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Question Number 158114 by cortano last updated on 31/Oct/21

{ ((x^2 −3xy+2y^2 =35)),((x^2 +y^2 = 13)) :} ⇒x=? ∧ y=?

$$\:\begin{cases}{{x}^{\mathrm{2}} −\mathrm{3}{xy}+\mathrm{2}{y}^{\mathrm{2}} =\mathrm{35}}\\{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\:\mathrm{13}}\end{cases} \\ $$$$\:\Rightarrow{x}=?\:\wedge\:{y}=?\: \\ $$

Commented by bobhans last updated on 31/Oct/21

(x−y)^2 +y^2 −xy=35 (x−y)^2 +2xy=13 ⇒y^2 −3xy=22 ; x=((y^2 −22)/(3y)) ⇒(((y^2 −22)/(3y)))^2 +y^2 =13 ⇒10y^4 −161y^2 +484=0 ⇒(y^2 −4)(10y^2 −121)=0 ⇒ { ((y^2 =4⇒ { ((y=−2⇒x=((4−22)/(−6))=3)),((y=2⇒x=((4−22)/6)=−3)) :})),((y^2 =((121)/(10))⇒ { ((y=−((11)/( (√(10))))⇒x=((((121)/(10))−22)/(−((33)/( (√(10))))))=(3/( (√(10)))))),((y=((11)/( (√(10))))⇒x=((((121)/(10))−22)/((33)/( (√(10)))))=−(3/( (√(10)))))) :})) :} (x,y)={(±3,∓2),(±(3/( (√(10)))),∓((11)/( (√(10)))))}

$$\:\left(\mathrm{x}−\mathrm{y}\right)^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\mathrm{xy}=\mathrm{35} \\ $$$$\:\left(\mathrm{x}−\mathrm{y}\right)^{\mathrm{2}} +\mathrm{2xy}=\mathrm{13} \\ $$$$\Rightarrow\mathrm{y}^{\mathrm{2}} −\mathrm{3xy}=\mathrm{22}\:;\:\mathrm{x}=\frac{\mathrm{y}^{\mathrm{2}} −\mathrm{22}}{\mathrm{3y}} \\ $$$$\Rightarrow\left(\frac{\mathrm{y}^{\mathrm{2}} −\mathrm{22}}{\mathrm{3y}}\right)^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} =\mathrm{13} \\ $$$$\Rightarrow\mathrm{10y}^{\mathrm{4}} −\mathrm{161y}^{\mathrm{2}} +\mathrm{484}=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{y}^{\mathrm{2}} −\mathrm{4}\right)\left(\mathrm{10y}^{\mathrm{2}} −\mathrm{121}\right)=\mathrm{0} \\ $$$$\Rightarrow\begin{cases}{\mathrm{y}^{\mathrm{2}} =\mathrm{4}\Rightarrow\begin{cases}{\mathrm{y}=−\mathrm{2}\Rightarrow\mathrm{x}=\frac{\mathrm{4}−\mathrm{22}}{−\mathrm{6}}=\mathrm{3}}\\{\mathrm{y}=\mathrm{2}\Rightarrow\mathrm{x}=\frac{\mathrm{4}−\mathrm{22}}{\mathrm{6}}=−\mathrm{3}}\end{cases}}\\{\mathrm{y}^{\mathrm{2}} =\frac{\mathrm{121}}{\mathrm{10}}\Rightarrow\begin{cases}{\mathrm{y}=−\frac{\mathrm{11}}{\:\sqrt{\mathrm{10}}}\Rightarrow\mathrm{x}=\frac{\frac{\mathrm{121}}{\mathrm{10}}−\mathrm{22}}{−\frac{\mathrm{33}}{\:\sqrt{\mathrm{10}}}}=\frac{\mathrm{3}}{\:\sqrt{\mathrm{10}}}}\\{\mathrm{y}=\frac{\mathrm{11}}{\:\sqrt{\mathrm{10}}}\Rightarrow\mathrm{x}=\frac{\frac{\mathrm{121}}{\mathrm{10}}−\mathrm{22}}{\frac{\mathrm{33}}{\:\sqrt{\mathrm{10}}}}=−\frac{\mathrm{3}}{\:\sqrt{\mathrm{10}}}}\end{cases}}\end{cases} \\ $$$$\:\left(\mathrm{x},\mathrm{y}\right)=\left\{\left(\pm\mathrm{3},\mp\mathrm{2}\right),\left(\pm\frac{\mathrm{3}}{\:\sqrt{\mathrm{10}}},\mp\frac{\mathrm{11}}{\:\sqrt{\mathrm{10}}}\right)\right\} \\ $$

