#### Answer

$n\le\dfrac{12}{11}$

#### Work Step by Step

Multiplying both sides by the $LCD=
12
$ and then using the properties of inequalities, the solution to the given inequality, $
\dfrac{3}{4}n+\dfrac{1}{6}n\le1
,$ is
\begin{array}{l}\require{cancel}
3(3n)+2(1n)\le12(1)
\\\\
9n+2n\le12
\\\\
11n\le12
\\\\
n\le\dfrac{12}{11}
.\end{array}