Solve 3x + 8 >2, when:

(i) x is an integer. (ii) x is a real number.

It is given in the question that,

3x + 8 >2

Subtracting 8 from both sides we get,

3x + 8 – 8 >2 – 8

3x >- 6

Dividing both sides by 3 we get,

x > -2

(i) When x is an integer

It can be clearly observed that the integer number greater than -2 are -1, 0, 1, 2,...

Thus, solution of 3x + 8 >2is -1, 0, 1, 2,… when x is an integer.

∴ {-1, 0, 1, 2,…} is the solution set.

(ii) When x is a real number.

It can be clearly observed that the solutions of 3x + 8 >2 will be given by x >-2 which states that all the real numbers that are greater than -2.

∴ x ∈ (-2, ∞) is the solution set.