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Question Number 162382 by HongKing last updated on 29/Dec/21

	Answered by Rasheed.Sindhi last updated on 29/Dec/21
	${: (( (x+(√(x^2 +1)) )(y+(√(y^2 +1)) )=2011)),(( x+y=((2010)/( (√(2011)))) )) } _(−) (x+(√(x^2 +1)) )∙(((y+(√(y^2 +1)) )(y−(√(y^2 +1)) ))/((y−(√(y^2 +1)) )))=2011 −((x+(√(x^2 +1)) )/(y−(√(y^2 +1)) ))=2011 ((x+(√(x^2 +1)) )/(y−(√(y^2 +1)) ))+1=−2011+1 ((x+y)/(y−(√(y^2 +1)) ))=−2010 (((2010)/( (√(2011)) ))/(y−(√(y^2 +1)) ))=−2010 (1/( (√(2011)) (y−(√(y^2 +1)) )))=−1 y−(√(y^2 +1)) =−(1/( (√(2011)))) 2y^2 +1−(1/(2011))=2y(√(y^2 +1)) y^2 +((1005)/(2011))=y(√(y^2 +1)) 2011y^2 +1005=2011y(√(y^2 +1)) 2011^2 y^4 ^(×) +2∙2011∙1005y^2 +1005^2 =2011^2 y^2 (y^2 +1) =2011^2 y^4 ^(×) +2011^2 y^2 (2011∙2010−2011^2 )y^2 =−1005^2 y^2 =((−1005^2 )/(2011(2010−2011)))=((1005^2 )/(2011)) y=±((1005)/( (√(2011)))) As the equations are symmetric in x & y So x=±((1005)/( (√(2011)))) (−((1005)/( (√(2011)))),−((1005)/( (√(2011))))) doesn′t satisfy x+y=((2010)/( (√(2011)))) Hence (x,y)=(((1005)/( (√(2011)))) , ((1005)/( (√(2011)))))$
	$\underline{\begin{matrix} (x + \sqrt{x^{2} + 1}) (y + \sqrt{y^{2} + 1}) = 2011 \\ x + y = \frac{2010}{\sqrt{2011}} \end{matrix}}}$ $(x + \sqrt{x^{2} + 1}) \cdot \frac{(y + \sqrt{y^{2} + 1}) (y - \sqrt{y^{2} + 1})}{(y - \sqrt{y^{2} + 1})} = 2011$ $- \frac{x + \sqrt{x^{2} + 1}}{y - \sqrt{y^{2} + 1}} = 2011$ $\frac{x + \sqrt{x^{2} + 1}}{y - \sqrt{y^{2} + 1}} + 1 = - 2011 + 1$ $\frac{x + y}{y - \sqrt{y^{2} + 1}} = - 2010$ $\frac{\frac{2010}{\sqrt{2011}}}{y - \sqrt{y^{2} + 1}} = - 2010$ $\frac{1}{\sqrt{2011} (y - \sqrt{y^{2} + 1})} = - 1$ $y - \sqrt{y^{2} + 1} = - \frac{1}{\sqrt{2011}}$ ${2 y}^{2} + 1 - \frac{1}{2011} = 2 y \sqrt{y^{2} + 1}$ $y^{2} + \frac{1005}{2011} = y \sqrt{y^{2} + 1}$ ${2011 y}^{2} + 1005 = 2011 y \sqrt{y^{2} + 1}$ $\overset{\times}{2011^{2} y^{4}} + 2 \cdot 2011 \cdot {1005 y}^{2} + 1005^{2}$ $= 2011^{2} y^{2} (y^{2} + 1)$ $= \overset{\times}{2011^{2} y^{4}} + 2011^{2} y^{2}$ $(2011 \cdot 2010 - 2011^{2}) y^{2} = - 1005^{2}$ $y^{2} = \frac{- 1005^{2}}{2011 (2010 - 2011)} = \frac{1005^{2}}{2011}$ $y = \pm \frac{1005}{\sqrt{2011}}$ $As the equations are symmetric in x & y$ $So$ $x = \pm \frac{1005}{\sqrt{2011}}$ $(- \frac{1005}{\sqrt{2011}}, - \frac{1005}{\sqrt{2011}}) {doesn}^{'} t satisfy x + y = \frac{2010}{\sqrt{2011}}$ $Hence$ $(x, y) = (\frac{1005}{\sqrt{2011}}, \frac{1005}{\sqrt{2011}})$
	Commented by HongKing last updated on 29/Dec/21

