Solve the following inequations:
(a) `x + 5 lt 7, x in R`
(b) `4x -3 ge 17, x in Z`
(c) `3x -2 lt 1, x in N`

Explanation

(a) Subtracting 5 from both sides, we get
`x + 5 - 5 lt 7 - 5 rArr x lt 2`
The set of all the number which are less than 2 is the solution set of the given inequation
(b) Adding 3 on both sides, we get
`4x - 3 + 3 ge 17 + 3`
`4x ge 20`
Dividing both sides by 4, we get `(4x)/(4) ge (20)/(4) rArr x ge 5`
The set of all the integers which are greater than or equal to 5 is the solution set of the given inequation.
(c) Adding 2 on both sides of `3x - 2 + 2 lt 1 + 2`, we get `3x lt 3`
Dividing both sides by 3, we get `(3x)/(3) lt (3)/(3) rArr x lt (3)/(3) rArr x lt 1`
There is no natural number which is less than 1
`:.` There is no solution for the given inequation in the set of natural numbers