Question

Solve the following system of inequality:
$\frac{x}{3}-\frac{4}{3}⩽\frac{4}{x},\frac{1}{x}>-1,x^2+3x+1>0$

Answer 1

$\frac{x}{3}-\frac{4}{3}\le \frac{4}{x},\frac{1}{x}>-1,x^2+3x+1>0$

$\cdot x^2-4x\le 12$

$\Rightarrow (x-2)(x+6)\le 0$-Equation 1

$\cdot x\ne 0$-Equation 2

$\cdot x^2+3x+1>0$

$\Rightarrow (x-\frac{-3\pm \sqrt{9-4}}{2})>0$

$\Rightarrow (x+\frac{3}{2}-\frac{\sqrt{5}}{2})(x+\frac{3}{2}-\frac{\sqrt{5}}{2})>0$-Equation 3

$\cdot \frac{1}{x}>-1$

$\Rightarrow 1>-x$

$\Rightarrow x>-1$-Eqaution 4

From equation $1$, $2$, $3$ &$4$and figure we get,

$\Rightarrow x\in (\frac{-3}{2}+\frac{\sqrt{5}}{2},2]-0$.