Solve: x^2 + y^2 = 13 ....... (i) 2x^2 + 3y = 2xy^2 ....... (ii)


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Question Number 95093 by I want to learn more last updated on 22/May/20

$Solve :$ $x^{2} + y^{2} = 13 . . . . . . . (i)$ ${2 x}^{2} + 3 y = {2 xy}^{2} . . . . . . . (ii)$
	Answered by behi83417@gmail.com last updated on 23/May/20

	$(2 \times i - ii) \Rightarrow {2 y}^{2} - 3 y = 26 - {2 xy}^{2}$ $\Rightarrow x = \frac{26 + 3 y - {2 y}^{2}}{{2 y}^{2}}$ $\Rightarrow {(\frac{26 + 3 y - {2 y}^{2}}{{2 y}^{2}})}^{2} + y^{2} = 13$ ${(26 + 3 y - {2 y}^{2})}^{2} + {4 y}^{6} = {52 y}^{4}$ $676 + {9 y}^{2} + {4 y}^{4} + 156 y - {104 y}^{2} - {12 y}^{3} + {4 y}^{6} = {52 y}^{4}$ $\Rightarrow {4 y}^{6} - {48 y}^{4} - {12 y}^{3} - {95 y}^{2} + 156 y + 676 = 0$ $(y - 2) (p (y)) = 0 \Rightarrow$ $\Rightarrow y = 2, \pm 3.6, - 1.6$ $\Rightarrow x = 3, \pm 0.43, 3.14$
	Commented by I want to learn more last updated on 23/May/20

	$I appreciate sir$
	Answered by 1549442205 last updated on 23/May/20
	$From (1)we have y^2 =13−x^2 ,putting into (ii) we get y=((−2x^3 −2x^2 +26x)/3).Squaring we get((4x^(6 ) +8x^5 +4x^4 +676x^2 −104x^4 −104x^3 )/9). Putting into (i)we get 4x^6 +8x^5 −100x^4 − 104x^3 +685x^2 −117=0.This equation has four real roots: x_1 =−0.4058529736 and 0.4337536916,x_3 =3 and Res x_4 =3.23939445877281 pectively,we get y_1 =−3.582636371 and y_2 =3.579365549,y_3 =2 and y_4 =1.583137246.Thus,our system of equations has four solutions: (x;y)∈{(x_1 ;y_1 );(x_2 ;y_2 );(x_3 ,y_3 );(x_4 ;y_4 )} [ (y/(x^2 +y^2 )) + (x/(x^2 +y^2 )) ] dx + [(y/(x^2 +y^2 ))−(x/(x^2 +y^2 )) ]dy=0$
	$From (1) we have y^{2} = 13 - x^{2}, putting into$ $(ii) we get y = \frac{- {2 x}^{3} - {2 x}^{2} + 26 x}{3} . Squaring$ $we get \frac{{4 x}^{6} + {8 x}^{5} + {4 x}^{4} + {676 x}^{2} - {104 x}^{4} - {104 x}^{3}}{9} .$ $Putting into (i) we get {4 x}^{6} + {8 x}^{5} - {100 x}^{4} -$ ${104 x}^{3} + {685 x}^{2} - 117 = 0 . This equation has four real roots :$ $x_{1} = - 0.4058529736 and 0.4337536916, x_{3} = 3 and Res$ $x_{4} = 3.23939445877281$ $pectively, we get y_{1} = - 3.582636371 and$ $y_{2} = 3.579365549, y_{3} = 2 and y_{4} = 1.583137246 . Thus, our system of$ $equations has four solutions :$ $(x; y) \in {(x_{1}; y_{1}); (x_{2}; y_{2}); (x_{3}, y_{3}); (x_{4}; y_{4})}$ $[\frac{y}{x^{2} + y^{2}} + \frac{x}{x^{2} + y^{2}}] dx + [\frac{y}{x^{2} + y^{2}} - \frac{x}{x^{2} + y^{2}}] dy = 0$
	Commented by I want to learn more last updated on 23/May/20

	$I appreciate sir$
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