# Number of solutions to the Diophantine equation $xy-6(x+y)= 0$ such that $x\leq y$

The question is to find number of integer solutions to the equation $$xy-6(x+y)= 0$$ Given that, $$x \leq y$$

So I proceeded as follows,

From the 1st equation I get $$x=\frac{6y}{y-6}$$

Putting it into the 2nd I got

$$\frac{6y}{y-6} \leq y$$ $$\implies \frac{y(y-12)}{y-6} \geq0$$

So, $$y \in [0,6)\cup[12,\infty)$$

How do I proceed from here? As there seem to be infinity possible values of y which may satisfy the inequality...

You already made good progress. When you got $$x=\frac{6y}{y-6}$$ you should write it as $$x=\frac{6y}{y-6}=\frac{6(y-6)+36}{y-6}=6+\frac{36}{y-6},$$ and since $$x$$ is an integer we must have that $$36/(y-6)$$ is an integer as well, in other words $$(y-6)\mid 36$$. This leaves finitely many values to check, which I'm sure you can do.
I just think this approach is easier: \begin{align} xy-6(x+y)&= 0 \\ xy-6(x+y)+36&=36 \\ (x-6)(y-6)&=36=d_1d_2 \\ \to \begin{cases} x&=6+d_1 \\ y&=6+d_2 \end{cases} \end{align}
Symmetric equation, so permuting $$x,y$$ won't give you new answers. Also, $$36=2^23^2$$ has $$9$$ divisors. They can be broken up into $$5$$ pairs of divisors, including one with two equal factors. Negating each pair of divisors also produces a new answer. This implies that you have $$10$$ integral solutions.