How to solve 2 absolute value inequalities #abs(x+1)+abs(x-1)<=2#?

1 Answer
Alan P.
Apr 7, 2015

With respect to the two absolute value expressions
#abs(x+1)# and #abs(x-1)#
there are two values of #x# for which the absolute value operation becomes significant:
#x= -1# and #x=+1#

This gives us three ranges to consider:
(1) #x<=-1#
(2)#-1 < x < +1# and
(3)# x >= +1#

Case 1 : #x <= -1#
Both #(x+1)# and and #(x-1)# are zero or negative
so the original inequality can be re-written as
#-(x+1)-(x-1) <= 2#
#-2x <= 2#
#x >= -1#
The only value that satisfies #(x <= -1)# and #(x>=-1)#
is #(x=-1)#

Case 2: #-1 < x < +1#
Of the two terms involving #x# only #(x-1)# will be negative and the inequality can be re-written as
#(x+1)-(x-1) <= 2#
#2 <= 2#
This is true for all values of #x# in the "Case 2" range.

Case 3: #x>=+1#
Both of the terms involving #x# are positive and the inequality can be re-written as
#(x+1)+(x-1)<=2#
#2x<=2#
#x<=1#

Solution
Combining all three cases we get
#-1 <= x <= +1#
as the solution range.

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