Question

Solution for inequality $\frac{3}{x+2}\ge \frac{1}{x-3}$ is given by
$\left(i\right)\left(-2,3\right)\cup \left[\frac{11}{2},\infty \right)$
$\left(ii\right)\left[-2,3\right]\cup \left(\frac{11}{2},\infty \right)$
$\left(iii\right)\left[-2,3\right]\cup \left[\frac{9}{2},\infty \right)$
$\left(iv\right)\left(-2,3\right)\cup \left(\frac{9}{2},\infty \right)$

Answer 1

$\frac{3}{x+2}\ge \frac{1}{x-3}$

$\Rightarrow \frac{3}{x+2}-\frac{1}{x-3}\ge 0$

$\Rightarrow \frac{3x-9-x-2}{(x+2)(x-3)}\ge 0$

$\Rightarrow \frac{2x-11}{(x+2)(x-3)}\ge 0$

$\Rightarrow 2\frac{x-\frac{11}{2}}{(x+2)(x-3)}\ge 0$

$\Rightarrow \frac{x-\frac{11}{2}}{(x+2)(x-3)}\ge 0$

$\Rightarrow x\in (-2,3)\cup [\frac{11}{2},\infty )$