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Question Number 154986 by mathdanisur last updated on 23/Sep/21

$$\mathrm{Solve}\mathrm{the}\mathrm{system}\mathrm{in}\mathbb{R}$$$$2\mathrm{x}=\frac{{\mathrm{y}}^2-4\mathrm{y}+1}{{\mathrm{y}}^2-\mathrm{y}+1}$$$$\mathrm{y}=\frac{-{\mathrm{x}}^2+6\mathrm{x}-1}{{3\mathrm{x}}^2-2\mathrm{x}+3}$$

Answered by Rasheed.Sindhi last updated on 25/Sep/21

$$2\mathrm{x}=\frac{{\mathrm{y}}^2-4\mathrm{y}+1}{{\mathrm{y}}^2-\mathrm{y}+1};\mathrm{y}=\frac{-{\mathrm{x}}^2+6\mathrm{x}-1}{{3\mathrm{x}}^2-2\mathrm{x}+3}$$$$2\mathrm{x}-1=\frac{{\mathrm{y}}^2-4\mathrm{y}+1}{{\mathrm{y}}^2-\mathrm{y}+1}-1$$$$=\frac{-3\mathrm{y}}{{\mathrm{y}}^2-\mathrm{y}+1}$$$$\frac{1}{2\mathrm{x}-1}-1=\frac{{\mathrm{y}}^2-\mathrm{y}+1}{-3\mathrm{y}}-1$$$$\frac{1-2\mathrm{x}+1}{2\mathrm{x}-1}=\frac{{\mathrm{y}}^2+2\mathrm{y}+1}{-3\mathrm{y}}$$$$\frac{2-2\mathrm{x}}{2\mathrm{x}-1}=\frac{{(\mathrm{y}+1)}^2}{-3\mathrm{y}}=\frac{{(\frac{-{\mathrm{x}}^2+6\mathrm{x}-1}{{3\mathrm{x}}^2-2\mathrm{x}+3}+1)}^2}{-3\left(\frac{-{\mathrm{x}}^2+6\mathrm{x}-1}{{3\mathrm{x}}^2-2\mathrm{x}+3}\right)}$$$$\frac{2\mathrm{x}-2}{2\mathrm{x}-1}=\frac{{\left(\frac{{2\mathrm{x}}^2+4\mathrm{x}+2}{{3\mathrm{x}}^2-2\mathrm{x}+3}\right)}^2}{\frac{3(-{\mathrm{x}}^2+6\mathrm{x}-1)}{{3\mathrm{x}}^2-2\mathrm{x}+3}}$$$$=\frac{4{(\mathrm{x}+1)}^4}{{({3\mathrm{x}}^2-2\mathrm{x}+3)}^2}\times \frac{{3\mathrm{x}}^2-2\mathrm{x}+3}{3(-{\mathrm{x}}^2+6\mathrm{x}-1)}$$$$\frac{\mathrm{x}-1}{2\mathrm{x}-1}=\frac{2{(\mathrm{x}+1)}^4}{3({3\mathrm{x}}^2-2\mathrm{x}+3)(-{\mathrm{x}}^2+6\mathrm{x}-1)}$$$$▸3(\mathrm{x}-1)({3\mathrm{x}}^2-2\mathrm{x}+3)(-{\mathrm{x}}^2+6\mathrm{x}-1)$$$$=2(2\mathrm{x}-1){(\mathrm{x}+1)}^4$$$$▸{9\mathrm{x}}^5-{69\mathrm{x}}^4+{114\mathrm{x}}^3-{114\mathrm{x}}^2+69\mathrm{x}-9$$$$+{4\mathrm{x}}^5+{14\mathrm{x}}^4+{16\mathrm{x}}^3+{4\mathrm{x}}^2-4\mathrm{x}-2=0$$$$▸{13\mathrm{x}}^5-{55\mathrm{x}}^4+{130\mathrm{x}}^3-{110\mathrm{x}}^2+65\mathrm{x}-11=0$$$$Nofactors$$$$Numericalmethodsbeapplied.$$

Commented by mathdanisur last updated on 25/Sep/21

$$\mathrm{My}\mathrm{dear}\mathbf{S}\mathrm{er}$$$$\mathbf{x}=\frac{\mathrm{u}-1}{\mathrm{u}+1}\mathrm{and}\mathbf{y}=\frac{\mathrm{v}-1}{\mathrm{v}+1}$$

Answered by Rasheed.Sindhi last updated on 24/Sep/21

$$2\mathrm{x}=\frac{{\mathrm{y}}^2-4\mathrm{y}+1}{{\mathrm{y}}^2-\mathrm{y}+1},\mathrm{y}=\frac{-{\mathrm{x}}^2+6\mathrm{x}-1}{{3\mathrm{x}}^2-2\mathrm{x}+3}$$$$Let\mathrm{x}=\mathrm{k}:$$$$2\mathrm{k}({\mathrm{y}}^2-\mathrm{y}+1)={\mathrm{y}}^2-4\mathrm{y}+1$$$$(2\mathrm{k}-1){\mathrm{y}}^2-2(\mathrm{k}-2)\mathrm{y}+2\mathrm{k}-1=0$$$$\mathrm{y}=\frac{2(\mathrm{k}-2)\pm \sqrt{4{(\mathrm{k}-2)}^2-4{(2\mathrm{k}-1)}^2}}{2(2\mathrm{k}-1)}$$$$\mathrm{y}=\frac{2(\mathrm{k}-2)\pm \sqrt{{4\mathrm{k}}^2-16\mathrm{k}+16-{16\mathrm{k}}^2+16\mathrm{k}+4}}{2(2\mathrm{k}-1)}$$$$\mathrm{y}=\frac{2(\mathrm{k}-2)\pm \sqrt{-{12\mathrm{k}}^2+20}}{2(2\mathrm{k}-1)}$$$$\mathrm{y}=\frac{2(\mathrm{k}-2)\pm 2\sqrt{5-{3\mathrm{k}}^2}}{2(2\mathrm{k}-1)}$$$$\mathrm{y}=\frac{(\mathrm{k}-2)\pm \sqrt{5-{3\mathrm{k}}^2}}{2\mathrm{k}-1}$$$$5-{3\mathrm{k}}^2⩾0$$$${\mathrm{k}}^2⩽\frac{5}{3}\Rightarrow {\mathrm{x}}^2⩽\frac{5}{3}\Rightarrow \mathrm{x}⩽\pm \sqrt{\frac{5}{3}}$$$$\mathrm{x}⩽-\sqrt{\frac{5}{3}}\vee \mathrm{x}⩽\sqrt{\frac{5}{3}}$$$$\mathrm{Continue}$$

Commented by mathdanisur last updated on 24/Sep/21

$$\mathrm{Thank}\mathrm{you},\mathrm{answer}\mathrm{how}\mathbf{S}\mathrm{er}$$

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