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

Solve the system: { ((x^2 y^2 + xy^2 + x + y = 0)),((x^2 y + xy + 1 = 0)) :}

$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{system}: \\ $$$$\begin{cases}{\mathrm{x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} \:+\:\mathrm{xy}^{\mathrm{2}} \:+\:\mathrm{x}\:+\:\mathrm{y}\:=\:\mathrm{0}}\\{\mathrm{x}^{\mathrm{2}} \mathrm{y}\:+\:\mathrm{xy}\:+\:\mathrm{1}\:=\:\mathrm{0}}\end{cases} \\ $$

Answered by dumitrel last updated on 18/Aug/21

y(x^2 y+xy+1)+x=0⇒y∙0+x=0⇒x=0 ⇒0+0+1=0⇒no solution

$${y}\left({x}^{\mathrm{2}} {y}+{xy}+\mathrm{1}\right)+{x}=\mathrm{0}\Rightarrow{y}\centerdot\mathrm{0}+{x}=\mathrm{0}\Rightarrow{x}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{0}+\mathrm{0}+\mathrm{1}=\mathrm{0}\Rightarrow{no}\:{solution} \\ $$

Commented by mathdanisur last updated on 18/Aug/21

Thank You Ser

$$\mathrm{Thank}\:\mathrm{You}\:\boldsymbol{\mathrm{S}}\mathrm{er} \\ $$

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

{ ((x^2 y^2 + xy^2 + x + y = 0)),(((x^2 y + xy + 1 = 0)×y)) :} { ((x^2 y^2 + xy^2 + x + y = 0.....(i))),((x^2 y^2 + xy^2 + y = 0..........(ii))) :} (i)−(ii):x=0 This doesn′t satisfy the second given equation→No solution

$$\:\begin{cases}{\mathrm{x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} \:+\:\mathrm{xy}^{\mathrm{2}} \:+\:\mathrm{x}\:+\:\mathrm{y}\:=\:\mathrm{0}}\\{\left(\mathrm{x}^{\mathrm{2}} \mathrm{y}\:+\:\mathrm{xy}\:+\:\mathrm{1}\:=\:\mathrm{0}\right)×\mathrm{y}}\end{cases} \\ $$$$\:\begin{cases}{\mathrm{x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} \:+\:\mathrm{xy}^{\mathrm{2}} \:+\:\mathrm{x}\:+\:\mathrm{y}\:=\:\mathrm{0}.....\left(\mathrm{i}\right)}\\{\mathrm{x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} \:+\:\mathrm{xy}^{\mathrm{2}} \:+\:\mathrm{y}\:=\:\mathrm{0}..........\left(\mathrm{ii}\right)}\end{cases} \\ $$$$\left(\mathrm{i}\right)−\left(\mathrm{ii}\right):\mathrm{x}=\mathrm{0} \\ $$$${This}\:{doesn}'{t}\:{satisfy}\:{the}\:{second} \\ $$$${given}\:{equation}\rightarrow{No}\:{solution} \\ $$

Commented by mathdanisur last updated on 18/Aug/21

Thank You Ser

$$\mathrm{Thank}\:\mathrm{You}\:\mathrm{Ser} \\ $$

Answered by Olaf_Thorendsen last updated on 18/Aug/21

in C, x = e^(i((2π)/3)) , y = 2 or e^(i((4π)/3)) , y = 2

$$\mathrm{in}\:\mathbb{C},\:{x}\:=\:{e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} ,\:{y}\:=\:\mathrm{2}\:\mathrm{or}\:{e}^{{i}\frac{\mathrm{4}\pi}{\mathrm{3}}} ,\:{y}\:=\:\mathrm{2} \\ $$

Commented by MJS_new last updated on 18/Aug/21

how?

$$\mathrm{how}? \\ $$

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