How do you solve #absx> x - 1 #?

1 Answer
May 24, 2015

When dealing with moduli it's often useful to split into cases where the enclosed value changes sign. For this example, that means examine the #x < 0# and #x >= 0# cases separately.


Case (a) : #x < 0#

If #x < 0# then #|x| = -x# and the inequality takes the form

#-x > x - 1#

Add #x + 1# to both sides to get:

#2x < 1#

Divide both sides by #2# to get:

#x < 1/2#

Since we're dealing with the case #x < 0# this condition is already satisfied.


Case (b) : #x >= 0#

If #x >= 0# then #|x| = x# and the inequality takes the form

#x > x - 1#

Subtract #x# from both sides to get:

#0 > -1#

...which is always true.


Putting these two cases together, we find that the original inequality is satisfied for all #x in RR#.