Okay. So our assignment is to get the domain, range, intercepts, etc. of the relation (also a function but not expressed as one)

$\displaystyle 5y(x-3)(x-1)=2(5x+3)$. I didn't have a problem at all till I realize that I got $\displaystyle (-\infty,-5]\cup[-1/5,0)\cup(0,+\infty)$

(I EXCLUDED ZERO because when I solved for x, y is in the DENOMINATOR) as the range and have an x-intercept $\displaystyle (-3/5,0)$

which is a contradiction with the range.

Now, it leaves me with a dilemma because I can't include zero in the range because x will be undefined (at least if you use quadratic formula by manipulating the given equation which I think alters its nature I presume), and I can't just simply remove the x-intercept because of the same reason.

Here's my question, do I include zero in the range OR remove the x-intercept OR just leave my answer as it is.

And please explain why.