Question

Solve $3x+8>2$ when

(i) Solve $3x+8>2$ when
$x$ is an integer

(ii) Solve $3x+8>2$ when
$x$ is a real number

Answer 1

(i) $x$ is an integer

Given : $3x+8>2$
$\Rightarrow 3x>2-8$
$\Rightarrow 3x>-6$
$\Rightarrow x>\frac{-6}{3}$
$\Rightarrow x>-2$
Since $x$ is an integer. i.e, $x$ can be $...,-2,-1,0,1,2,3,\mathrm{4...}$

We need to find values of $x$ which is greater than $-2$.
i.e., $x$ can be $-1,0,1,2,3,4,...=\{-1,0,1,2,3,\mathrm{4....}\}$

(ii) $x$ is a real number
Given : $3x+8>2$
$\Rightarrow 3x>2-8$
$\Rightarrow 3x>-6$
$\Rightarrow x>\frac{-6}{3}$
$\Rightarrow x>-2$
Since $x$ is real number which is greater than $-2$.
Thus, $x\in (-2,\infty )$