### Instruction: Solve the inequality
\[\dfrac{x+1}{x+2}>\dfrac{3x+4}{2x+9}.\].### Response: I notice that this is a rational inequality, where the numerator and denominator of each fraction are linear expressions.
To solve this, I want to clear the fractions by multiplying both sides by the least common multiple of the denominators.
The least common multiple of $x+2$ and $2x+9$ is $(x+2)(2x+9)$, so I multiply both sides by that.
I get \[(x+2)(2x+9)\dfrac{x+1}{x+2}>(x+2)(2x+9)\dfrac{3x+4}{2x+9}.\]
I can simplify this by canceling out the common factors of $x+2$ and $2x+9$ in the numerators and denominators.
I get \[(x+1)(2x+9)>(x+2)(3x+4).\]
Now I have a quadratic inequality, which I can solve by expanding, collecting like terms, and finding the zeros.
Expanding the brackets, I get \[2x^2+11x+9>3x^2+10x+8.\]
Subtracting $3x^2+10x+8$ from both sides, I get \[-x^2+x+1>0.\]
To find the zeros of this quadratic expression, I can use the quadratic formula.
I get \[x=\dfrac{-1\pm\sqrt{1^2-4(-1)(1)}}{2(-1)}=\dfrac{-1\pm\sqrt{5}}{-2}.\]
These are the two values of $x$ that make the quadratic expression zero.
To determine the sign of the quadratic expression in between and outside these values, I can use a sign chart or a test point.
I will use a test point.
For example, when $x=0$, the expression is $-0^2+0+1=1>0$.
This means that the expression is positive for all values of $x$ between the two zeros.
So the inequality is satisfied when \[\dfrac{-1-\sqrt{5}}{-2}<x<\dfrac{-1+\sqrt{5}}{-2}.\]