Solve as directed: `abs(x-3) lt 11` , in N and in R.

Explanation

`abs(x-3) lt 11`
`rArr -1 lt x - 3 lt 11
`rArr -11 + 3 lt x - 3 + 3 lt 11 + 3`
` rArr -8 lt x lt 14 `
If x `in` N the solution set is S = {1,2,3,4,5,…….12,13} If x `in` R then the solution set is : S -= {x:x `in` R and -8 `lt x lt 14}
=(-8,14)