# How do you solve (x+5)/(x+2)<0?

Jun 30, 2016

$x \in \left(- 5 , - 2\right]$

#### Explanation:

$\frac{x + 5}{x + 2} < 0$

A very stodgy mechanical way of looking at this is to say that the expression will be negative if either (x+5) < 0 or (x+2) < 0, but not both

so we look at these pairs

PAIR A
(x+5) < 0 and (x+2) > 0

this requires x < -5 and x > -2, so no solution

PAIR B
(x+5) > 0 and (x+2) < 0

this requires x > -5 and x < -2, so this solution works

further refining this approach, if x = 5, then the numerator is zero, not <0. so we must exclude x = 5

if x = -2, we have a singularity $\frac{3}{{0}^{-}} = - \infty$

so the complete answer appears to be $x \in \left(- 5 , - 2\right]$

the obvious temptation here must to be to cross multiply ie to say that

if $\frac{x + 5}{x + 2} < 0$

then

$\frac{x + 5}{x + 2} \cdot \left(x + 2\right) < 0 \cdot \left(x + 2\right)$

$\setminus \implies x + 5 < 0 , q \quad x < - 5$

but that doesn't work with inequalities. worth thinking about.