Solve the following inequation: $\;$ $\dfrac{1}{5} \leq \dfrac{3 x}{10} + 1 < 1 \dfrac{3}{5}, \; x \in R$
Write the solution set and represent the solution set on a number line.
Consider $\;$ $\dfrac{1}{5} \leq \dfrac{3 x}{10} + 1$
i.e. $\;$ $\dfrac{1}{5} \leq \dfrac{3x + 10}{10}$
i.e. $\;$ $1 \leq \dfrac{3x + 10}{2}$
i.e. $\;$ $2 \leq 3x + 10$
i.e. $\;$ $- 8 \leq 3x$
i.e. $- \dfrac{8}{3} \leq x$ $\;\;\; \cdots \; (1)$
$\implies$ $x \geq - \dfrac{8}{3}$
Consider $\;$ $\dfrac{3 x}{10} + 1 < 1 \dfrac{3}{5}$
i.e. $\;$ $\dfrac{3x + 10}{10} < \dfrac{8}{5}$
i.e. $\;$ $\dfrac{3x + 10}{2} < 8$
i.e. $\;$ $3x + 10 < 16$
i.e. $\;$ $3x < 6$ $\implies$ $x < 2$ $\;\;\; \cdots \; (2)$
$\therefore \;$ We have from equations $(1)$ and $(2)$, $\;$ $- \dfrac{8}{3} \leq x < 2$
$\therefore \;$ The solution set of the given inequation is: $\;$ $\left\{x \mid -\dfrac{8}{3} \leq x < 2, \; x \in R \right\}$
