How do you solve #abs(2x)<2+ abs3#?

1 Answer
May 26, 2015

First of all, simplify the inequality.

Since #|3|=3#, we can rewrite this inequality as
#|2x| < 5#

Both sides of an inequality can be divided by a positive constant without changing the sign of inequality, and the new inequality will be equivalent to an old one.
Let's divide out equation by #2#. The result is:
#|x| < 2.5#

The solution to this inequality is, obviously,
#-2.5 < x < 2.5#

The above solution can be obtained using the following logic:
By definition, #|x|# is defined as
#|x| = x# for all #x>=0# and
#|x| = -x# for all #x<0#.
Therefore, we can look for solutions of our inequality for #x>=0# and, separately, for #x<0#.

Case 1. Assume #x>=0#.
Then #|x|=x# and our inequality looks like #x < 2.5#.
Combining this with a requirement #x>=0#, we have the following solutions:
#0<=x<2.5#

Case 2. Assume #x < 0#.
Then #|x|=-x# and our inequality looks like #-x < 2.5# or #x > -2.5#..
Combining this with a requirement #x < 0#, we have the following solutions:
#-2.5 < x < 0#

Combining two areas that represent solutions,
#0 <= x < 2.5# and #-2.5 < x < 0#,
we obtain one interval for #x# - the solution to our inequality:
#-2.5 < x < 2.5#.