# 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.

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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.

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That will definitely work, but I can guarantee you that this teacher will not be willing to teach that method. This is in a US high school. It is kept very simple... –  Ezea Jan 26 '13 at 22:43
@only_gad Multiplying by the denominator is quite ugly as you'd need to care about all the possible sign changes. See my edit. –  Calvin Lin Jan 26 '13 at 22:45
@only_gad You can choose to teach something that's the correct approach, or you can choose to teach something that's wrong and hence require a lot of case checking. Interesting that your opinion of US high schools is that math must be kept very simple, to the extent of being more complicated than necessary. –  Calvin Lin Jan 26 '13 at 22:50
She would claim that this example would support her method because you achieve 0 test points from multiplying and 1 test point from the original equation with which you could then determine the correct interval. –  Ezea Jan 26 '13 at 22:54
The way that I feel so far is that it is taught in such a way where you memorize operations, but are never asked to understand anything you're doing. I plan to teach a much different way when I have my own classroom, but as I am underneath someone, I have to be careful in my arguments. Its a very scary thing now that I am on this side of the classroom... –  Ezea Jan 26 '13 at 22:56

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 $x>-3$.

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Thank you for this. This is really a fascinating explanation. I know that this type of solution would be met by great resistance among my peers, but it does provide an interesting way to look at it. I generally teach the students to think of these as functions compared to the x-axis when we set one side to 0. We can solve using test points because if we know where the function is equal to 0 and its asymptotes, we know that it must be positive or negative over those intervals. It seems your explanation is much more fundamental to the concept of an inequality, however. –  Ezea Jan 27 '13 at 5:37
@only_gad: You’re welcome. Yes, it’s intended to emphasize the arithmetic/algebra of inequalities at a very basic level. I’m honestly surprised that it would meet with resistance: it’s one of the classic standard methods, and perhaps the most elementary of them. (But even after a lifetime of teaching I’m still faintly surprised to find a teacher objecting to any mathematically legitimate approach that isn’t absurdly inefficient, and I consider such a teacher at best marginally competent.) –  Brian M. Scott Jan 27 '13 at 5:53
I would have to agree with you. At the high school level, all that is required to teach mathematics is a degree in education and passing a 90 question test with questions written in standardized testing formats. For an example, I could take the English test and become a high school English teacher if I pass even though I would have no clue what I was doing. –  Ezea Jan 27 '13 at 15:16
Sorry, that comment wasn't clear. It should have said, all that is required to teach any class in a high-school is... –  Ezea Jan 27 '13 at 17:13