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

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Answer 1 · 2016-06-30T21:19:49

#x in (-5, -2]#

#(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 #3/(0^-) = - oo#

so the complete answer appears to be #x in (-5, -2]#

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

if #(x+5)/(x+2)<0#

then

#(x+5)/(x+2) * (x+2) < 0 * (x+2) #

#\implies x+5 <0, qquad x < -5#

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