Question

Solve the following system of inequality:
$\frac{1}{3x}<1,x+\frac{4}{3}⩾\frac{4}{3x},9x^2-9x+1<0$

Answer 1

$\frac{1}{3x}<1,x+\frac{4}{3}\ge \frac{4}{3x},9x^2-9x+1<0$

$\Rightarrow \cdot x\ne 0$-Equation 1

$\cdot \frac{1}{3x}<1$

$\Rightarrow x>\frac{1}{3}$-Equation 2

$\cdot x+\frac{4}{3}\ge \frac{4}{3x}$

$\Rightarrow 3x^2+4x-4\ge 0$

$\Rightarrow 3x^2+6x-2x-4\ge 0$

$\Rightarrow (3x-2)(x+2)\ge 0$

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

$\cdot 9x^2-9x+1<0$

$\Rightarrow [x-\frac{9\pm \sqrt{81-36}}{2\left(9\right)})<0$

$\Rightarrow [x-\frac{1}{2}\pm \frac{\sqrt{4}5}{18}]<0$-Equation 4

From Equation 1,2,3 & 4

Range is $x\in \varphi$