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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

take y=mx

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

Answered by MJS_new last updated on 13/Oct/21

(√x)(1+(1/(x+y)))=2 ⇒ (1/(x+y))=(2/( (√x)))−1 (√y)(1−(1/(x+y)))=3 ⇒ (1/(x+y))=1−(3/( (√y))) (2/( (√x)))−1=1−(3/( (√y))) ⇒ y=((9x)/(4((√x)−1)^2 )) insert in 1^(st) equation and transforming ⇒ x^2 −4x^(3/2) +((33)/4)x−((17)/2)x^(1/2) +1=0 (x−2x^(1/2) +4)(x−2x^(1/2) +(1/4))=0 ⇒ x=−2±2(√3)i∨x=(7/4)±(√3) ⇒ y=(3/2)∓((3(√3))/2)i∨y=((21)/4)±3(√3) testing ⇒ both complex solutions hold only one real solution x=(7/4)+(√3)∧y=((21)/4)+3(√3)

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

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

Thank you Sir

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

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