8. Problem: Find the set of solutions for $x$ in the inequality $\frac{x+1}{x+2}>\frac{3 x+4}{2 x+9}$ when $x \neq-2, x \neq \frac{9}{2}$.
Solution: There are 3 possible cases of $x$ : 1) $\left.\left.-\frac{9}{2}\frac{3 x+4}{2 x+9} \\
\Rightarrow 2 x^{2}+11 x+9 & >3 x^{2}+10 x+8 \\
\Rightarrow 0 & >x^{2}-x-1
\end{aligned}
$$

The inequality holds for all $\frac{1-\sqrt{5}}{2}0$. Then the following operations will change the direction of the inequality:
$$
\begin{aligned}
\frac{x+1}{x+2} & >\frac{3 x+4}{2 x+9} \\
\Rightarrow 2 x^{2}+11 x+9 & <3 x^{2}+10 x+8 \\
\Rightarrow 0 & <x^{2}-x-1
\end{aligned}
$$

The inequality holds for all $x<\frac{1-\sqrt{5}}{2}$ and $\frac{1+\sqrt{5}}{2}<x$. The initial condition was $\frac{-9}{2} \leq x \leq-2$. Hence the intersection of these conditions yields all $x$ such that $\frac{-9}{2} \leq x \leq-2$. Then all possible cases of $x$ are
$$
\frac{-9}{2} \leq x \leq-2 \cup \frac{1-\sqrt{5}}{2}<x<\frac{1+\sqrt{5}}{2} \text {. }
$$