Solving rational inequalities The problem is relatively simple, but I am a student teacher and the students were working on solving rational inequalities. 
Such as $\frac{x+1}{x+3} \leq 1$.
I recommended that they move everything to one side and find a common denominator, and then determine what x values will make the function equal to 0 and the vertical asymptotes. 
My mentor teacher, however, suggested that they multiply both sides by the denominator to simplify and then simply refer to the original problem to obtain the vertical asymptotes. From what I can tell, her method seems to arrive at the correct answer.
I'm worried that there is the possibility that multiplying by this denominator could have consequences for certain problems since there is no way of knowing beforehand if it is positive or negative and therefore could change the direction of the inequality. 
Can anyone clear this up? Will her method always arrive at the correct answer? If not, could you please provide an example where the wrong answer will be reached? Thanks for any help, I just want to make sure that this is taught correctly to the students so that they understand what they are doing.
 A: You multiply by the square of the denominator, which is always non-negative, and so the sign stays the same.

For example, in $ \frac {x+1}{x+3} \leq 1$, we get that $ (x+1)(x+3) \leq (x+3)^2$, which gives us $ 0 \leq (x+3)(x+3-x-1) \Rightarrow 0 \leq (x+3) \times 2$, which has solution set $ -3 \leq x$.
If you simply multiply by $x+3$ (without caring for the sign), you get $x+1 \leq x+3$, or that $0 \leq 2$ which is always true, and thus that the original inequality holds for all real numbers.

Reasoning behind method.
Polynomials and rational functions are mostly continuous. Hence, to see when $f(x) \geq g(x)$, it suffices to check various points of crossover (i.e. $f(x) = g(x)$) or jump over (i.e. vertical asymptotes for rational functions). As such, we look at the equality case and zero denominator respectively, to deal with the potential regions. No further regions can be introduced.
A: I would simply do this problem by cases, since there are only two of them. Obviously the lefthand side is undefined at $x=-3$.


*

*If $x>-3$, the inequality is $x+1\le x+3$, or $1\le 3$, which is true for all $x$; after incorporating the restriction on $x$ that puts us in this case in the first place, we see that all $x>3$ are solutions.

*If $x<-3$, the inequality is $x+1\ge x+3$, or $1\ge 3$, which is never true. Thus, the solutions found in the first case are the only solutions.
If your mentor teacher got the right answer, she must have been keeping track of the cases, though you didn’t say so. (Or she made some compensating error.) At some point they need to learn to work with separate cases, and it might as well be now, when the work is still fairly simple.
I might also show them the benefits of being able easily to do simple algebraic rearrangements:
$$\frac{x+1}{x+3}=1-\frac2{x+3}\;,$$
so the inequality reduces to 
$$\frac2{x+3}\ge 0\;,$$
which is trivially the case if and only if the denominator is positive, i.e., if and only if $x>-3$.
