### Instruction: Solve the inequality
\[\dfrac{x+1}{x+2}>\dfrac{3x+4}{2x+9}.\].### Response: I notice that this is a rational inequality, which means that the fractions can change sign when their denominators are zero or undefined.
So, before I do anything else, I want to find out when that happens.
I set each denominator equal to zero and solve for x.
For the denominator x+2, I get x=-2.
For the denominator 2x+9, I get x=-9/2.
These are the values of x that make the fractions undefined, so they are not part of the solution set.
They also divide the number line into three intervals: \[(-\infty, -9/2), \quad (-9/2, -2), \quad (-2, \infty).\]
I will test each interval to see if it satisfies the inequality or not.
To do that, I will pick any point in each interval and plug it into the inequality.
For example, in the interval (-\infty, -9/2), I can pick x=-5.
When I plug x=-5 into the inequality, I get \[\dfrac{-5+1}{-5+2}>\dfrac{3(-5)+4}{2(-5)+9},\] which simplifies to \[\dfrac{-4}{-3}>\dfrac{-11}{-1},\] which is false, since -4/-3 is less than -11/-1.
This means that the interval (-\infty, -9/2) is not part of the solution set.
I repeat this process for the other two intervals.
In the interval (-9/2, -2), I can pick x=-3.
When I plug x=-3 into the inequality, I get \[\dfrac{-3+1}{-3+2}>\dfrac{3(-3)+4}{2(-3)+9},\] which simplifies to \[\dfrac{-2}{-1}>\dfrac{-5}{3},\] which is true, since -2/-1 is greater than -5/3.
This means that the interval (-9/2, -2) is part of the solution set.
In the interval (-2, \infty), I can pick x=0.
When I plug x=0 into the inequality, I get \[\dfrac{0+1}{0+2}>\dfrac{3(0)+4}{2(0)+9},\] which simplifies to \[\dfrac{1}{2}>\dfrac{4}{9},\] which is also true, since 1/2 is greater than 4/9.
This means that the interval (-2, \infty) is also part of the solution set.
Therefore, the solution set for the inequality is the union of the two intervals where it is true: \[(-9/2, -2) \cup (-2, \infty).\]