please-help-me-How-can-resolve-this-system-1-2-x-y-1-x-2-y-2-1-

Question Number 12534 by JAZAR last updated on 24/Apr/17

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

{\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 .