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

1 Answer
Apr 27, 2017

Answer:

#(-\infty,-9)\cup (15,\infty)#

Explanation:

First, subtract two from both sides to get
#\abs{\frac{x}{3}-1} >4#

The absolute value can be split into
#\frac{x}{3}-1>4#
#-(\frac[x][3]-1)>4#

Simplifying for the first equation we get the following
#\frac{x}{3}>5#
#x>15#

For the second equation
#1-\frac{x}{3}>4#
#-\frac[x][3]>3#
#\frac[x][3]<-3#
#x<-9#

So all values less than -9 and greater than 15 satisfy the inequality. This can be written as #(-\infty,-9)\cup (15,\infty)#.

Open brackets are used since infinity is not included as a point and 9 and 15 do not satisfy the inequality.