Ans: (a) $-8 \le x < \dfrac{-11}{5}$ (b) $-3$
- Consider $\dfrac{3-x}{2} > 2x + 7$.
$\begin{array}{rcl}
\dfrac{3-x}{2} & > & 2x + 7 \\
3 – x & > & 4x + 14 \\
-5x & > & 11 \\
x & < & \dfrac{-11}{5} \end{array}$Consider $x+8 \ge 0$.
$\begin{array}{rcl}
x + 8 & \ge & 0 \\
x & \ge & -8
\end{array}$Therefore, $x < \dfrac{-11}{5}$ and $x \ge -8$.
Hence, the overall solution is $-8 \le x < \dfrac{-11}{5}$.
- The greatest integer satisfying inequalities in (a) is $-3$.
