How do you solve the inequality # |x + 2| ≤ 3# ?

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Answer 1 · 2016-02-08T22:09:01

#x in [-5, 1]#

#|x+2|<=3#

We must divide the equation in two parts:

If #x+2>=0#:

The module of a positive number is the same number, the equation gives:

#x+2<=3# and #x+2>=0#:
#x<=1# and #x>=-2#:

#x in [-2, 1]#

If #x+2<=0#:

The module of a negative number is its inverse , the equation gives:

#-(x+2)<=3 # and #x+2<=0#:

#-x-2<=3 # and #x<=-2#:

#-x<=5 # and #x<=-2#:

Now we must multiply x by -1. This will change the comparision sign.

#x>=-5 # and #x<=-2#,

#x in [-5, -2]#

The solution will be the union of the two intervals:

#x in [-5, 1]#