Ans: (a) $1 < x \le 5$ (b) $4$
- Consider $7(x-2) \le \dfrac{11x+8}{3}$, we have
$\begin{array}{rcl}
7(x-2) & \le & \dfrac{11x+8}{3} \\
21x – 42 & \le & 11x + 8 \\
10x & \le & 50 \\
x & \le & 5
\end{array}$Consider $6-x<5$, we have
$\begin{array}{rcl}
6 – x & < & 5 \\ -x & < & -1 \\ x & > & 1
\end{array}$Therefore, the overall solution is $1 < x \le 5$.
- $2$, $3$, $4$ and $5$ satisfy both inequalities in (a). Therefore $4$ integers satisfy both inequalities in (a).
