How do you solve #| 2x+6| +4x< 3#?

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Answer 1 · 2017-04-20T19:36:02

#-11/2 < x < -1/2#

#abs(2x+6)+4x-3 < 0#

or

#abs(2x+6)+2(2x+6)-15 < 0#

or for #x ne -3# we have

#1+2(2x+6)/abs(2x+6)-15/abs(2x+6) < 0#

or

#1 pm 2 < 15/abs(2x+6)#

then

#max(-1,3) < 15/abs(2x+6)# or

#abs(2x+6) < 15/3=5#

and finally

#-5 < 2x+6 < 5#

with

#-11/2 < x < -1/2#