How do you solve #(x+1)/(x+3)<2#?

1 Answer
Dec 20, 2016

Answer:

There are two cases to investigate

Explanation:

(1)
#x> -3->x+3>0#
We may multiply both sides by #x+3# and the #<# sign stays:
#x+1<2(x+3)->x+1<2x+6-># subtract #x# and #6#
#cancelx-cancelx+1-6<2x-x+cancel6-cancel6->#
#-5 < x->x> -5#
This, with the first assumption, leads to #x> -3#
(2)
#x<-3->x+3<0#
Multiply by #x+3# but the #<#sign becomes #>#
#x+1>2x+6-># subtract #x# and #6#
#cancelx-cancelx+1-6>2x-x+cancel6-cancel6->#
#-5 > x->x< -5#

Combined: #x< -5orx> -3#

graph{(x+1)/(x+3) [-10, 10, -5, 5]}