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

Question

1654 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-02T09:06:27

See below

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)#