<s>[INST] Solution of Inequality $ \frac{1}{x-6}\le 3$ 
Solve the inequality: $\displaystyle \frac{1}{x-6}\le 3$

solution:
\begin{align*}\frac{1}{x-6}& \le 3 \\ x-6& \le \frac{1}{3}
\\x& \le 6+\frac{1}{3}\\
x&\le19/3\end{align*}
but, for values of $x\le 6$ also, inequality holds true as left hand side provides some negative value. but, i don't find this from the solution. what am i missing?
 [/INST] Find  the critical values: The values where the domain is undefined and when the function $f(x)= \frac{1}{x-6} -3$ is equal to $0$.
Here, the critical values are $6, \frac{19}{3} $
Use these values to divide the $x-axis$ into intervals: $(-\infty, 6), (6, \frac{19}{3}), (\frac{19}{3}, \infty) $
Test each intervals by plugging in the number within the interval and see if the value is lesser or equal than 3. If it is, then it's answer. If not, then you can eliminate that interval. 
For example: $(-\infty, 6]$ pick $x=0$ then $ \frac{1}{0-6} =  \frac{1}{-6} < 3$ so  this is an answer. 
$[6, \frac{19}{3}]$, pick $x= \frac{19}{3} $, then $\frac{1}{\frac{19}{3}-6} = 3$ So this is not an answer
Similarly, $[\frac{19}{3}, \infty) $, pick $x=7$, you can see that $\frac{1}{7-6} =1 < 3 $. Thus, this is an answer! 
Therefore, the answer are the intervals: $(-\infty, 6)$ and $(\frac{19}{3}, \infty) $
Another way to do this is by graphing! Graph the function $f(x) = \frac{1}{x-6}$. This is  the graph of the funcion  $f(x) = \frac{1}{x}$ shifted right by 6 units. From the graph, you can see where the values are below the line $y=3$. 
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