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Question Number 99807 by PengagumRahasiamu last updated on 23/Jun/20

Answered by ajfour last updated on 23/Jun/20

$$x=54,y=24,z=6$$$$x+y+z=84.$$$$$$

Commented by PengagumRahasiamu last updated on 23/Jun/20

How to solve this?

Answered by ajfour last updated on 24/Jun/20

$$x-y+\sqrt{z}(\sqrt{x}-\sqrt{y})=42-6and$$$$y-z+\sqrt{x}(\sqrt{y}-\sqrt{z})=6+30$$$$\Rightarrow$$$$(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y}+\sqrt{z})=36..\left(i\right)$$$$(\sqrt{y}-\sqrt{z})(\sqrt{x}+\sqrt{y}+\sqrt{z})=36..\left(ii\right)$$$$\Rightarrow \sqrt{x}-\sqrt{y}=\sqrt{y}-\sqrt{z}$$$$or\sqrt{\mathit{x}}+\sqrt{\mathit{z}}=2\sqrt{\mathit{y}}$$$$Adding\left(i\right),\left(ii\right)$$$$(\sqrt{x}-\sqrt{z})\left(3\sqrt{y}\right)=36+36$$$$\Rightarrow (\sqrt{x}-\sqrt{z})(\sqrt{x}+\sqrt{z})=\frac{72\times 2}{3}$$$$\Rightarrow x-z=48....\left(I\right)$$$$Nowy-\sqrt{xz}=6$$$$\Rightarrow {\left(\frac{\sqrt{x}+\sqrt{z}}{2}\right)}^2-\sqrt{xz}=6$$$$\Rightarrow {(\sqrt{x}-\sqrt{z})}^2=24$$$$\sqrt{x}-\sqrt{z}=2\sqrt{6}$$$$andusingz=x-48\left[see\left(I\right)\right]$$$$\sqrt{x}-\sqrt{x-48}=2\sqrt{6}.....\left(iv\right)$$$$Also\frac{48}{\sqrt{x}+\sqrt{x-48}}=2\sqrt{6}$$$$\Rightarrow \sqrt{x}+\sqrt{x-48}=4\sqrt{6}...\left(v\right)$$$$Adding\left(iv\right),\left(v\right)$$$$2\sqrt{x}=6\sqrt{6}or$$$$x=54,z=6,y=24.$$$$\left(Onesetofpossiblesolution\right)!$$

Answered by 1549442205 last updated on 24/Jun/20

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