Number of possible integral solutions of an equation. The number of integer solutions of the following equation
$(x \leq y)$
$xy -6(x+y) = 0$
I have concluded that $xy$ is a multiple of $6$ and that $x$ is negative and $y$ is positive.
After that, I obtained these solutions with trial and error: $(-3,2) , (-6,3), (-12,4), (-18,4)$.
Now the obvious question is how long do I continue? And isn't there any better process than this clumsy mess?
And @Dhruv in the comments below proves my analysis wrong as well.
 A: This was a fun question, thank you :)  
First you factor using Simon's favorite factoring trick (explanation below):
Multiplying out $$xy-6(x+y)=0,$$
you get:
$$xy-6x-6y=0.$$
Adding $-6\cdot-6=36$ to both sides, you end up with:
$$xy-6x-6y+36=36\implies (x-6)(y-6)=36.$$
Now factor $36$ for the posisble values of $(x-6)$ and $(y-6)$.
Hint: $36$ can be broken up as $1\cdot36$, $2\cdot18$, $3\cdot12$, $4\cdot9$, and $6\cdot6$ (and their negative counterparts).
Another hint: since $x\leq y$, we know that $(x-6)\le(y-6)$.
Another hint: plug in the factors of $36$ for $(x-6)$ and $(y-6)$:
$(x-6)(y-6)=36$
$(1)(36)=36$
$(2)(18)=36$
$(3)(12)=36$
$(4)(9)=36$
$(6)(6)=36$
$(-36)(-1)=36$
$(-18)(-2)=36$
$(-12)(-3)=36$
$(-9)(-4)=36$
$(-6)(-6)=36$
Solution:
We count the amount of listings above to get our answer $\boxed{10}.$
If we want the individual values of $x$ and $y$, we plug them into the equations above, for example, for $(-9)(-4)$, we would do:
$x-6=-9\implies x=-3$ and $y-6=-4\implies y=2$.
However, your assumption about a positive $y$ and negative $x$ is false. For example, for $(6)(6)$, we get $x=y=0$, where neither $y$ is positive nor $x$ is negative.
Hope this helps :)
