Solve-x-2-y-2-13-i-2x-2-3y-2xy-2-ii-

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

Solve :

x^{2} + y^{2} = 13 \dots \dots . (i)

{2 x}^{2} + 3 y = {2 xy}^{2} \dots \dots . (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 = \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