How do you solve #abs(x + 2) +abs(x - 2)= 4#?

When dealing with moduli, it often helps to split into cases at the points where the enclosed value changes sign.

#x+2# changes sign at the point #x=-2#

#x-2# changes sign at the point #x=2#

Case (a): #x < -2#

Multiply both sides of #x < -2# by #-2# and reverse the inequality (because the multiplier is negative) to get:

#-2x > 4# (used below)

If #x < -2#, then #(x + 2) < 0# and #(x - 2) < 0#

#|x+2| = -(x+2) = -x-2#
#|x-2| = -(x-2) = -x+2#

#|x+2|+|x-2| = (-x-2)+(-x+2) = -2x > 4#

Case (b): #-2 <= x <= 2#

#(x+2) >= 0# and #(x-2) <= 0#

#|x+2| = x+2#
#|x-2| = -(x-2) = -x+2#

#|x+2|+|x-2| = (x+2)+(-x+2) = 4#

Case (c): #x > 2#

#(x+2) > 0# and #(x-2) > 0#

#|x+2| = x+2#
#|x-2| = x-2#

#|x+2|+|x-2| = (x+2) + (x-2) = 2x > 4#

In summary, we find that the inequality #|x+2|+|x-2|=4# is satisfied for all #x# in the range #-2 <= x <= 2#.

In general you can perform any of the following operations on an inequality and retain its truth:
(1) Add or subtract the same value on both sides.
(2) Multiply or divide by the same positive value on both sides.
(3) Multiply or divide by the same negative value on both sides and reverse the inequality ( #<# becomes #>#, #>=# becomes #<=#, etc.).