Commented by cortano last updated on 31/Oct/21

the solution is (±3,∓2) ,(±(3/( (√(10)))),∓((11)/( (√(10)))))

$${the}\:{solution}\:{is}\:\left(\pm\mathrm{3},\mp\mathrm{2}\right)\:,\left(\pm\frac{\mathrm{3}}{\:\sqrt{\mathrm{10}}},\mp\frac{\mathrm{11}}{\:\sqrt{\mathrm{10}}}\right) \\ $$

Answered by ajfour last updated on 31/Oct/21

(x/y)=t t^2 −3t+2=((35)/y^2 ) t^2 +1=((13)/y^2 ) ⇒ 13t^2 −39t+26=35(t^2 +1) ⇒ 22t^2 +39t+9=0 ⇒ t=−((39)/(44))±(√((39^2 −9×88)/(44^2 ))) ⇒ t=((−39±27)/(44))=−(3/2), −(3/(11)) ⇒ y^2 =((13)/(t^2 +1)) x^2 =t^2 y^2 =13−y^2 ⇒ x^2 =13−((13)/(t^2 +1)) =((13t^2 )/(t^2 +1)) .

$$\frac{{x}}{{y}}={t} \\ $$$${t}^{\mathrm{2}} −\mathrm{3}{t}+\mathrm{2}=\frac{\mathrm{35}}{{y}^{\mathrm{2}} } \\ $$$${t}^{\mathrm{2}} +\mathrm{1}=\frac{\mathrm{13}}{{y}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\:\mathrm{13}{t}^{\mathrm{2}} −\mathrm{39}{t}+\mathrm{26}=\mathrm{35}\left({t}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\Rightarrow\:\:\mathrm{22}{t}^{\mathrm{2}} +\mathrm{39}{t}+\mathrm{9}=\mathrm{0} \\ $$$$\Rightarrow\:\:{t}=−\frac{\mathrm{39}}{\mathrm{44}}\pm\sqrt{\frac{\mathrm{39}^{\mathrm{2}} −\mathrm{9}×\mathrm{88}}{\mathrm{44}^{\mathrm{2}} }} \\ $$$$\Rightarrow\:\:{t}=\frac{−\mathrm{39}\pm\mathrm{27}}{\mathrm{44}}=−\frac{\mathrm{3}}{\mathrm{2}},\:−\frac{\mathrm{3}}{\mathrm{11}} \\ $$$$\Rightarrow\:\:{y}^{\mathrm{2}} =\frac{\mathrm{13}}{{t}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\:\:\:{x}^{\mathrm{2}} ={t}^{\mathrm{2}} {y}^{\mathrm{2}} =\mathrm{13}−{y}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:{x}^{\mathrm{2}} =\mathrm{13}−\frac{\mathrm{13}}{{t}^{\mathrm{2}} +\mathrm{1}}\:=\frac{\mathrm{13}{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} +\mathrm{1}}\:\:\:. \\ $$

Answered by Rasheed.Sindhi last updated on 31/Oct/21

{ ((x^2 −3xy+2y^2 =35)),((x^2 +y^2 = 13)) :} { ((13x^2 −39xy+26y^2 =455)),((35x^2 +35y^2 =455)) :} 22x^2 +39xy+9y^2 =0 (2x+3y)(11x+3y)=0 { ((2x+3y=0⇒x=−((3y)/2))),((11x+3y=0⇒x=−((3y)/(11)))) :} x=−((3y)/2): x^2 +y^2 =13⇒(−((3y)/2))^2 +y^2 =13 9y^2 +4y^2 =52⇒13y^2 =52⇒y=±2 x=−((3(±2))/2)=∓3 (x,y)=(3,−2)(−3,2) x=−((3y)/(11)):(−((3y)/(11)))^2 +y^2 =13 ⇒9y^2 +121y^2 =1573 130y^2 =1573⇒y^2 =((121)/(10)) ⇒y=±((11(√(10)))/( 10)) x=−((3y)/(11))=−(3/(11))(±((11(√(10)))/( 10)))=∓((3(√(10)))/(10)) (x,y)=(((3(√(10)))/(10)),−((11(√(10)))/( 10))),(−((3(√(10)))/(10)),((11(√(10)))/( 10))) determinant ((((x,y)=(3,−2)(−3,2),(((3(√(10)))/(10)),−((11(√(10)))/( 10))),(−((3(√(10)))/(10)),((11(√(10)))/( 10))))))

