# Solve 3x + 8 > 2, when

Question:

Solve $3 x+8>2$, when

(i) x is an integer

(ii) x is a real number

Solution:

The given inequality is $3 x+8>2$.

$3 x+8>2$

$\Rightarrow 3 x+8-8>2-8$

$\Rightarrow 3 x>-6$

$\Rightarrow \frac{3 x}{3}>\frac{-6}{3}$

$\Rightarrow x>-2$

(i) The integers greater than $-2$ are $-1,0,1,2, \ldots$

Thus, when $x$ is an integer, the solutions of the given inequality are

$-1,0,1,2 \ldots$

Hence, in this case, the solution set is $\{-1,0,1,2, \ldots\}$.

(ii) When $x$ is a real number, the solutions of the given inequality are all the real numbers, which are greater than $-2$.

Thus, in this case, the solution set is $(-2, \infty)$.