Solve-the-system-y-2x-1-x-2-z-2y-1-y-2-x-2z-1-z-2-

Question Number 151043 by mathdanisur last updated on 17/Aug/21

Solve the system :

{\begin{cases} y = \frac{2 x}{1 - x^{2}} \\ z = \frac{2 y}{1 - y^{2}} \\ x = \frac{2 z}{1 - z^{2}} \end{cases}

Answered by john_santu last updated on 18/Aug/21

x = y = z \Rightarrow x = \frac{2 x}{1 - x^{2}}

\Rightarrow x (1 - \frac{2}{1 - x^{2}}) = 0; x = 0

\Rightarrow 1 - x^{2} = 2; x^{2} = - 1; x = \pm i

s o l u t i o n (x, y, z) = (0, 0, 0), (\pm i, \pm i, \pm i)

Answered by MJS_new last updated on 18/Aug/21

z = \frac{2 y}{(1 - y) (1 + y)} \land y = \frac{2 x}{(1 - x) (1 + x)} \Rightarrow

z = \frac{4 x (1 - x) (1 + x)}{(x^{2} - 2 x - 1) (x^{2} + 2 x - 1)}

\Rightarrow

x = \frac{8 x (1 - x) (1 + x) (x^{2} - 2 x - 1) (x^{2} + 2 x - 1)}{x^{8} - 28 x^{6} + 70 x^{4} - 38 x^{2} + 1}

x^{9} - 20 x^{7} + 14 x^{5} + 28 x^{3} - 7 x = 0

x_{1} = 0

x^{8} - 20 x^{6} + 14 x^{4} + 28 x^{2} - 7 = 0

(x^{2} + 1) (x^{6} - 21 x^{4} + 35 x^{2} - 7) = 0

x_{2, 3} = \pm i

{(x^{2})}^{3} - 21 {(x^{2})}^{2} + 35 x^{2} - 7 = 0

x^{2} = u + 7

u^{3} - 112 u - 448 = 0

Trigonometric Solution

u_{1} = - \frac{8 \sqrt{21}}{3} \cos \frac{π + 2 \arcsin \frac{3 \sqrt{21}}{14}}{6} \approx - 6.76808589

u_{2} = - \frac{8 \sqrt{21}}{3} \sin \frac{\arcsin \frac{3 \sqrt{21}}{14}}{3} \approx - 5.42758347

u_{3} = \frac{8 \sqrt{21}}{3} \sin \frac{π + \arcsin \frac{3 \sqrt{21}}{14}}{3} \approx 12.1956694

\Rightarrow

x_{4, 5} \approx \pm . 481574619

x_{6, 7} \approx \pm 1.25396034

x_{8, 9} \approx \pm 4.38128627

we get the following triplets (\begin{matrix} x \\ y \\ x \end{matrix})

(\begin{matrix} - 4.38 \\ . 482 \\ 1.25 \end{matrix}), (\begin{matrix} - 1.25 \\ 4.38 \\ - . 482 \end{matrix}), (\begin{matrix} - . 482 \\ - 1.25 \\ 4.38 \end{matrix}), (\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}), (\begin{matrix} . 482 \\ 1.25 \\ - 4.38 \end{matrix}), (\begin{matrix} 1.25 \\ - 4.38 \\ . 482 \end{matrix}), (\begin{matrix} 4.38 \\ - . 482 \\ - 1.25 \end{matrix}), (\begin{matrix} - i \\ - i \\ - i \end{matrix}), (\begin{matrix} i \\ i \\ i \end{matrix})

Facebook Tweet Pin