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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)

$${please}\:{help}\:{me}\:.{How}\:{can}\:{resolve}\:{this}\:{system}? \\ $$$$\left\{_{\left(\mathrm{1}+\sqrt{\mathrm{2}}\right){x}+{y}=\mathrm{1}} ^{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{1}} \right. \\ $$

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.

$$\begin{cases}{{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:=\:\mathrm{1}\:\:\:\:\:\left({i}\right)}\\{{x}\:+\:{x}\sqrt{\mathrm{2}}\:+\:{y}\:=\:\mathrm{1}\:\:\:\:\left({ii}\right)}\end{cases} \\ $$$$\left({ii}\right)\::\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\:{x}\:+\:{x}\sqrt{\mathrm{2}}\:=\:\mathrm{1}\:−\:{y}\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({x}\:+\:{x}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \:=\:\left(\mathrm{1}\:−\:{y}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} \:+\:\mathrm{2}\sqrt{\mathrm{2}}{x}^{\mathrm{2}} \:+\:\mathrm{2}{x}^{\mathrm{2}} \:=\:\mathrm{1}\:−\:\mathrm{2}{y}\:+\:{y}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} \left(\mathrm{1}\:+\:\mathrm{2}\sqrt{\mathrm{2}}\:+\:\mathrm{2}\right)\:=\:\mathrm{1}\:−\:\mathrm{2}{y}\:+\:{y}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} \:=\:\frac{\mathrm{1}\:−\:\mathrm{2}{y}\:+\:{y}^{\mathrm{2}} }{\mathrm{2}\sqrt{\mathrm{2}}\:+\:\mathrm{3}} \\ $$$$\left({i}\right)\::\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}\:−\:\mathrm{2}{y}\:+\:{y}^{\mathrm{2}} }{\mathrm{2}\sqrt{\mathrm{2}}\:+\:\mathrm{3}}\:+\:{y}^{\mathrm{2}} \:=\:\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}\:−\:\mathrm{2}{y}\:+\:{y}^{\mathrm{2}} \:+\left(\mathrm{2}\sqrt{\mathrm{2}}\:+\:\mathrm{3}\right){y}^{\mathrm{2}} }{\mathrm{2}\sqrt{\mathrm{2}}\:+\:\mathrm{3}}\:=\:\mathrm{1} \\ $$$$\:\:\:\:\mathrm{1}\:−\:\mathrm{2}{y}\:+\:{y}^{\mathrm{2}} \left(\mathrm{1}\:+\:\mathrm{2}\sqrt{\mathrm{2}}\:+\:\mathrm{3}\right)\:=\:\mathrm{2}\sqrt{\mathrm{2}}\:+\:\mathrm{3} \\ $$$$\:\:\:\left(\mathrm{4}\:+\:\mathrm{2}\sqrt{\mathrm{2}}\right){y}^{\mathrm{2}} \:−\:\mathrm{2}{y}\:+\:\mathrm{1}\:−\:\mathrm{2}\sqrt{\mathrm{2}}\:−\:\mathrm{3}\:=\:\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\left(\mathrm{4}\:+\:\mathrm{2}\sqrt{\mathrm{2}}\right){y}^{\mathrm{2}} \:−\:\mathrm{2}{y}\:−\:\mathrm{2}\sqrt{\mathrm{2}}\:−\:\mathrm{2}\:=\:\mathrm{0} \\ $$$$\mathrm{divide}\:\mathrm{by}\:\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{2}\:+\:\sqrt{\mathrm{2}}\right){y}^{\mathrm{2}} \:−\:{y}\:−\:\sqrt{\mathrm{2}}\:−\:\mathrm{1}\:=\:\mathrm{0} \\ $$$$\mathrm{solving},\:\mathrm{you}\:\mathrm{have}\:\mathrm{two}\:\mathrm{values}\:\mathrm{for}\:{y}:\:{y}_{\mathrm{1}} ,\:{y}_{\mathrm{2}} \\ $$$$\mathrm{substitute}\:\mathrm{in}\:\mathrm{the}\:\left({i}\right)\:\boldsymbol{\mathrm{or}}\:\left({ii}\right)\:\mathrm{and}\:\mathrm{you}\:\mathrm{find} \\ $$$${x}_{\mathrm{1}} \:\mathrm{and}\:{x}_{\mathrm{2}} \:\mathrm{too}. \\ $$$$ \\ $$

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