Given: #|x + 3| + |x - 2| = 4#
Subtract #|x+3|# from both sides:
#|x - 2| = 4-|x + 3|" [1]"#
Take a minute to look at application of the definition:
#|A| = {(A; A >=0),(-A; A < 0):}#
#|x - 2| = {(x - 2; x -2 >=0),(2-x; x -2 < 0):}#
#|x + 3| = {(x + 3; x +3 >=0),(-x-3; x +3 < 0):}#
Simplifying the inequalities:
#|x - 2| = {(x - 2; x >= 2),(2-x; x < 2):}#
#|x + 3| = {(x + 3; x >=-3),(-x-3; x < -3):}#
Substitute the case #x >=2# into equation [1]:
#x - 2 = 4-(x + 3);x >=2#
#x - 2 = 4-x - 3;x >=2#
#2x = 4 - 3+ 2;x >=2#
#2x = 3;x >=2#
#x = 3/2; x >=2 larr# This solution is outside the domain, therefore, it is invalid.
Substitute the case #-3<=x<2# into equation [1]:
#2-x = 4-(x + 3);-3<=x<2#
#2-x = 4-x - 3;-3<=x<2#
#2 = 1;-3<=x<2 larr# no solution
Substitute the case #x < -3# into equation [1]:
#2-x = 4-(-x-3);x < -3#
#2-x = 4+x+3;x < -3#
#-2x = 4+3-2;x < -3#
#-2x = 5;x < -3#
#x = -5/2;x < -3 larr# This solution is outside the domain, therefore, it is invalid.
The overall answer is that there is no solution.
I checked this with WolframAlpha and it agreed with me.
