{ ((x^2 = x − 6y)),((y^2 = y − 6x)) :} ⇒ x + y = ?


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Question Number 207092 by hardmath last updated on 06/May/24
${ ((x^2 = x − 6y)),((y^2 = y − 6x)) :} ⇒ x + y = ?$
${\begin{cases} x^{2} = x - 6 y \\ y^{2} = y - 6 x \end{cases} \Rightarrow x + y = ?$
	Answered by A5T last updated on 06/May/24
	$x^2 −y^2 =x−y−6y+6x=7x−7y ⇒(x−y)(x+y)=7(x−y) x−y≠0 ⇒x+y=7 when x=y; x^2 =y^2 =−5x⇒x=y=0 or x=y=−5 ⇒x+y=0 or x+y=−10 ⇒x+y∈{−10,0,7}$
	$x^{2} - y^{2} = x - y - 6 y + 6 x = 7 x - 7 y$ $\Rightarrow (x - y) (x + y) = 7 (x - y)$ $x - y \neq 0 \Rightarrow x + y = 7$ $w h e n x = y; x^{2} = y^{2} = - 5 x \Rightarrow x = y = 0 o r x = y = - 5$ $\Rightarrow x + y = 0 o r x + y = - 10$ $\Rightarrow x + y \in {- 10, 0, 7}$
	Answered by mr W last updated on 06/May/24

	$(i) + (i i) :$ $x^{2} + y^{2} = - 5 (x + y) \Rightarrow x + y ⩽ 0$ $(i) - (i i) :$ $(x + y) (x - y) = 7 (x - y)$ $x \neq y : \Rightarrow x + y = 7 > 0 \Rightarrow r e j e c t e d$ $x = y : x^{2} = - 5 x \Rightarrow x = 0 o r - 5$ $\Rightarrow x + y = 0 o r - 10$
	Commented by Frix last updated on 06/May/24

	$But the solutions of the system are$ $x = y = 0 \Rightarrow x + y = 0$ $x = y = - 5 \Rightarrow x + y = - 10$ $x = \frac{7}{2} \pm \frac{\sqrt{119}}{2} i \land y = conj x \Rightarrow x + y = 7$
	Commented by mr W last updated on 06/May/24

	$i l i v e i n R w o r l d :)$
	Commented by Frix last updated on 06/May/24
	...but the real world is complex... ��
	Commented by mr W last updated on 06/May/24

	$i n d e e d! t h e r e a l w o r l d i s t o o$ $c o m p l e x t o b e r e a l t o m e . . .$
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