRiccardo.Alestra
5,95212153
5,95212153
add a comment |
add a comment |
$begingroup$
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 $y=frac{1}{x}$.
Remark : you cannot have complex solutions because $x>y$ doesn't make sense in $mathbb{C}$.
answered Nov 30 '18 at 11:15
Jean MarieJean Marie
28.8k41949
$endgroup$
add a comment |
$begingroup$
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 $y=frac{1}{x}$.
Remark : you cannot have complex solutions because $x>y$ doesn't make sense in $mathbb{C}$.
answered Nov 30 '18 at 11:15
Jean MarieJean Marie
28.8k41949
$endgroup$
add a comment |
$begingroup$
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 $y=frac{1}{x}$.
Remark : you cannot have complex solutions because $x>y$ doesn't make sense in $mathbb{C}$.
answered Nov 30 '18 at 11:15
Jean MarieJean Marie
28.8k41949
$endgroup$
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 $y=frac{1}{x}$.
Remark : you cannot have complex solutions because $x>y$ doesn't make sense in $mathbb{C}$.
answered Nov 30 '18 at 11:15
Jean MarieJean Marie
28.8k41949
edited Nov 30 '18 at 12:29
answered Nov 30 '18 at 11:15
Jean MarieJean Marie
28.8k41949
answered Nov 30 '18 at 11:15
Jean MarieJean Marie
28.8k41949
answered Nov 30 '18 at 11:15
Jean MarieJean Marie
28.8k41949
28.8k41949
add a comment |
add a comment |
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$begingroup$
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