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Question Number 207116 by hardmath last updated on 06/May/24

$$\{\begin{array}{l}{\mathrm{x}}^2=\mathrm{x}-2\mathrm{y}\\ {\mathrm{y}}^2=\mathrm{y}-2\mathrm{x}\end{array}\Rightarrow \mathrm{x}-\mathrm{y}=?$$

Answered by A5T last updated on 06/May/24

$$x^2-y^2=x-y+2(x-y)=3(x-y)$$$$x-y\ne 0\Rightarrow x+y=3$$$$x^2+y^2=x+y-2(x+y)\Rightarrow x^2+y^2=-x-y$$$$\Rightarrow {(x+y)}^2-2xy=-3\Rightarrow 2xy=12\Rightarrow xy=6$$$${(x-y)}^2={(x+y)}^2-4xy=9-24=-15$$$$\Rightarrow x-y=\underset{-}{+}i\sqrt{15}orx-y=0$$

Answered by MATHEMATICSAM last updated on 07/May/24

$$x^2=x-2y...\left(\mathrm{i}\right)$$$$y^2=y-2x...\left(\mathrm{ii}\right)$$$$....................$$$$\left(\mathrm{i}\right)-\left(\mathrm{ii}\right)$$$$x^2-y^2=x-y+2(x-y)$$$$\Rightarrow (x+y)(x-y)=3(x-y)$$$$\Rightarrow x+y=3\mathrm{when}x-y\ne 0$$$$\left(\mathrm{i}\right)+\left(\mathrm{ii}\right)$$$$x^2+y^2=x+y-2(x+y)$$$$\Rightarrow {(x+y)}^2-2xy=-(x+y)$$$$\Rightarrow 3^2-2xy=-3$$$$\Rightarrow 9+3=2xy$$$$\Rightarrow xy=6$$$$\mathrm{Now}{(x-y)}^2={(x+y)}^2-4xy$$$$=3^2-4\times 6=9-24=-15$$$$\therefore x-y=\pm i\sqrt{15}$$$$$$$$\therefore \mathit{x}-\mathit{y}=\pm \mathit{i}\sqrt{15}\mathbf{or}\mathit{x}-\mathit{y}=0\left(\mathbf{Ans}\right)$$

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