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Question-112050

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Question Number 112050 by naka3546 last updated on 05/Sep/20
Answered by ajfour last updated on 05/Sep/20
(x/y)=t ((t^2 −7t−9)/(t^2 +5t+11))=(9/5) t^2 +20t+36=0 (t+2)(t+18)=0 t=(x/y)=−2, −18 now t^2 y^2 −7ty^2 −9y^2 −9=0 y^2 =(9/(t^2 −7t−9)) ; x=ty y_1 ^2 =1 , y_2 ^2 =(1/(49)) (x,y)≡(2,−1); (−2,1); (((18)/7),((−1)/7)); (((−18)/7),(1/7))
$$\frac{{x}}{{y}}={t} \\ $$$$\frac{{t}^{\mathrm{2}} −\mathrm{7}{t}−\mathrm{9}}{{t}^{\mathrm{2}} +\mathrm{5}{t}+\mathrm{11}}=\frac{\mathrm{9}}{\mathrm{5}} \\ $$$${t}^{\mathrm{2}} +\mathrm{20}{t}+\mathrm{36}=\mathrm{0} \\ $$$$\left({t}+\mathrm{2}\right)\left({t}+\mathrm{18}\right)=\mathrm{0}\:\:\: \\ $$$${t}=\frac{{x}}{{y}}=−\mathrm{2},\:−\mathrm{18} \\ $$$${now} \\ $$$${t}^{\mathrm{2}} {y}^{\mathrm{2}} −\mathrm{7}{ty}^{\mathrm{2}} −\mathrm{9}{y}^{\mathrm{2}} −\mathrm{9}=\mathrm{0} \\ $$$${y}^{\mathrm{2}} =\frac{\mathrm{9}}{{t}^{\mathrm{2}} −\mathrm{7}{t}−\mathrm{9}}\:\:;\:\:{x}={ty} \\ $$$${y}_{\mathrm{1}} ^{\mathrm{2}} =\mathrm{1}\:\:,\:\:\:\:\:{y}_{\mathrm{2}} ^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{49}} \\ $$$$\left({x},{y}\right)\equiv\left(\mathrm{2},−\mathrm{1}\right);\:\left(−\mathrm{2},\mathrm{1}\right);\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\frac{\mathrm{18}}{\mathrm{7}},\frac{−\mathrm{1}}{\mathrm{7}}\right);\:\left(\frac{−\mathrm{18}}{\mathrm{7}},\frac{\mathrm{1}}{\mathrm{7}}\right)\: \\ $$
Answered by MJS_new last updated on 05/Sep/20
(ii) − (i) 12xy+20y^2 =−4 x=−((5y^2 +1)/(3y)) insert in (i) snd transform y^4 −((50)/(49))y^2 +(1/(49))=0 ⇒ y=±1∨y=±(1/7) ⇒ x=∓2∨x=∓((18)/7)
$$\left({ii}\right)\:−\:\left({i}\right) \\ $$$$\mathrm{12}{xy}+\mathrm{20}{y}^{\mathrm{2}} =−\mathrm{4} \\ $$$${x}=−\frac{\mathrm{5}{y}^{\mathrm{2}} +\mathrm{1}}{\mathrm{3}{y}} \\ $$$$\mathrm{insert}\:\mathrm{in}\:\left({i}\right)\:\mathrm{snd}\:\mathrm{transform} \\ $$$${y}^{\mathrm{4}} −\frac{\mathrm{50}}{\mathrm{49}}{y}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{49}}=\mathrm{0} \\ $$$$\Rightarrow\:{y}=\pm\mathrm{1}\vee{y}=\pm\frac{\mathrm{1}}{\mathrm{7}} \\ $$$$\Rightarrow\:{x}=\mp\mathrm{2}\vee{x}=\mp\frac{\mathrm{18}}{\mathrm{7}} \\ $$
Answered by 1549442205PVT last updated on 06/Sep/20
