Why is considering only quadratic in one of the variables of a two variable quadratic sufficient for calculating roots Find the $positive$ integral solutions to $7x^2-2xy+3y^2-27=0$
My solution:
Assuming the quadratic in $x$ , if we assume one root to be integral , the other has to be rational (as y must be an integer to satisfy the condition so ,product of roots is rational)
For the roots to be rational the $discriminant$ has to be a perfect square. We get the discriminant($\Delta$) as
$\Delta=4(189-20y^2)$ which has to be a perfect square.
So we get $y^2$=$1,9$ hence $y$ as $1,3$
putting the values back we get the pair $(x,y)=(2,1)$
If we again make a quadratic in $y$ we get the same solution. Hence considering a quadratic in $x$ only is sufficient.
My Question:
I don't get the fact or intuition behind as to why does considering the quadratic in either $x$ or $y$ is self sufficient as it is not symmetric. If someone could provide me the intuition or proof behind as to why both of them lead to the same results , it would be of great help.
Note: All the similar questions I have encountered can be solved by considering only quadratic in either $x$ or $y$ only, so i assume it is general. 
Thanks.

You are looking for solutions among $(x,y)$ where $x$ is restricted to be integer. But clearly, the integer solutions in $y$ all belong to this subset and you won't miss any.
