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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))

xy=tt2−7t−9t2+5t+11=95t2+20t+36=0(t+2)(t+18)=0t=xy=−2,−18nowt2y2−7ty2−9y2−9=0y2=9t2−7t−9;x=tyy12=1,y22=149(x,y)≡(2,−1);(−2,1);(187,−17);(−187,17)

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)

(ii)−(i)12xy+20y2=−4x=−5y2+13yinsertin(i)sndtransformy4−5049y2+149=0⇒y=±1∨y=±17⇒x=∓2∨x=∓187

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))}

Fromi)andii)weget5(x2−7xy−9y2)=9(x2+5xy+11y2)⇔4x2+80xy+144y2=0⇔x2+20xy+36y2=0⇔((x+2y)(x+18y)=0i)x+2y=0⇔x=−2y.Replaceinto(i)weget4y2+14y2−9y2=9⇔y2=1⇔y=±1⇒x=∓2ii)x+18y=0⇔x=−18y.Replaceinto(i)weget324y2+126y2−9y2=9⇔441y2=9⇔y2=9441⇔y=±17⇒x=∓187.Thusthegivensystemhasfourroots(x,y)∈{(1,−2),(−1,2),(17,−187),(−17,187)}

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