Solve-the-system-x-2-y-2-xy-2-x-y-0-x-2-y-xy-1-0-

Question Number 151115 by mathdanisur last updated on 18/Aug/21

Solve the system :

{\begin{cases} x^{2} y^{2} + {xy}^{2} + x + y = 0 \\ x^{2} y + xy + 1 = 0 \end{cases}

Answered by dumitrel last updated on 18/Aug/21

y (x^{2} y + x y + 1) + x = 0 \Rightarrow y \cdot 0 + x = 0 \Rightarrow x = 0

\Rightarrow 0 + 0 + 1 = 0 \Rightarrow n o s o l u t i o n

Commented by mathdanisur last updated on 18/Aug/21

Thank You \boldsymbol S er

Answered by Rasheed.Sindhi last updated on 18/Aug/21

{\begin{cases} x^{2} y^{2} + {xy}^{2} + x + y = 0 \\ (x^{2} y + xy + 1 = 0) \times y \end{cases}

{\begin{cases} x^{2} y^{2} + {xy}^{2} + x + y = 0 \dots . . (i) \\ x^{2} y^{2} + {xy}^{2} + y = 0 \dots \dots \dots . (ii) \end{cases}

(i) - (ii) : x = 0

T h i s {d o e s n}^{'} t s a t i s f y t h e s e c o n d

g i v e n e q u a t i o n \to N o s o l u t i o n

Commented by mathdanisur last updated on 18/Aug/21

Thank You Ser

Thank You Ser

Answered by Olaf_Thorendsen last updated on 18/Aug/21

in C, x = e^{i \frac{2 π}{3}}, y = 2 or e^{i \frac{4 π}{3}}, y = 2

in C, x = e^{i \frac{2 π}{3}}, y = 2 or e^{i \frac{4 π}{3}}, y = 2

Commented by MJS_new last updated on 18/Aug/21

how ?

how ?

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