please help me .How can resolve this system? {_((1+(√2))x+y=1) ^(x^2 +y^2 =1)


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Question Number 12534 by JAZAR last updated on 24/Apr/17
$please help me .How can resolve this system? {_((1+(√2))x+y=1) ^(x^2 +y^2 =1)$
$p l e a s e h e l p m e . H o w c a n r e s o l v e t h i s s y s t e m ?$ ${_{(1 + \sqrt{2}) x + y = 1}^{x^{2} + y^{2} = 1}$
	Answered by geovane10math last updated on 25/Apr/17
	${ ((x^2 + y^2 = 1 (i))),((x + x(√2) + y = 1 (ii))) :} (ii) : [ x + x(√2) = 1 − y (x + x(√2))^2 = (1 − y)^2 x^2 + 2(√2)x^2 + 2x^2 = 1 − 2y + y^2 x^2 (1 + 2(√2) + 2) = 1 − 2y + y^2 x^2 = ((1 − 2y + y^2 )/(2(√2) + 3)) (i) : ((1 − 2y + y^2 )/(2(√2) + 3)) + y^2 = 1 ((1 − 2y + y^2 +(2(√2) + 3)y^2 )/(2(√2) + 3)) = 1 1 − 2y + y^2 (1 + 2(√2) + 3) = 2(√2) + 3 (4 + 2(√2))y^2 − 2y + 1 − 2(√2) − 3 = 0 (4 + 2(√2))y^2 − 2y − 2(√2) − 2 = 0 divide by 2 (2 + (√2))y^2 − y − (√2) − 1 = 0 solving, you have two values for y: y_1 , y_2 substitute in the (i) or (ii) and you find x_1 and x_2 too.$
	${\begin{cases} x^{2} + y^{2} = 1 (i) \\ x + x \sqrt{2} + y = 1 (i i) \end{cases}$ $(i i) : [x + x \sqrt{2} = 1 - y$ ${(x + x \sqrt{2})}^{2} = {(1 - y)}^{2}$ $x^{2} + 2 \sqrt{2} x^{2} + 2 x^{2} = 1 - 2 y + y^{2}$ $x^{2} (1 + 2 \sqrt{2} + 2) = 1 - 2 y + y^{2}$ $x^{2} = \frac{1 - 2 y + y^{2}}{2 \sqrt{2} + 3}$ $(i) : \frac{1 - 2 y + y^{2}}{2 \sqrt{2} + 3} + y^{2} = 1$ $\frac{1 - 2 y + y^{2} + (2 \sqrt{2} + 3) y^{2}}{2 \sqrt{2} + 3} = 1$ $1 - 2 y + y^{2} (1 + 2 \sqrt{2} + 3) = 2 \sqrt{2} + 3$ $(4 + 2 \sqrt{2}) y^{2} - 2 y + 1 - 2 \sqrt{2} - 3 = 0$ $(4 + 2 \sqrt{2}) y^{2} - 2 y - 2 \sqrt{2} - 2 = 0$ $divide by 2$ $(2 + \sqrt{2}) y^{2} - y - \sqrt{2} - 1 = 0$ $solving, you have two values for y : y_{1}, y_{2}$ $substitute in the (i) or (i i) and you find$ $x_{1} and x_{2} too .$
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