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Question:

Solve: $$\frac{5x-6}{x+6}<1$$

My attempt:

$$\frac{5x-6-x-6}{x+6}<0$$ $$\Rightarrow \frac{4x-12}{x+6}<0$$ $$\Rightarrow \frac{x-3}{x+6}<0$$ $$\Rightarrow (x-3)(x+6) < 0$$

Thus, it implies that the answer is $-6 < x < 3$. However, in my textbook, it is given as wrong answer.

Can please someone tell what is the correct procedure for this.

Thanks.

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If you're really unsure when something like that happens, you can always graph your rational function. But consider that in $ \ \frac{x-3}{x+6} \ $ , there is a vertical asymptote at $ \ x \ = \ -6 \ $ and an $ \ x-$ intercept at $ \ x = 3 \ $ . So the real numbers are divided into intervals by these points, $ \ x \ < \ -6 \ , \ -6 \ < \ x \ < \ 3 \ , $ and $ \ x \ > \ 3 \ $ . Also, your rational function has a horizontal asymptote of $ \ y \ = \ 1 \ $ .

So the function is positive for "a lot" of the real numbers to start with. We can also use what we know about working with signed numbers. For large negative numbers, both the numerator and denominator are negative, and for large positive numbers, they are both positive. So for $ \ x \ < \ -6 \ $ and $ \ x \ > \ 3 \ $ , we can conclude that the ratio is positive. It will only be in the interval $ \ -6 \ < \ x \ < \ 3 \ $ that the numerator is negative, while the denominator is positive, so the ratio is negative in this interval.

So you are correct. Large introductory course textbook answers can be wrong anywhere from about $ \ \frac{1}{3} $ % to $ \ 2 $ % of the time, depending upon how carefully the submitted answers (of multiple authorship) have been combed through...

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  • $\begingroup$ Thanks for the detailed method! And yes, you are right, the textbook is indeed large, so the author might have missed it. $\endgroup$ – Gaurang Tandon May 18 '14 at 19:47
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    $\begingroup$ The authors (probably of just about all of the omnibus texts, and not just in math) generally hire teaching assistants to work out answers, which are then collected into the backs of these books. They do not check what can be two to three thousand problem answers for accuracy... $\endgroup$ – colormegone May 18 '14 at 19:54
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Your answer is correct, the textbook is wrong.

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  • $\begingroup$ Ok, thanks for clarification! $\endgroup$ – Gaurang Tandon May 18 '14 at 18:49
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Hint: There are two cases, either $x < -6$ or $x > -6$. You can multiply both sides of the inequality by $x + 6$ and depending on which case you are considering, either it will reverse the inequality or leave it intact. Then you can rewrite the inequality with $x$ on one side and a number on the other side by adding/subtracting/dividing.

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  • $\begingroup$ I am sorry but I do not understand you. The denominator has $(x+6)$, not $x-6$. $\endgroup$ – Gaurang Tandon May 18 '14 at 18:40
  • $\begingroup$ @GaurangTandon Sorry, I should have written $-6$. The point is that there are two cases, either $x+6$ is positive or negative and it will flip the inequality or not depending on what case you're in. $\endgroup$ – user2566092 May 18 '14 at 18:59
  • $\begingroup$ But, then, is my answer wrong ? What is the correct answer. $\endgroup$ – Gaurang Tandon May 18 '14 at 19:03

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