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Question Number 141372 by iloveisrael last updated on 18/May/21

Answered by EDWIN88 last updated on 18/May/21

(1) xy + (x/y) = 5 (2) xy−(x/y) = (1/5) { ((2xy = ((26)/5)→xy=((13)/5))),((((2x)/y) = ((24)/5)→(x/y) = ((12)/5))) :} ⇒ xy((x/y))= ((13×12)/(25)) ⇒x^2 = ((12×13)/(25)) → { ((x=((2(√(39)))/5)→y=((13)/(5(((2(√(39)))/5))))=((√(39))/6))),((x=−((2(√(39)))/5)→y=−((13)/(5(((2(√(39)))/5))))=−((√(39))/6))) :}

$$\left(\mathrm{1}\right)\:\mathrm{xy}\:+\:\frac{\mathrm{x}}{\mathrm{y}}\:=\:\mathrm{5} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{xy}−\frac{\mathrm{x}}{\mathrm{y}}\:=\:\frac{\mathrm{1}}{\mathrm{5}} \\ $$$$\:\begin{cases}{\mathrm{2xy}\:=\:\frac{\mathrm{26}}{\mathrm{5}}\rightarrow\mathrm{xy}=\frac{\mathrm{13}}{\mathrm{5}}}\\{\frac{\mathrm{2x}}{\mathrm{y}}\:=\:\frac{\mathrm{24}}{\mathrm{5}}\rightarrow\frac{\mathrm{x}}{\mathrm{y}}\:=\:\frac{\mathrm{12}}{\mathrm{5}}}\end{cases} \\ $$$$\Rightarrow\:\mathrm{xy}\left(\frac{\mathrm{x}}{\mathrm{y}}\right)=\:\frac{\mathrm{13}×\mathrm{12}}{\mathrm{25}} \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{2}} \:=\:\frac{\mathrm{12}×\mathrm{13}}{\mathrm{25}}\:\rightarrow\begin{cases}{\mathrm{x}=\frac{\mathrm{2}\sqrt{\mathrm{39}}}{\mathrm{5}}\rightarrow\mathrm{y}=\frac{\mathrm{13}}{\mathrm{5}\left(\frac{\mathrm{2}\sqrt{\mathrm{39}}}{\mathrm{5}}\right)}=\frac{\sqrt{\mathrm{39}}}{\mathrm{6}}}\\{\mathrm{x}=−\frac{\mathrm{2}\sqrt{\mathrm{39}}}{\mathrm{5}}\rightarrow\mathrm{y}=−\frac{\mathrm{13}}{\mathrm{5}\left(\frac{\mathrm{2}\sqrt{\mathrm{39}}}{\mathrm{5}}\right)}=−\frac{\sqrt{\mathrm{39}}}{\mathrm{6}}}\end{cases} \\ $$

Answered by MJS_new last updated on 18/May/21

xy+(x/y)=5 xy−(x/y)=(1/5) y=px px^2 +(1/p)=5 ⇒ x^2 =((5p−1)/p^2 ) px^2 −(1/p)=(1/5) ⇒ x^2 =((p+5)/(5p^2 )) ((5p−1)/p^2 )=((p+5)/(5p^2 )) ⇒ p=(5/(12)) x^2 =((156)/(25)) x=±((2(√(39)))/5)∧y=±((√(39))/6)

$${xy}+\frac{{x}}{{y}}=\mathrm{5} \\ $$$${xy}−\frac{{x}}{{y}}=\frac{\mathrm{1}}{\mathrm{5}} \\ $$$${y}={px} \\ $$$${px}^{\mathrm{2}} +\frac{\mathrm{1}}{{p}}=\mathrm{5}\:\Rightarrow\:{x}^{\mathrm{2}} =\frac{\mathrm{5}{p}−\mathrm{1}}{{p}^{\mathrm{2}} } \\ $$$${px}^{\mathrm{2}} −\frac{\mathrm{1}}{{p}}=\frac{\mathrm{1}}{\mathrm{5}}\:\Rightarrow\:{x}^{\mathrm{2}} =\frac{{p}+\mathrm{5}}{\mathrm{5}{p}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{5}{p}−\mathrm{1}}{{p}^{\mathrm{2}} }=\frac{{p}+\mathrm{5}}{\mathrm{5}{p}^{\mathrm{2}} }\:\Rightarrow\:{p}=\frac{\mathrm{5}}{\mathrm{12}} \\ $$$${x}^{\mathrm{2}} =\frac{\mathrm{156}}{\mathrm{25}} \\ $$$${x}=\pm\frac{\mathrm{2}\sqrt{\mathrm{39}}}{\mathrm{5}}\wedge{y}=\pm\frac{\sqrt{\mathrm{39}}}{\mathrm{6}} \\ $$

Answered by Rasheed.Sindhi last updated on 18/May/21

≾An Alternate Way≿ x(y+(1/y))=5.........(i) x(y−(1/y))=1/5.......(ii) (i)/(ii): ((y^2 +1)/(y^2 −1))=25 25y^2 −25−y^2 −1=0 24y^2 =26⇒y=±(√((13)/(12)))=((±(√(39)))/6) x(((±(√(39)))/6)+(6/(±(√(39)))))=5 x(((39+36)/(±6(√(39)))))=5 x=5×((±6(√(39)))/(75))=((±2(√(39)))/5) (x,y)=(((±2(√(39)))/5),((±(√(39)))/6))

$$\precsim\mathcal{A}{n}\:\mathcal{A}{lternate}\:\mathcal{W}{ay}\succsim \\ $$$${x}\left({y}+\frac{\mathrm{1}}{{y}}\right)=\mathrm{5}.........\left({i}\right) \\ $$$${x}\left({y}−\frac{\mathrm{1}}{{y}}\right)=\mathrm{1}/\mathrm{5}.......\left({ii}\right) \\ $$$$\left({i}\right)/\left({ii}\right):\:\frac{{y}^{\mathrm{2}} +\mathrm{1}}{{y}^{\mathrm{2}} −\mathrm{1}}=\mathrm{25} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{25}{y}^{\mathrm{2}} −\mathrm{25}−{y}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:\:\:\mathrm{24}{y}^{\mathrm{2}} =\mathrm{26}\Rightarrow{y}=\pm\sqrt{\frac{\mathrm{13}}{\mathrm{12}}}=\frac{\pm\sqrt{\mathrm{39}}}{\mathrm{6}} \\ $$$${x}\left(\frac{\pm\sqrt{\mathrm{39}}}{\mathrm{6}}+\frac{\mathrm{6}}{\pm\sqrt{\mathrm{39}}}\right)=\mathrm{5} \\ $$$${x}\left(\frac{\mathrm{39}+\mathrm{36}}{\pm\mathrm{6}\sqrt{\mathrm{39}}}\right)=\mathrm{5} \\ $$$${x}=\mathrm{5}×\frac{\pm\mathrm{6}\sqrt{\mathrm{39}}}{\mathrm{75}}=\frac{\pm\mathrm{2}\sqrt{\mathrm{39}}}{\mathrm{5}} \\ $$$$\left({x},{y}\right)=\left(\frac{\pm\mathrm{2}\sqrt{\mathrm{39}}}{\mathrm{5}},\frac{\pm\sqrt{\mathrm{39}}}{\mathrm{6}}\right) \\ $$

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