how does k=0 still make and expression undefined when there are still whole numbers

Do you mean (-10k+13)/(6k+4)? or is one term -10k, another term (13/6)*k or 13/(6k), and the last term 4? I'm going to assume you mean (-10k+13)/(6k+4) based on what you said about monomial factors cancelling out.

When you simplify an expression, you are finding an EQUIVALENT expression, but it's not EXACTLY the same. You still have to take into account the original expression.

So, for example, if the original problem was [(k)(-10k+13)] / [(k)(6k+4)], then think about what would happen if k=0?
Just looking at the denominator, (k)(6k+4), it would become (0)(6*0+4), and, of course, anything times 0 is 0, so it would make our WHOLE denominator 0, which is undefined.

So even though you can cancel the individual factors, you still have to take into account the original expression. Hope this helps.

so obvious when you put it like that. That was very helpful, thank you.