	$cool my dear Sir thank you so much$
	Commented by mr W last updated on 29/Dec/21

	$h o w d i d y o u g e t t h i s s t e p s i r ?$ $f r o m \frac{x + \sqrt{x^{2} + 1}}{y - \sqrt{y^{2} + 1}} + 1 = - 2011 + 1$ $t o \frac{x + y}{y - \sqrt{y^{2} + 1}} = - 2010$
	Commented by mr W last updated on 29/Dec/21
	$for more generality can we change the question to { (((x+(√(x^2 +1)))(y+(√(y^2 +1)))=10)),((x+y=4)) :}$
	$f o r m o r e g e n e r a l i t y c a n w e c h a n g e$ $t h e q u e s t i o n t o$ ${\begin{cases} (x + \sqrt{x^{2} + 1}) (y + \sqrt{y^{2} + 1}) = 10 \\ x + y = 4 \end{cases}$
	Commented by Rasheed.Sindhi last updated on 29/Dec/21

	$Mistake sir! I considered \sqrt{x^{2} + 1} & \sqrt{y^{2} + 1}$ $same! I was very happy for eliminating x,$ $but this is not so easy .$ 😭😭😭
	Commented by mr W last updated on 29/Dec/21

	$t h a n k s f o r r e p l y i n g s i r .$ ${i t}^{'} s r e a l l y n o t e a s y t o s e p a r a t e x, y$ $f r o m e a c h o t h e r .$
	Commented by Rasheed.Sindhi last updated on 29/Dec/21

	$I w o n d e r s i r, t h a t H o n g K i n g s i r$ $a l s o {h a s n}^{'} t n o t i c e d i t! P e r h a p s t h e$ $a n s w e r i s a c c i d e n t l y c o r r e c t o r h e$ ${h a d n}^{'} t a n s w e r o f t h e q u e s t i o n!$ $A n y w a y t h a n k s f o r y o u r k e e n o b s e r v a t i o n!$
	Commented by mr W last updated on 29/Dec/21

	$t h e q u e s t i o n i s n o t g e n e r a l e n o u g h .$ $b o t h s y m m e t r i c c u r v e s t a n g e n t e a c h$ $o t h e r, s o x = y f u l f i l l s t h e s o l u t i o n .$
	Commented by Rasheed.Sindhi last updated on 29/Dec/21

	$E x a c l y s i r, a n d t h i s i s t h e r e a s o n$ $I r e a c h e d c o r r e c t a n s w e r b y$ $c o n s i d e r i n g \sqrt{x^{2} + 1} = \sqrt{y^{2} + 1}!$ $T h a n k s a g a i n s i r!$
	Commented by mr W last updated on 29/Dec/21
	��
	Commented by Ar Brandon last updated on 29/Dec/21

	${It}^{'} s really not easy to separate x and y$ $from each other . They are deeply in love$ 🤗🤭
	Answered by MJS_new last updated on 29/Dec/21

	$(x + \sqrt{x^{2} + 1}) (y + \sqrt{y^{2} + 1}) = a$ $x + y = b$ $let u = x + \sqrt{x^{2} + 1} \land v = y + \sqrt{y^{2} + 1}$ $u v = a$ $\frac{1}{2} (u - \frac{1}{u} + v - \frac{1}{v}) = b$ $\Leftrightarrow$ $u v = a$ $u + v = \frac{2 a b}{a - 1}$ $\Leftrightarrow$ $v = \frac{a}{u}$ $u^{2} - \frac{2 a b}{a - 1} u + a = 0$ $\Leftrightarrow$ $u = \frac{a b \pm \sqrt{- a (a^{2} - (b^{2} + 2) a + 1)}}{a - 1}$ $v = \frac{a b \mp \sqrt{- a (a^{2} - (b^{2} + 2) a + 1)}}{a - 1}$ $\Rightarrow$ $x = \frac{b}{2} \pm \frac{(a + 1)}{2 a (a - 1)} \sqrt{- a (a^{2} - (b^{2} + 2) a + 1)}$ $y = \frac{b}{2} \mp \frac{(a + 1)}{2 a (a - 1)} \sqrt{- a (a^{2} - (b^{2} + 2) a + 1)}$
	Commented by mr W last updated on 29/Dec/21

	$g r e a t!$
	Commented by Tawa11 last updated on 30/Dec/21

	$Great sir$
	Commented by HongKing last updated on 31/Dec/21

	$cool my dear Sir$
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