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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$

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
$$
x = 4+2sqrt 2, y = 4-2sqrt 2.
$$

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

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 $mathbb{C}$.