assuming a positive in an inequality why can you not multiply out $x-4$ from both sides in:
 $$\frac{3}{x-4}>1$$
can we not assume $x-4$ is positive due to the fact that a positive integer divided by $x-4$ is greater than 1?
(I understand you can multiply both sides by $(x-4)^2$ as it guarantees to be positive)
 A: You can do that if you're careful.  You'll need to consider two separate cases:
Case 1:  $x-4 > 0$, i.e., $x > 4$.


*

*This gives us $3 > x-4$, and so $7 > x$.  Since $x>4$ also then all together we get $4 < x < 7$ from this case.


Case 2:  $x-4 < 0$, i.e., $x < 4$.


*

*This gives us $3 < x-4$, i.e., $x>7$.  But $x>7$ is not possible because this is the case where $x<4$.  We can't have both $x<4$ and $x>7$ true at the same time.  Thus we get nothing from this case.


So all together we get $4 < x < 7$.
This is fine for such a relatively simple problem, but this method doesn't scale well.  How would you do this for something like $\dfrac{x-1}{(x+2)(x+7)} > 1$?
A better way to approach rational inequality problems (better because it scales well for more difficult problems) is to do the following:


*

*First get $0$ on one side and everything else on the other side.


*

*This gives us $\dfrac3{x-4} - 1 > 0$.


*Now get a common denominator on the nonzero side.


*

*This gives us $\dfrac3{x-4} - \dfrac{x-4}{x-4} > 0$, which simplifies to $\dfrac{7-x}{x-4} > 0$


*Now find what some textbooks call your "key values" (or, tragically, "critical values") which are the values of $x$ that make the numerator and denominator zero.


*

*For the problem we're doing, these values are $x=4$ and $x=7$.


*Set up a number line with your key values labeled.


*

*We'll have this: 
<---------|-------|---------> x
          4       7



*So now you have three intervals:  $(-\infty, 4)$, $(4,7)$, and $(7,+\infty)$.  Choose one number from each interval and plug your choice into $\dfrac{7-x}{x-4}$ to determine its sign.


*

*From $(-\infty, 4)$ I'll choose $x=0$.  $\dfrac{7-0}{0-4} = -\dfrac74 < 0$.

*From $(4,7)$ I'll choose $x=5$.  $\dfrac{7-5}{5-4} = 2 > 0$.

*From $(7,+\infty)$ I'll choose $x=8$.  $\dfrac{7-8}{8-4} = -\dfrac14 < 0$.


*Put your findings on your number line from step 4, so that the number line becomes a sign chart, tracking the sign of $\dfrac{7-x}{x-4}$ in each interval:
         -        +        -
    <---------|-------|---------> x
              4       7


*From the sign chart, we see that the only interval where $\dfrac{7-x}{x-4} > 0$ is true (and therefore also the original inequality $\dfrac3{x-4} > 1$ is true) is $(4,7)$, i.e., $4 < x < 7$.

Yes, this way is more work for something like $\dfrac3{x-4} > 1$ (which is why I said the shortcut method is fine for it, as long as you're careful).  But again, it's important to know this more standard method because the shortcut way of doing $\dfrac3{x-4} > 1$ doesn't scale well for more complicated rational inequality problems, even one with two or more linear factors involving $x$.  Technically the shortcut way is still possible but it becomes extremely tedious and very difficult to properly keep track of everything, especially if there are three or more linear factors involving $x$.  The more standard approach outlined above is significantly more straightforward with the tradeoff that there's potentially a little bit of messy algebra.
A: In the present case we can multiply "out" $x-4$; for, that $\frac{3}{x-4} > 1$ implies $x-4 > 0$. 
To prevent random people struggling with their own misconception, let me provide a detailed explanation. The original inequality implies $x-4 > 0$, i.e. $x > 4$. Then $3 > x-4$, i.e. $x < 7$. So $4 < x < 7$. 
If you are dealing with something like $\frac{y}{x}  > 1$, then be cautioned; this implies that either $x, y > 0$ or $x,y < 0$ and hence multiplication by $x$ could flip the inequality sign.
