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...