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Question Number 92426 by Jidda28 last updated on 06/May/20

$$\mathrm{If}{\mathrm{x}}^2+2\mathrm{xy}=0\mathrm{find}\mathrm{y}$$

Commented by Jidda28 last updated on 07/May/20

$$\mathrm{thank}\mathrm{you}\mathrm{so}\mathrm{much}\mathrm{sir}$$

Commented by Rio Michael last updated on 06/May/20

$$\mathrm{sir}\mathrm{please}\mathrm{reduce}\mathrm{your}\mathrm{font}\mathrm{sizes}\mathrm{alittle}$$$$\mathbf{solution}$$$$x^2+2xy=0$$$$\Rightarrow x(x+2y)=0$$$$\mathrm{either}x=0\mathrm{or}x+2y=0$$$$y=-\frac{x}{2}$$$$\Rightarrow y=0$$

Commented by Jidda28 last updated on 06/May/20

$$\mathrm{ok}\mathrm{sir}\mathrm{thank}\mathrm{for}\mathrm{the}\mathrm{solution}$$

Commented by MJS last updated on 07/May/20

$$\mathrm{no}.$$$$x^2+2xy=0$$$$(x+2y)x=0$$$$\Rightarrow x+2y=0\vee x=0$$$$\Rightarrow y=-\frac{x}{2}\vee x=0$$$$\mathrm{this}\mathrm{means}\mathrm{if}x=0\Rightarrow y\in \mathbb{C}$$$$\mathrm{but}\mathrm{if}x\ne 0\Rightarrow y=-\frac{x}{2}$$$$\mathrm{you}\mathrm{reduce}\mathrm{this}\mathrm{to}y=0\mathrm{but}\mathrm{then}\mathrm{you}\mathrm{lose}$$$$\mathrm{\infty }\mathrm{solutions}\mathrm{of}\mathrm{the}\mathrm{original}\mathrm{equation}$$

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