<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] If $x < 6$, then $x - 6$ is negative, so multiplying both sides by $x-6$ reverses the inequality.  Thus $1 \ge 3(x - 6)$, or $19 \ge 3x$, which is true for all $x < 6$.
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