Combining the two fractions on the right side gives $\frac{2(3x+4)}{3(2x+9)}+\dfrac{1}{3} = \frac{6x+8}{6x+27}+\frac{1}{3}$. Combining over a common denominator gives $\frac{6x+8}{6x+27}+\frac{1}{3} = \frac{6x+8}{6x+27}+\frac{1}{3} \cdot \frac{9}{9} = \frac{6x+8}{6x+27}+\frac{9}{27} = \frac{6x+8}{6x+27}+\frac{1}{3}$ $= \frac{6x+8+2(6x+27)}{9 \cdot 3}=\frac{14x+62}{9 \cdot 3} = \frac{14x+62}{27}$.
Our inequality becomes $\frac{x+1}{x+2}>\frac{14x+62}{27}$, and we get rid of the fraction by multiplying both sides by $27(x+2)$, $27(x+2) \cdot \frac{x+1}{x+2} > 27(x+2) \cdot \frac{14x+62}{27}$. This gives $[27(x+2)] \frac{x+1}{\cancel{x+2}} > [27(x+2)] \frac{14x+62}{\cancel{27}}$ $\Rightarrow 27(x+1) > (x+2)(14x+62)$ $\Rightarrow27x+27 > (x+2)(14x+62)$.
Expanding the right side gives $27x+27 > 14x^2+78x+124x+124$, which simplifies to $27x+27 > 14x^2+202x+124$ and $0 > 14x^2+175x+97$. Trying to factor this quadratic limited by the Rational Root Theorem, we find that $x = 1$ does not work but $x = -\frac{97}{14}$ does. Therefore, since $\frac{x+1}{x+2}$ is defined when $x \neq -2$, and after testing we find that $x > -\frac{97}{14}$ for other real numbers $x$, our answer is all $x$ such that $x \in \boxed{\left(-\frac{97}{14}, -2\right)}$.
Solve the inequality
\[\dfrac{x+1}{x+2}>\dfrac{3x+4}{2x+9}+\dfrac{1}{3}.\]
1 answer