Q. 44.0( 41 Votes )
Solve 3x + 8 >2, when:
(i) x is an integer. (ii) x is a real number.
Answer :
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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