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Question Number 161564 by HongKing last updated on 19/Dec/21

$$\mathrm{Find}\mathrm{all}\mathrm{values}\mathrm{x};\mathrm{y};\mathrm{z}>0\mathrm{such}\mathrm{that}:$$$$\{\begin{array}{l}\mathrm{x}+\mathrm{y}+2\mathrm{z}=6\\ \frac{3}{\mathrm{y}}\cdot (\frac{2}{\mathrm{x}}+\frac{1}{\mathrm{y}})=4\cdot {(\frac{2}{\mathrm{x}+\mathrm{y}}+\frac{1}{2\mathrm{y}})}^2\\ \mathrm{x}+2^{\mathbf{y}}+{\log }_2\mathbf{z}=4\end{array}$$

Commented by Rasheed.Sindhi last updated on 20/Dec/21

$$(\mathrm{x},\mathrm{y},\mathrm{z})=(1,1,2)$$

Answered by Rasheed.Sindhi last updated on 20/Dec/21

$$\underset{―}{\mathit{A}\mathit{s}\mathit{s}\mathit{u}\mathit{m}\mathit{i}\mathit{n}\mathit{g}\mathbf{x},\mathbf{y},\mathbf{z}\in {\mathbb{Z}}^{+}}$$$$\mathrm{x}+\mathrm{y}+2\mathrm{z}=6$$$$\mathrm{x}+\mathrm{y}=6-2\mathrm{z}\Rightarrow \mathrm{z}=1,2$$$$\mathrm{x}+\mathrm{y}=2(3-\mathrm{z})\Rightarrow \mathrm{x}+\mathrm{y}iseven.$$$$\mathrm{x}+\mathrm{y}=2=1+1\mid \mathrm{x}+\mathrm{y}=4=2+2=1+3=3+1$$$$(\mathrm{x},\mathrm{y},\mathrm{z})=(1,1,2),(2,2,1),(1,3,1),(3,1,1)$$$$(1,1,2)satisfyingallthethree$$$$equations.$$$$\therefore (\mathrm{x},\mathrm{y},\mathrm{z})=(1,1,2)$$

Commented by HongKing last updated on 20/Dec/21

$$\mathrm{very}\mathrm{nice}\mathrm{dear}\mathrm{Sir}\mathrm{thank}\mathrm{you}$$

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