How do you solve #1/2abs(2x + 1 )> 2 #?

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Answer 1 · 2018-03-11T21:14:08

#x > 3/2 or x < -5/2#

in interval notation

#(-oo, -5/3) uu (3/2, oo)#

#1/2|2x+1|>2#

#|2x+1|>4#

#2x+1>4# or #2x+1<-4# so

#-4>2x+1>4#

#-5>2x>3#

#-5/2>x>3/2#

#x > 3/2 or x < -5/2#

#(-oo, -5/3)uu(3/2, oo)#