Question

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

Answer 1

`(-\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.