I think you are still in doubt , isn't it , I saw you peeking at this again and again
$\displaystyle \frac{1}{x-6} +1 \ge 0$
$\displaystyle \frac{x-5}{x-6} \ge 0$
---Now if $\displaystyle x> 6 $the denominator on LHS is positive
Hence $\displaystyle x\ge 5$ but in the initial condition we have taken it greater than 6
Hence this is correct for $\displaystyle x>6$
---Now when $\displaystyle 5< x<6$ the denominator on LHS is negative
So there will be a change in inequality sign when we will multiply it by
(x-6) on both sides
Hence $\displaystyle x-5 < 0 $
$\displaystyle
\implies x < 5$ but this is against initial condtion hence no answer in this range of x
----When $\displaystyle x \le 5$ the
denominator on LHS is negative
So on multiplying the sign of inequality changes
Hence $\displaystyle x-5 \le 0 $
thus $\displaystyle x \le 5$ is an answer
Now for the complete answer we have to take a combination of all these solution sets(answer as in my last post)
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I am not sure if you were unclear about it, but I think if you were this will be helpful ,ask incase if you still find any trouble
(remember the change of inequality sign on multiplying with a negative number on both sides)