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Question Number 111611 by mathdave last updated on 04/Sep/20

$$solveforxadyin$$$$5x(1+\frac{1}{x^2+y^2})=12.........\left(1\right)$$$$5y(1-\frac{1}{x^2+y^2})=4..........\left(2\right)$$

Answered by Her_Majesty last updated on 04/Sep/20

$$y=px$$$$\left(1\right)5x^2-12x+\frac{5}{p^2+1}=0$$$$\left(2\right)5x^2-\frac{4}{p}x-\frac{5}{p^2+1}=0$$$$\left(1\right)-\left(2\right)\Rightarrow x=\frac{5p}{2(3p-1)(p^2+1)}$$$$\left(1\right)+\left(2\right)\Rightarrow x=\frac{2(3p+1)}{5p}$$$$\Rightarrow \frac{5p}{2(3p-1)(p^2+1)}=\frac{2(3p+1)}{5p}$$$$p^4+\frac{7}{36}{p}^2-\frac{1}{9}=0$$$$\Rightarrow p=\pm \frac{1}{2}\vee p=\pm \frac{2}{3}\mathrm{i}$$$$\Rightarrow (x;y)\in \{(\frac{2}{5};-\frac{1}{5});(2;1);(\frac{6}{5}\pm \frac{3}{5}\mathrm{i};\frac{2}{5}\mp \frac{4}{5}\mathrm{i})\}$$

Answered by mathdave last updated on 04/Sep/20

$$solutionto$$$$5x(1+\frac{1}{x^2+y^2})=12........\left(1\right)and$$$$5y(1-\frac{1}{x^2+y^2})=4.........\left(2\right)$$$$letdecomposetheequationstomakethemsimilar$$$$thatis$$$$5x(1+\frac{1}{x^2+y^2})=5\bullet (2.4)........\left(3\right)$$$$5y(1-\frac{1}{x^2+y^2})=5\bullet (0.8)..........\left(4\right)$$$$comparingdecomposedequations$$$$5x=5.........\left(5\right)$$$$x=1$$$$substitutethevalueofxin\left(6\right)below$$$$1+\frac{1}{x^2+y^2}=2.4..........\left(6\right)$$$$1+\frac{1}{1+y^2}=2.4\Rightarrow 2+y^2=2.4+2.4y^2$$$$-1.4y^2=0.4$$$$y=0.535i(3d.p)$$$$andfromequation\left(4\right)comparingsides$$$$5y=5........\left(7\right)$$$$y=1$$$$substitutingforthevalueofybelow$$$$1-\frac{1}{x^2+y^2}=0.8.........\left(8\right)$$$$x^2+1-1=0.8x^2+0.8$$$$0.2x^2=0.8$$$$x=2$$$$\because (x_1,y_1)and(x_2,y_2)=(1,0.535i)and(2,1)$$$$bymathdave\left(04/09/2020\right)$$

Commented by Tawa11 last updated on 06/Sep/21

$$\mathrm{great}\mathrm{sir}$$

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