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.


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