How do you solve and write the following in interval notation: #-1/6+|2−x/3|>1/2#?

1 Answer
Apr 19, 2018

Answer:

#x in[-oo,4)andx in(8,+oo]# or #x notin(4,8)#

Explanation:

First we rearrange to get the #abs(f(x))# part on its own by adding #1/6# to both sides.

#abs(2-x/3)>2/3#

Due to the nature of #abs()# we can take the inside to be positive or negative, since it turn either into a positive number.

#2-x/3>2/3# or #-2+x/3>2/3#
#x/3<2-2/3# or #x/3>2/3+2#
#x/3<4/3# or #x/3>8/3#
#x<4# or #x>8#

So, we have #x in[-oo,4)andx in(8,+oo]# or #x notin(4,8)#