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.
On
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.
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...