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Question Number 156593 by Ar Brandon last updated on 13/Oct/21

Commented by cortano last updated on 13/Oct/21

$$\mathrm{take}\mathrm{y}=\mathrm{mx}$$

Answered by MJS_new last updated on 13/Oct/21

$$\sqrt{x}(1+\frac{1}{x+y})=2\Rightarrow \frac{1}{x+y}=\frac{2}{\sqrt{x}}-1$$$$\sqrt{y}(1-\frac{1}{x+y})=3\Rightarrow \frac{1}{x+y}=1-\frac{3}{\sqrt{y}}$$$$\frac{2}{\sqrt{x}}-1=1-\frac{3}{\sqrt{y}}\Rightarrow y=\frac{9x}{4{(\sqrt{x}-1)}^2}$$$$\mathrm{insert}\mathrm{in}{1}^{\mathrm{st}}\mathrm{equation}\mathrm{and}\mathrm{transforming}\Rightarrow$$$$x^2-4x^{3/2}+\frac{33}{4}x-\frac{17}{2}{x}^{1/2}+1=0$$$$(x-2x^{1/2}+4)(x-2x^{1/2}+\frac{1}{4})=0$$$$\Rightarrow$$$$x=-2\pm 2\sqrt{3}\mathrm{i}\vee x=\frac{7}{4}\pm \sqrt{3}$$$$\Rightarrow$$$$y=\frac{3}{2}\mp \frac{3\sqrt{3}}{2}\mathrm{i}\vee y=\frac{21}{4}\pm 3\sqrt{3}$$$$\mathrm{testing}\Rightarrow$$$$\mathrm{both}\mathrm{complex}\mathrm{solutions}\mathrm{hold}$$$$\mathrm{only}\mathrm{one}\mathrm{real}\mathrm{solution}x=\frac{7}{4}+\sqrt{3}\wedge y=\frac{21}{4}+3\sqrt{3}$$

Commented by Ar Brandon last updated on 13/Oct/21

$$\mathrm{Thank}\mathrm{you}\mathrm{Sir}$$

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