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?

Hint:
$$\frac{1}{x-6} \leq 3 \iff \frac{1}{x-6} - 3 \leq 0 \iff \frac{19-3x}{x-6} \leq 0 \iff \frac{3x-19}{x-6} \geq 0$$
Can you take it from here?
