There are 3 possible cases of $x$: 1) $-\frac{9}{2} < x$, 2) $\frac{9}{2} \leq x \leq -2$, 3) $-2 < x$. For the cases (1) and (3), $x+2$ and $2x+9$ are both positive or negative, so the following operation can be carried out without changing the inequality sign: $$\begin{aligned} \frac{x+1}{x+2} & > \frac{3x+4}{2x+9} \\ \Rightarrow 2x^{2} + 11x + 9 & > 3x^{2} + 10x + 8 \\ \Rightarrow 0 & > x^{2} - x - 1 \end{aligned}$$ The inequality holds for all $\frac{1-\sqrt{5}}{2} < x < \frac{1+\sqrt{5}}{2}$. The initial conditions were $-\frac{9}{2} < x$ or $-2 < x$. The intersection of these three conditions occurs when $\frac{1-\sqrt{5}}{2} < x < \frac{1+\sqrt{5}}{2}$. Case (2) is $\frac{9}{2} \leq x \leq -2$. For all $x$ satisfying these conditions, $x+2 < 0$ and $2x+9 > 0$. Then the following operations will change the direction of the inequality: $$\begin{aligned} \frac{x+1}{x+2} & > \frac{3x+4}{2x+9} \\ \Rightarrow 2x^{2} + 11x + 9 & < 3x^{2} + 10x + 8 \\ \Rightarrow 0 & < x^{2} - x - 1 \end{aligned}$$ 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 $\frac{-9}{2} \leq x \leq -2 \cup \frac{1-\sqrt{5}}{2} < x < \frac{1+\sqrt{5}}{2}$.