Answer
$x\lt\dfrac{10}{3}$
Work Step by Step
Using the properties of inequalities, the solution to the given inequality, $
6x\lt20
,$ is
\begin{array}{l}\require{cancel}
x\lt\dfrac{20}{6}
\\\\
x\lt\dfrac{\cancel{2}\cdot10}{\cancel{2}\cdot3}
\\\\
x\lt\dfrac{10}{3}
.\end{array}