Elementary Algebra

$x\lt0$
Multiplying both sides by the $LCD= 28$ and then using the properties of inequalities, the solution to the given inequality, $\dfrac{x+3}{4}+\dfrac{x-5}{7}\lt\dfrac{1}{28} ,$ is \begin{array}{l}\require{cancel} 7(x+3)+4(x-5)\lt1(1) \\\\ 7x+21+4x-20\lt1 \\\\ 7x+4x\lt1-21+20 \\\\ 11x\lt0 \\\\ x\lt\dfrac{0}{11} \\\\ x\lt0 .\end{array}