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Question Number 115909 by Eric002 last updated on 29/Sep/20

$$solvethesystemofequations$$$$x+\frac{3x-y}{x^2+y^2}=3,y-\frac{x+3y}{x^2+y^2}=0$$

Answered by MJS_new last updated on 29/Sep/20

$$\mathrm{let}y=px$$$$\left(1\right)x+\frac{3-p}{(1+p^2)x}=3$$$$\left(2\right)px-\frac{1+3p}{(1+p^2)x}=0$$$$\left(1\right)x^2=3x+\frac{p-3}{1+p^2}$$$$\left(2\right)x^2=\frac{3p+1}{(1+p^2)p}$$$$\Rightarrow$$$$3x=\frac{3p+1}{(1+p^2)p}-\frac{p-3}{1+p^2}$$$$x=-\frac{p^2-6p-1}{3(1+p^2)p}$$$$\left(1\right)x^2-3x-\frac{p-3}{1+p^2}=0$$$$\Rightarrow$$$$p^4+\frac{21}{26}{p}^3-\frac{7}{26}{p}^2-\frac{3}{26}p-\frac{1}{26}=0$$$$(p+1)(p-\frac{1}{2})(p^2+\frac{4}{13}p+\frac{1}{13})=0$$$$p_1=-1$$$$p_2=\frac{1}{2}$$$$p_{3,4}=-\frac{2}{13}\pm \frac{3}{13}\mathrm{i}$$$$\Rightarrow$$$$x_1=1\wedge y_1=-1$$$$x_2=2\wedge y_2=1$$$$x_3=\frac{3}{2}-\mathrm{i}\wedge y_3=\frac{1}{2}\mathrm{i}$$$$x_4=\frac{3}{2}+\mathrm{i}\wedge y_4=-\frac{1}{2}\mathrm{i}$$

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