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Question Number 133431 by Algoritm last updated on 22/Feb/21

Answered by benjo_mathlover last updated on 22/Feb/21

$${\mathrm{x}}^2\mathrm{y}-5\mathrm{xy}-\mathrm{y}={\mathrm{x}}^2-3\mathrm{x}-1$$$$(\mathrm{y}-1){\mathrm{x}}^2-(5\mathrm{y}-3)\mathrm{x}+1-\mathrm{y}=0$$$$\mathrm{x}=\frac{5\mathrm{y}-3\pm \sqrt{{(5\mathrm{y}-3)}^2-4(\mathrm{y}-1)(1-\mathrm{y})}}{2(\mathrm{y}-1)}$$$${\mathrm{f}}^{-1}\left(\mathrm{y}\right)=\frac{5\mathrm{y}-3\pm \sqrt{{25\mathrm{y}}^2-30\mathrm{y}+9+{4\mathrm{y}}^2-8\mathrm{y}+4}}{2(\mathrm{y}-1)}$$$${\mathrm{f}}^{-1}\left(\mathrm{y}\right)=\frac{5\mathrm{y}-3\pm \sqrt{{29\mathrm{y}}^2-38\mathrm{y}+13}}{2(\mathrm{y}-1)}$$$${\mathrm{f}}^{-1}\left(\mathrm{x}\right)=\frac{5\mathrm{x}-3\pm \sqrt{{29\mathrm{x}}^2-38\mathrm{x}+13}}{2\mathrm{x}-2}$$

Answered by Dwaipayan Shikari last updated on 22/Feb/21

$$x^2(y-1)-x(5y-3)+1-y=0$$$$\Rightarrow x=\frac{5y-3\pm \sqrt{25y^2+9-30y+4y^2-8y+4}}{2(1-y)}$$$$\Rightarrow x=\frac{5y-3\pm \sqrt{29y^2-38y+13}}{2(y-1)}y=f\left(x\right)$$$$f^{-1}\left(x\right)=\frac{5x-3\pm \sqrt{29x^2-38x+13}}{2(x-1)}$$

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