Find the solution set of the in equation

$\frac{x+1}{x+2}<1$

$\Rightarrow \frac{x+1}{x+2}-1<0$

$\Rightarrow \frac{x+1-x-2}{x+2}<0$

$\Rightarrow \frac{-1}{x+2}<0$

We have to find values of $x$ for which $\frac{-1}{x+2}<0$ that is $\frac{-1}{x+2}$ is negative

The numerator of $\frac{-1}{x+2}$ is $-1$ which is negative hence for $\frac{-1}{x+2}$ to be negative $x+2$ must be positive (otherwise if $x+2$ is negative then negative upon negative will be positive)

That is x + 2 should be greater than 0

$\Rightarrow x+2>0$

$\Rightarrow x>-2$

Hence $x$ should be greater than $-2$ for $\frac{-1}{x+2}<0$

$x>-2$ means $x$ can take values from $-2$ to $\infty$ hence $x \in(-2, \infty)$

Hence the solution set for $\frac{x+1}{x+2}<0$ is $(-2, \infty)$