Solve 3x+8>2, when

(i) x is an integer

(ii) x is a real number

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Given Linear inequality: 3x+8>2

The given inequality can also be written as:

3x+8 -8 > 2 -8 …(1)

In the above inequality, -8 is multiplied on both the sides, as it does not change the definition of the given expression.

Now, simplify the expression (1)

⇒ 3x > -6

Now, both the sides, divide it by 3

⇒ 3x/3 > -6/3

⇒ x > -2

(i) x is an integer

Hence, the integers greater than -2 are -1,0,1,2,…etc

Thus, when x is an integer, the solutions of the given inequality are -1,0,1,2,…

Hence, the solution set for the given linear inequality is {-1,0,1,2,…}

(ii) x is a real number

If x is a real number, the solutions of the given inequality are all the real numbers, which

are greater than 2.

Therefore, in the case of x is a real number, the solution set is (-2, ∞)

Thanks