How do you solve #abs(2x-1)<3x+3#?

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Answer 1 · 2018-06-30T14:32:42

The solution is #x in (-2/5, +oo)#

This is an inequality with absolute values, there are #2# cases to consider in the intervals

#(-oo,1/2)# and #(1/2,+oo)#

In the interval #(-oo,1/2)#

#(2x-1-3x-3<0)#

#=>#, #x> -4#

This point #!in# in the interval

In the interval #(1, +oo/2)#

#(-2x+1-3x-3<0):}#

#=>#, #5x> -2#

#=>#, #x> -2/5#

The solution is #x in (-2/5, +oo)#

graph{|2x-1|-3x-3 [-8.08, 11.92, -7.52, 2.48]}