Solve the inequation $\;$ $-3 + x \leq \dfrac{8x}{3} + 2 \leq \dfrac{14}{3} + 2x; \; x \in I$ $\;$ and represent the solution set on a number line.
Consider $\;$ $-3 + x \leq \dfrac{8x}{3} + 2$
i.e. $\;$ $-3 - 2\leq \dfrac{8x}{3} - x$
i.e. $\;$ $-5 \leq \dfrac{5x}{3}$
i.e. $\;$ $-1 \leq \dfrac{x}{3}$
i.e. $\;$ $-3 \leq x$ $\;\;\; \cdots \; (1)$
Consider $\;$ $\dfrac{8x}{3} + 2 \leq \dfrac{14}{3} + 2x$
i.e. $\;$ $\dfrac{8x}{3} - 2x \leq \dfrac{14}{3} - 2$
i.e. $\;$ $\dfrac{2x}{3} \leq \dfrac{8}{3}$
i.e. $\;$ $2x \leq 8$
i.e. $\;$ $x \leq 4$ $\;\;\; \cdots (2)$
$\therefore \;$ We have from equations $(1)$ and $(2)$, $\;$ $- 3 \leq x \leq 4$
$\therefore \;$ The solution set of the given inequation is: $\;$ $\left\{x \mid -3 \leq x \leq 4, \; x \in I \right\}$
