Easy Inequality......

$\begin{array}{l} \displaystyle\frac{{x + 5}}{{2x - 1}} \le 3 \\ \Leftrightarrow \displaystyle\frac{{x + 5 - 3(2x - 1)}}{{2x - 1}} \le 0 \\ \Leftrightarrow \displaystyle\frac{{ - 5x + 8}}{{2x - 1}} \le 0 \\ \end{array}$

We have this table:

\begin{array}{*{20}{c}} \hline & x & & {\left( { - \infty ,\displaystyle\frac{1}{2}} \right)} & {\displaystyle\frac{1}{2}} & {\left( {\displaystyle\frac{1}{2},\displaystyle\frac{8}{5}} \right)} & {\displaystyle\frac{8}{5}} & {\left( {\displaystyle\frac{8}{5}, + \infty } \right)} & \\ \hline & { - 5x + 8} & & + & + & + & 0 & - & \\ \hline & {2x - 1} & & - & 0 & + & + & + & \\ \hline & {\displaystyle\frac{{ - 5x + 8}}{{2x - 1}}} & & - & {||} & + & 0 & - & \\ \hline \end{array}

As shown in this table, the solution set is $\left( { - \infty ,\displaystyle\frac{1}{2}} \right) \cup \left[ {\displaystyle\frac{8}{5}, + \infty } \right)$

Observe that when the denominator is an algebraic expression, we need to consider 2 cases :- denominator positive and denominator negative. After considering both cases, we need to take the most appropriate solution set of the variable that satisfy both cases.