Solve 3x + 8 >2, when

(i) x is an integer.

(ii) x is a real number.

Asked by Abhisek | 1 year ago |  50

##### Solution :-

(i) Given that, 3x + 8 > 2

Now by subtracting 8 from both sides we get,

3x + 8 – 8 > 2 – 8

The above inequality becomes,

3x > – 6

Again by dividing both sides by 3 we get,

$$\dfrac{3x}{3}> \dfrac{-6}{3}$$

Hence x > -2

When x is an integer then

It is clear 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.

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

(ii) Given that, 3x + 8 > 2

Now by subtracting 8 from both sides we get,

3x + 8 – 8 > 2 – 8

The above inequality becomes,

3x > – 6

Again by dividing both sides by 3 we get,

$$\dfrac{3x}{3}> \dfrac{-6}{3}$$

Hence x > -2

When x is a real number.

It is clear 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.

Therefore the solution set is x ∈ (-2, ∞)

Answered by Pragya Singh | 1 year ago

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