## Elementary Algebra

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