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Question Number 88921 by M±th+et£s last updated on 13/Apr/20

$$findx,y$$$$x-2y-\sqrt{xy}=0$$$$\sqrt{x-1}-\sqrt{2y-1}=1$$

Answered by behi83417@gmail.com last updated on 14/Apr/20

$$\{\begin{array}{l}{(\mathrm{x}-2\mathrm{y})}^2=\mathrm{xy}\Rightarrow {\mathrm{x}}^2-4\mathrm{xy}+{4\mathrm{y}}^2=\mathrm{xy}\\ \sqrt{\mathrm{x}-1}-\sqrt{2\mathrm{y}-1}=1\end{array}$$$$\Rightarrow \left(\frac{\mathrm{x}}{\mathrm{y}}\right)+4\left(\frac{\mathrm{y}}{\mathrm{x}}\right)-5=0\Rightarrow \frac{\mathrm{x}}{\mathrm{y}}=1,4$$$$1)\mathrm{x}=\mathrm{y}\Rightarrow \sqrt{\mathrm{x}-1}-\sqrt{2\mathrm{y}-1}=1$$$$\Rightarrow \mathrm{x}-1-2\sqrt{\mathrm{x}-1}+1=2\mathrm{x}-1$$$$\Rightarrow 2\sqrt{\mathrm{x}-1}=1-\mathrm{x}\Rightarrow 4\mathrm{x}-4={\mathrm{x}}^2-2\mathrm{x}+1$$$$\Rightarrow {\mathrm{x}}^2-6\mathrm{x}+5=0\Rightarrow [\mathbf{x}=\mathbf{y}=1,5]$$$$2)\mathrm{x}=4\mathrm{y}\Rightarrow 4\mathrm{y}-1-2\sqrt{4\mathrm{y}-1}+1=2\mathrm{y}-1$$$$\Rightarrow 2\mathrm{y}+1=2\sqrt{4\mathrm{y}-1}\Rightarrow {4\mathrm{y}}^2+4\mathrm{y}+1=16\mathrm{y}-4$$$$\Rightarrow {4\mathrm{y}}^2-12\mathbf{y}+5=0\Rightarrow \mathbf{y}=\frac{6\pm \sqrt{36-20}}{4}=\frac{5}{2},\frac{1}{2}$$$$\Rightarrow \mathrm{x}=4\mathrm{y}=10,2$$$$\mathrm{thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{sir}\mathrm{john}.$$$$\mathrm{typo}\mathrm{is}\mathrm{fixed}.$$

Commented by M±th+et£s last updated on 14/Apr/20

$$1-2-\sqrt{1}\ne 0??$$

Commented by john santu last updated on 14/Apr/20

$$mrBehiittypo$$

Answered by john santu last updated on 14/Apr/20

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