solve-the-system-of-equations-x-3x-y-x-2-y-2-3-y-x-3y-x-2-y-2-0-

Question Number 115909 by Eric002 last updated on 29/Sep/20

s o l v e t h e s y s t e m o f e q u a t i o n s

x + \frac{3 x - y}{x^{2} + y^{2}} = 3, y - \frac{x + 3 y}{x^{2} + y^{2}} = 0

Answered by MJS_new last updated on 29/Sep/20

let y = p x

(1) x + \frac{3 - p}{(1 + p^{2}) x} = 3

(2) p x - \frac{1 + 3 p}{(1 + p^{2}) x} = 0

(1) x^{2} = 3 x + \frac{p - 3}{1 + p^{2}}

(2) x^{2} = \frac{3 p + 1}{(1 + p^{2}) p}

\Rightarrow

3 x = \frac{3 p + 1}{(1 + p^{2}) p} - \frac{p - 3}{1 + p^{2}}

x = - \frac{p^{2} - 6 p - 1}{3 (1 + p^{2}) p}

(1) x^{2} - 3 x - \frac{p - 3}{1 + p^{2}} = 0

\Rightarrow

p^{4} + \frac{21}{26} p^{3} - \frac{7}{26} p^{2} - \frac{3}{26} p - \frac{1}{26} = 0

(p + 1) (p - \frac{1}{2}) (p^{2} + \frac{4}{13} p + \frac{1}{13}) = 0

p_{1} = - 1

p_{2} = \frac{1}{2}

p_{3, 4} = - \frac{2}{13} \pm \frac{3}{13} i

\Rightarrow

x_{1} = 1 \land y_{1} = - 1

x_{2} = 2 \land y_{2} = 1

x_{3} = \frac{3}{2} - i \land y_{3} = \frac{1}{2} i

x_{4} = \frac{3}{2} + i \land y_{4} = - \frac{1}{2} i

Facebook Tweet Pin