$$\:\begin{cases}{{x}^{\mathrm{2}} −\mathrm{3}{xy}+\mathrm{2}{y}^{\mathrm{2}} =\mathrm{35}}\\{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\:\mathrm{13}}\end{cases} \\ $$$$\begin{cases}{\mathrm{13}{x}^{\mathrm{2}} −\mathrm{39}{xy}+\mathrm{26}{y}^{\mathrm{2}} =\mathrm{455}}\\{\mathrm{35}{x}^{\mathrm{2}} +\mathrm{35}{y}^{\mathrm{2}} =\mathrm{455}}\end{cases} \\ $$$$\mathrm{22}{x}^{\mathrm{2}} +\mathrm{39}{xy}+\mathrm{9}{y}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left(\mathrm{2}{x}+\mathrm{3}{y}\right)\left(\mathrm{11}{x}+\mathrm{3}{y}\right)=\mathrm{0} \\ $$$$\begin{cases}{\mathrm{2}{x}+\mathrm{3}{y}=\mathrm{0}\Rightarrow{x}=−\frac{\mathrm{3}{y}}{\mathrm{2}}}\\{\mathrm{11}{x}+\mathrm{3}{y}=\mathrm{0}\Rightarrow{x}=−\frac{\mathrm{3}{y}}{\mathrm{11}}}\end{cases} \\ $$$${x}=−\frac{\mathrm{3}{y}}{\mathrm{2}}:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{13}\Rightarrow\left(−\frac{\mathrm{3}{y}}{\mathrm{2}}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{13} \\ $$$$\mathrm{9}{y}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} =\mathrm{52}\Rightarrow\mathrm{13}{y}^{\mathrm{2}} =\mathrm{52}\Rightarrow{y}=\pm\mathrm{2} \\ $$$${x}=−\frac{\mathrm{3}\left(\pm\mathrm{2}\right)}{\mathrm{2}}=\mp\mathrm{3} \\ $$$$\left({x},{y}\right)=\left(\mathrm{3},−\mathrm{2}\right)\left(−\mathrm{3},\mathrm{2}\right) \\ $$$${x}=−\frac{\mathrm{3}{y}}{\mathrm{11}}:\left(−\frac{\mathrm{3}{y}}{\mathrm{11}}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{13} \\ $$$$\Rightarrow\mathrm{9}{y}^{\mathrm{2}} +\mathrm{121}{y}^{\mathrm{2}} =\mathrm{1573} \\ $$$$\:\:\:\:\:\mathrm{130}{y}^{\mathrm{2}} =\mathrm{1573}\Rightarrow{y}^{\mathrm{2}} =\frac{\mathrm{121}}{\mathrm{10}} \\ $$$$\Rightarrow{y}=\pm\frac{\mathrm{11}\sqrt{\mathrm{10}}}{\:\mathrm{10}} \\ $$$${x}=−\frac{\mathrm{3}{y}}{\mathrm{11}}=−\frac{\mathrm{3}}{\mathrm{11}}\left(\pm\frac{\mathrm{11}\sqrt{\mathrm{10}}}{\:\mathrm{10}}\right)=\mp\frac{\mathrm{3}\sqrt{\mathrm{10}}}{\mathrm{10}} \\ $$$$\left({x},{y}\right)=\left(\frac{\mathrm{3}\sqrt{\mathrm{10}}}{\mathrm{10}},−\frac{\mathrm{11}\sqrt{\mathrm{10}}}{\:\mathrm{10}}\right),\left(−\frac{\mathrm{3}\sqrt{\mathrm{10}}}{\mathrm{10}},\frac{\mathrm{11}\sqrt{\mathrm{10}}}{\:\mathrm{10}}\right) \\ $$$$\begin{array}{|c|}{\left({x},{y}\right)=\left(\mathrm{3},−\mathrm{2}\right)\left(−\mathrm{3},\mathrm{2}\right),\left(\frac{\mathrm{3}\sqrt{\mathrm{10}}}{\mathrm{10}},−\frac{\mathrm{11}\sqrt{\mathrm{10}}}{\:\mathrm{10}}\right),\left(−\frac{\mathrm{3}\sqrt{\mathrm{10}}}{\mathrm{10}},\frac{\mathrm{11}\sqrt{\mathrm{10}}}{\:\mathrm{10}}\right)}\\\hline\end{array} \\ $$

Commented by Rasheed.Sindhi last updated on 31/Oct/21

Corrected! Only values of x & y were exchanged.Sorry for this mistake.

$${Corrected}!\:{Only}\:{values}\:{of}\:{x}\:\&\:{y}\:{were} \\ $$$${exchanged}.{Sorry}\:{for}\:{this}\:{mistake}. \\ $$

Commented by cortano last updated on 31/Oct/21

ok

$${ok} \\ $$

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