Algebra

Find all real x for

        I                II

$\color{blue}2 \cdot \frac{x - 5}{x - 3} > \frac{2x - 5}{x + 2} + 20\\ .\\ f(x)=2 \cdot \frac{x - 5}{x - 3} - \frac{2x - 5}{x + 2} -20>0\\ 2(x-5)(x+2)-(2x-5)(x-3)-20(x-3)(x+2)>0\\ (2x-10)(x+2)-(2x^2-6x-5x+15)-20(x^2+2x-3x-6)>0$

$2x^2+4x-10x-20-2x^2+11x-15-20x^2-40x+60x+120>0\\ f(x)=-20x^2+25x+85>0$

$x_{1,2} = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x_{1,2} = {-25 \pm \sqrt{25^2+4\cdot 20\cdot 85} \over -40}\\ x_0\in \{-1.5292,2.7792\}\\$

$The\ extreme\ points\ of\ the\ hyperbolas\ I\ and\ II\ are\ x_I=3\ and\ x_{II}=-2.$

$In\ the\ calculated\ range\ is\\ (2 \cdot \frac{x - 5}{x - 3} ){\color{red}\ <}\ (\frac{2x - 5}{x + 2} + 20)\\ x\in \mathbb R\ (-1.52921099245< x< 2.77921099245)\\$

$The\ extreme\ points\ of\ the\ hyperbolas\ I\ and\ II\ are\ x_I=3\ and\ x_{II}=-2.\\ \color{blue }This\ gives:\\ (2 \cdot \frac{x - 5}{x - 3} )\ {\color{red}>}\ (\frac{2x - 5}{x + 2} + 20)\\ \color{blue}x\in \mathbb R(-2 <-1.5292)\\ and\\ \color{blue}x=\mathbb R(2.7792 <3)$

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$This\ gives:\\ (2 \cdot \frac{x - 5}{x - 3} )\ {\color{red}>}\ (\frac{2x - 5}{x + 2} + 20)$

$x\in \mathbb R$(-2 < x < -1.5292)

and

$x\in \mathbb R$(2.7792 < x < 3)

Sorry. It didn't display correctly in answer 1#.

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