From i)and ii)we get 5(x^2 −7xy−9y^2 )=9(x^2 +5xy+11y^2 ) ⇔4x^2 +80xy+144y^2 =0 ⇔x^2 +20xy+36y^2 =0⇔((x+2y)(x+18y)=0 i)x+2y=0⇔x=−2y.Replace into (i) we get 4y^2 +14y^2 −9y^2 =9 ⇔y^2 =1⇔y=±1⇒x=∓2 ii)x+18y=0⇔x=−18y.Replace into (i)we get 324y^2 +126y^2 −9y^2 =9 ⇔441y^2 =9⇔y^2 =(9/(441))⇔y=±(1/7) ⇒x=∓((18)/7).Thus the given system has four roots (x,y)∈{(1,−2),(−1,2),((1/7),−((18)/7)),(((−1)/7),((18)/7))}
$$\left.\mathrm{F}\left.\mathrm{rom}\:\mathrm{i}\right)\mathrm{and}\:\mathrm{ii}\right)\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{5}\left(\mathrm{x}^{\mathrm{2}} −\mathrm{7xy}−\mathrm{9y}^{\mathrm{2}} \right)=\mathrm{9}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{5xy}+\mathrm{11y}^{\mathrm{2}} \right) \\ $$$$\Leftrightarrow\mathrm{4x}^{\mathrm{2}} +\mathrm{80xy}+\mathrm{144y}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Leftrightarrow\mathrm{x}^{\mathrm{2}} +\mathrm{20xy}+\mathrm{36y}^{\mathrm{2}} =\mathrm{0}\Leftrightarrow\left(\left(\mathrm{x}+\mathrm{2y}\right)\left(\mathrm{x}+\mathrm{18y}\right)=\mathrm{0}\right. \\ $$$$\left.\mathrm{i}\right)\mathrm{x}+\mathrm{2y}=\mathrm{0}\Leftrightarrow\mathrm{x}=−\mathrm{2y}.\mathrm{Replace}\:\mathrm{into}\:\left(\mathrm{i}\right) \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{4y}^{\mathrm{2}} +\mathrm{14y}^{\mathrm{2}} −\mathrm{9y}^{\mathrm{2}} =\mathrm{9} \\ $$$$\Leftrightarrow\mathrm{y}^{\mathrm{2}} =\mathrm{1}\Leftrightarrow\mathrm{y}=\pm\mathrm{1}\Rightarrow\mathrm{x}=\mp\mathrm{2} \\ $$$$\left.\mathrm{ii}\right)\mathrm{x}+\mathrm{18y}=\mathrm{0}\Leftrightarrow\mathrm{x}=−\mathrm{18y}.\mathrm{Replace}\:\mathrm{into} \\ $$$$\left(\mathrm{i}\right)\mathrm{we}\:\mathrm{get}\:\mathrm{324y}^{\mathrm{2}} +\mathrm{126y}^{\mathrm{2}} −\mathrm{9y}^{\mathrm{2}} =\mathrm{9} \\ $$$$\Leftrightarrow\mathrm{441y}^{\mathrm{2}} =\mathrm{9}\Leftrightarrow\mathrm{y}^{\mathrm{2}} =\frac{\mathrm{9}}{\mathrm{441}}\Leftrightarrow\mathrm{y}=\pm\frac{\mathrm{1}}{\mathrm{7}} \\ $$$$\Rightarrow\mathrm{x}=\mp\frac{\mathrm{18}}{\mathrm{7}}.\mathrm{Thus}\:\mathrm{the}\:\mathrm{given}\:\mathrm{system}\:\mathrm{has} \\ $$$$\mathrm{four}\:\mathrm{roots} \\ $$$$\left(\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{y}}\right)\in\left\{\left(\mathrm{1},−\mathrm{2}\right),\left(−\mathrm{1},\mathrm{2}\right),\left(\frac{\mathrm{1}}{\mathrm{7}},−\frac{\mathrm{18}}{\mathrm{7}}\right),\left(\frac{−\mathrm{1}}{\mathrm{7}},\frac{\mathrm{18}}{\mathrm{7}}\right)\right\} \\ $$

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