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Question Number 205211 by cortano12 last updated on 13/Mar/24

	Answered by Berbere last updated on 13/Mar/24
	${ ((5x^2 (y^2 −1)=4x(x^2 +y^2 ))),((5y^2 (x^2 +1)=3y(x^2 +y^2 ))) :} ⇒5y^2 +5x^2 =(x^2 +y^2 )(3y−4x) ⇒(x^2 +y^2 )(3y−4x−5)=0 x=y=0 or 3y−4x=5 x=((3y−5)/4);(5/4)(3y−5)(y^2 −1)=4y^2 +(1/4)(3y−5)^2 5(3y−5)(y^2 −1)=16y^2 +(3y−5)^2 15y^3 −25y^2 −15y+25=25y^2 +25−30y y(3y^2 −10y^2 +3y)=0 y=0;y=((10−8)/6)=(1/3);y=3 y=3⇒x=1;y=(1/3)⇒x=−1 S={(−1,(1/3));(1,3);(0,0)}$
	${\begin{cases} 5 x^{2} (y^{2} - 1) = 4 x (x^{2} + y^{2}) \\ 5 y^{2} (x^{2} + 1) = 3 y (x^{2} + y^{2}) \end{cases}$ $\Rightarrow 5 y^{2} + 5 x^{2} = (x^{2} + y^{2}) (3 y - 4 x)$ $\Rightarrow (x^{2} + y^{2}) (3 y - 4 x - 5) = 0$ $x = y = 0 o r 3 y - 4 x = 5$ $x = \frac{3 y - 5}{4}; \frac{5}{4} (3 y - 5) (y^{2} - 1) = 4 y^{2} + \frac{1}{4} {(3 y - 5)}^{2}$ $5 (3 y - 5) (y^{2} - 1) = 16 y^{2} + {(3 y - 5)}^{2}$ $15 y^{3} - 25 y^{2} - 15 y + 25 = 25 y^{2} + 25 - 30 y$ $y (3 y^{2} - 10 y^{2} + 3 y) = 0$ $y = 0; y = \frac{10 - 8}{6} = \frac{1}{3}; y = 3$ $y = 3 \Rightarrow x = 1; y = \frac{1}{3} \Rightarrow x = - 1$ $S = {(- 1, \frac{1}{3}); (1, 3); (0, 0)}$
	Commented by Berbere last updated on 13/Mar/24

	$(1) * x; (2) * y$
	Commented by cortano12 last updated on 13/Mar/24

	$oh no . it 5 x (y^{2} - 1), not {5 x}^{2} (y^{2} - 1)$
	Commented by A5T last updated on 13/Mar/24

	$I f x, y \in C$ $x^{2} + y^{2} = 0 \Rightarrow 5 x (y^{2} - 1) = 0 \land 5 y (x^{2} + 1) = 0$ $\Rightarrow x = 0 o r y = \underline{+} 1 \land y = 0 o r x = \underline{+} i$ $(x, y) = (0, 0), (i, 1), (- i, 1), (i, - 1), (- i, - 1),$
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