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Question Number 173978 by dragan91 last updated on 22/Jul/22

$$\mathrm{Solve}\mathrm{system}\mathrm{of}\mathrm{equations}:$$$$\mathrm{x}+\frac{3\mathrm{x}-\mathrm{y}}{{\mathrm{x}}^2+{\mathrm{y}}^2}=3$$$$\mathrm{y}-\frac{\mathrm{x}+3\mathrm{y}}{{\mathrm{x}}^2+{\mathrm{y}}^2}=0$$$$$$

Answered by Frix last updated on 22/Jul/22

$$y=ax$$$$\left(\mathrm{i}\right)x+\frac{3x-ax}{x^2+a^2{x}^2}=3\iff x+\frac{3-a}{(a^2+1)x}=3$$$$\left(\mathrm{j}\right)ax-\frac{3+3ax}{x^2+a^2{x}^2}=0\iff ax-\frac{3a+1}{(a^2+1)x}=0$$$$\left(\mathrm{i}\right)x^2-3x-\frac{a-3}{a^2+1}=0$$$$\left(\mathrm{j}\right)x^2-\frac{3a+1}{a(a^2+1)}=0$$$$(\mathrm{j}-\mathrm{i})\Rightarrow x=-\frac{a^2-6a-1}{3a(a^2+1)}$$$$\mathrm{inserting}\mathrm{in}\left(\mathrm{i}\right)\mathrm{or}\left(\mathrm{j}\right)\mathrm{and}\mathrm{transforming}\mathrm{we}\mathrm{get}$$$$a^4+\frac{21a^3}{26}-\frac{7a^2}{26}-\frac{3a}{26}-\frac{1}{26}=0$$$$(a+1)(a-\frac{1}{2})(a^2+\frac{4a}{13}+\frac{1}{13})=0$$$$a=-1\vee a=\frac{1}{2}\vee a=-\frac{2\pm 3\mathrm{i}}{13}$$$$\mathrm{we}\mathrm{get}$$$$x=1\wedge y=-1\vee x=2\wedge y=1\vee x=\frac{3}{2}\pm \mathrm{i}\wedge y=\mp \frac{i}{2}\vee$$

Commented by Tawa11 last updated on 22/Jul/22

$$\mathrm{Great}\mathrm{sir}$$

Commented by dragan91 last updated on 22/Jul/22

$$y=ax$$$$\left(\mathrm{i}\right)x+\frac{3x-ax}{x^2+a^2{x}^2}=3\iff x+\frac{3-a}{(a^2+1)x}=3$$$$\left(\mathrm{j}\right)ax-\frac{x+3ax}{x^2+a^2{x}^2}=0\iff ax-\frac{3a+1}{(a^2+1)x}=0$$$$\left(\mathrm{i}\right)x^2-3x-\frac{a-3}{a^2+1}=0$$$$\left(\mathrm{j}\right)x^2-\frac{3a+1}{a(a^2+1)}=0$$$$(\mathrm{j}-\mathrm{i})\Rightarrow x=-\frac{a^2-6a-1}{3a(a^2+1)}$$$$\mathrm{inserting}\mathrm{in}\left(\mathrm{i}\right)\mathrm{or}\left(\mathrm{j}\right)\mathrm{and}\mathrm{transforming}\mathrm{we}\mathrm{get}$$$$a^4+\frac{21a^3}{26}-\frac{7a^2}{26}-\frac{3a}{26}-\frac{1}{26}=0$$$$(a+1)(a-\frac{1}{2})(a^2+\frac{4a}{13}+\frac{1}{13})=0$$$$a=-1\vee a=\frac{1}{2}\vee a=-\frac{2\pm 3\mathrm{i}}{13}$$$$\mathrm{we}\mathrm{get}$$$$x=1\wedge y=-1\vee x=2\wedge y=1\vee x=\frac{3}{2}\pm \mathrm{i}\wedge y=\mp \frac{i}{2}\vee$$

Commented by dragan91 last updated on 22/Jul/22

$$\mathrm{By}\mathrm{the}\mathrm{way}\mathrm{nice}\mathrm{solution}$$

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