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
⇒ x + 2 > 0
⇒ 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 ∞ hence x ∈ (-2, ∞)
Hence the solution set for \(\frac{x+1}{x+ 2} < 0\) is (-2, ∞)