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Question Number 161159 by daus last updated on 13/Dec/21

Answered by Rasheed.Sindhi last updated on 13/Dec/21

x=((2+y^2 )/y) ;y≠0 (((2+y^2 )/y))^2 −5(((2+y^2 )/y))y+6y^2 =0 (2+y^2 )^2 −10y^2 −5y^4 +6y^4 =0 4+4y^2 +y^4 −10y^2 −5y^4 +6y^4 =0 y^4 −3y^2 +2=0 (y^2 −1)(y^2 −2)=0 y=±1,±(√2) x=((2+y^2 )/y)=((2+(±1)^2 )/(±1))=±3 x=((2+(±(√2))^2 )/(±(√2)))(4/(±(√2))).((±(√2))/( ±(√2)))=±2(√2) (x,y)=(±3,±1),(±2(√2) ,±(√2) )

$${x}=\frac{\mathrm{2}+{y}^{\mathrm{2}} }{{y}}\:;{y}\neq\mathrm{0} \\ $$$$\left(\frac{\mathrm{2}+{y}^{\mathrm{2}} }{{y}}\right)^{\mathrm{2}} −\mathrm{5}\left(\frac{\mathrm{2}+{y}^{\mathrm{2}} }{{y}}\right){y}+\mathrm{6}{y}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left(\mathrm{2}+{y}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{10}{y}^{\mathrm{2}} −\mathrm{5}{y}^{\mathrm{4}} +\mathrm{6}{y}^{\mathrm{4}} =\mathrm{0} \\ $$$$\mathrm{4}+\mathrm{4}{y}^{\mathrm{2}} +{y}^{\mathrm{4}} −\mathrm{10}{y}^{\mathrm{2}} −\mathrm{5}{y}^{\mathrm{4}} +\mathrm{6}{y}^{\mathrm{4}} =\mathrm{0} \\ $$$${y}^{\mathrm{4}} −\mathrm{3}{y}^{\mathrm{2}} +\mathrm{2}=\mathrm{0} \\ $$$$\left({y}^{\mathrm{2}} −\mathrm{1}\right)\left({y}^{\mathrm{2}} −\mathrm{2}\right)=\mathrm{0} \\ $$$${y}=\pm\mathrm{1},\pm\sqrt{\mathrm{2}} \\ $$$${x}=\frac{\mathrm{2}+{y}^{\mathrm{2}} }{{y}}=\frac{\mathrm{2}+\left(\pm\mathrm{1}\right)^{\mathrm{2}} }{\pm\mathrm{1}}=\pm\mathrm{3} \\ $$$${x}=\frac{\mathrm{2}+\left(\pm\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }{\pm\sqrt{\mathrm{2}}}\frac{\mathrm{4}}{\pm\sqrt{\mathrm{2}}}.\frac{\pm\sqrt{\mathrm{2}}}{\:\pm\sqrt{\mathrm{2}}}=\pm\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\left({x},{y}\right)=\left(\pm\mathrm{3},\pm\mathrm{1}\right),\left(\pm\mathrm{2}\sqrt{\mathrm{2}}\:,\pm\sqrt{\mathrm{2}}\:\right) \\ $$

Commented by daus last updated on 13/Dec/21

thanks

$${thanks}\: \\ $$

Answered by Rasheed.Sindhi last updated on 13/Dec/21

x^2 −5xy+6y^2 =0_((i)) ∧ xy−y^2 =2_((ii)) ; x,y=? (i):x^2 −5xy+6y^2 =0 (x−2y)(x−3y)=0 x=2y ∣ x=3y xy−y^2 ^((ii): ) =2⇒ { ((2y^2 −y^2 =2⇒y=±(√2) ...(iii))),((3y^2 −y^2 =2⇒y=±1...(iv))) :} (iii)⇒x=2y=2(±(√2) )=±2(√2) (iv)⇒x=3y=3(±1)=±3 (x,y)=(±2(√2) ,±(√2) ),(±3,±1)

$$\underset{\left({i}\right)} {\underbrace{{x}^{\mathrm{2}} −\mathrm{5}{xy}+\mathrm{6}{y}^{\mathrm{2}} =\mathrm{0}}}\:\wedge\:\underset{\left({ii}\right)} {\underbrace{{xy}−{y}^{\mathrm{2}} =\mathrm{2}}};\:{x},{y}=? \\ $$$$\left({i}\right):{x}^{\mathrm{2}} −\mathrm{5}{xy}+\mathrm{6}{y}^{\mathrm{2}} =\mathrm{0}\: \\ $$$$\:\:\:\:\:\:\:\:\left({x}−\mathrm{2}{y}\right)\left({x}−\mathrm{3}{y}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:{x}=\mathrm{2}{y}\:\mid\:{x}=\mathrm{3}{y} \\ $$$$\overset{\left({ii}\right):\:\:\:\:\:\:\:\:\:\:\:\:} {{xy}−{y}^{\mathrm{2}} }=\mathrm{2}\Rightarrow\begin{cases}{\mathrm{2}{y}^{\mathrm{2}} −{y}^{\mathrm{2}} =\mathrm{2}\Rightarrow{y}=\pm\sqrt{\mathrm{2}}\:...\left({iii}\right)}\\{\mathrm{3}{y}^{\mathrm{2}} −{y}^{\mathrm{2}} =\mathrm{2}\Rightarrow{y}=\pm\mathrm{1}...\left({iv}\right)}\end{cases} \\ $$$$\left({iii}\right)\Rightarrow{x}=\mathrm{2}{y}=\mathrm{2}\left(\pm\sqrt{\mathrm{2}}\:\right)=\pm\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\left({iv}\right)\Rightarrow{x}=\mathrm{3}{y}=\mathrm{3}\left(\pm\mathrm{1}\right)=\pm\mathrm{3} \\ $$$$\left({x},{y}\right)=\left(\pm\mathrm{2}\sqrt{\mathrm{2}}\:,\pm\sqrt{\mathrm{2}}\:\right),\left(\pm\mathrm{3},\pm\mathrm{1}\right) \\ $$

Commented by daus last updated on 13/Dec/21

thanks

$${thanks} \\ $$

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