## Elementary Algebra

$n\le-\dfrac{180}{13}$
Multiplying both sides by the $LCD= 30$ and then using the properties of inequalities, the solution to the given inequality, $\dfrac{1}{2}n-\dfrac{1}{3}n-4\ge\dfrac{3}{5}n+2 ,$ is \begin{array}{l}\require{cancel} 15(1n)-10(1n)-30(4)\ge6(3n)+30(2) \\\\ 15n-10n-120\ge18n+60 \\\\ 15n-10n-18n\ge60+120 \\\\ -13n\ge180 \\\\ n\le\dfrac{180}{-13} \\\\ n\le-\dfrac{180}{13} .\end{array}