How do you solve #abs(4x+2)<6#?

1 Answer
Jul 2, 2018

Answer:

See below

Explanation:

Use definition of absolute value

#absx=x# if #x>=0# and #absx=-x# if x<0

In our case

first case: #abs(4x+2)=4x+2# if #4x+2>=0# but this last it's the same that: #4x>=-2# or #x>=-1/2#

If #x>=-1/2# then #abs(4x+2)=4x+2<6# or #4x<4# or #x<1#

Then #abs(4x+2)<6# if #x in [-1/2,1]#

Second case: #abs(4x+2)=-(4x+2)# if #4x+2<0#. But this it's the same that #x<-1/2#

If #x<-1/2# then #abs(4x+2)=-(4x+2)=-4x-2<6# or #-8<4x# or #-2< x#

then #abs(4x+2)<6# if #x in (-2, -1/2)#