Question Number 99807 by PengagumRahasiamu last updated on 23/Jun/20
Answered by ajfour last updated on 23/Jun/20
$${x}=\mathrm{54},\:{y}=\mathrm{24},\:{z}=\mathrm{6} \\ $$$${x}+{y}+{z}=\mathrm{84}. \\ $$$$ \\ $$
Commented by PengagumRahasiamu last updated on 23/Jun/20
How to solve this?
Answered by ajfour last updated on 24/Jun/20
![x−y+(√z)((√x)−(√y))=42−6 and y−z+(√x)((√y)−(√z))=6+30 ⇒ ((√x)−(√y))((√x)+(√y)+(√z))=36 ..(i) ((√y)−(√z))((√x)+(√y)+(√z))=36 ..(ii) ⇒ (√x)−(√y) = (√y)−(√z) or (√x)+(√z)=2(√y) Adding (i),(ii) ((√x)−(√z))(3(√y))=36+36 ⇒ ((√x)−(√z))((√x)+(√z))=((72×2)/3) ⇒ x−z = 48 ....(I) Now y−(√(xz)) = 6 ⇒ ((((√x)+(√z))/2))^2 −(√(xz)) = 6 ⇒ ((√x)−(√z))^2 =24 (√x) −(√z) = 2(√6) and using z=x−48 [see (I)] (√x)−(√(x−48)) = 2(√6) .....(iv) Also ((48)/( (√x)+(√(x−48)))) = 2(√6) ⇒ (√x)+(√(x−48)) = 4(√6) ...(v) Adding (iv), (v) 2(√x) = 6(√6) or x=54, z=6 , y=24. (One set of possible solution)!](https://www.tinkutara.com/question/Q99912.png)
$${x}−{y}+\sqrt{{z}}\left(\sqrt{{x}}−\sqrt{{y}}\right)=\mathrm{42}−\mathrm{6}\:\:{and} \\ $$$${y}−{z}+\sqrt{{x}}\left(\sqrt{{y}}−\sqrt{{z}}\right)=\mathrm{6}+\mathrm{30} \\ $$$$\Rightarrow \\ $$$$\:\:\left(\sqrt{{x}}−\sqrt{{y}}\right)\left(\sqrt{{x}}+\sqrt{{y}}+\sqrt{{z}}\right)=\mathrm{36}\:\:\:..\left({i}\right) \\ $$$$\:\left(\sqrt{{y}}−\sqrt{{z}}\right)\left(\sqrt{{x}}+\sqrt{{y}}+\sqrt{{z}}\right)=\mathrm{36}\:\:\:\:..\left({ii}\right) \\ $$$$\Rightarrow\:\:\:\sqrt{{x}}−\sqrt{{y}}\:=\:\sqrt{{y}}−\sqrt{{z}} \\ $$$${or}\:\:\:\sqrt{\boldsymbol{{x}}}+\sqrt{\boldsymbol{{z}}}=\mathrm{2}\sqrt{\boldsymbol{{y}}} \\ $$$${Adding}\:\left({i}\right),\left({ii}\right) \\ $$$$\:\:\left(\sqrt{{x}}−\sqrt{{z}}\right)\left(\mathrm{3}\sqrt{{y}}\right)=\mathrm{36}+\mathrm{36} \\ $$$$\Rightarrow\:\:\:\left(\sqrt{{x}}−\sqrt{{z}}\right)\left(\sqrt{{x}}+\sqrt{{z}}\right)=\frac{\mathrm{72}×\mathrm{2}}{\mathrm{3}} \\ $$$$\Rightarrow\:\:\:\:\:{x}−{z}\:=\:\mathrm{48}\:\:\:\:….\left({I}\right) \\ $$$$\:\:\:{Now}\:\:\:\:{y}−\sqrt{{xz}}\:=\:\mathrm{6} \\ $$$$\Rightarrow\:\:\:\:\:\:\left(\frac{\sqrt{{x}}+\sqrt{{z}}}{\mathrm{2}}\right)^{\mathrm{2}} −\sqrt{{xz}}\:=\:\mathrm{6} \\ $$$$\Rightarrow\:\:\:\:\:\left(\sqrt{{x}}−\sqrt{{z}}\right)^{\mathrm{2}} =\mathrm{24} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\sqrt{{x}}\:−\sqrt{{z}}\:=\:\mathrm{2}\sqrt{\mathrm{6}} \\ $$$${and}\:{using}\:\:\:{z}={x}−\mathrm{48}\:\:\:\:\left[{see}\:\left({I}\right)\right] \\ $$$$\:\:\:\:\:\sqrt{{x}}−\sqrt{{x}−\mathrm{48}}\:=\:\mathrm{2}\sqrt{\mathrm{6}}\:\:\:\:\:…..\left({iv}\right) \\ $$$$\:\:\:\:\:{Also}\:\:\:\frac{\mathrm{48}}{\:\sqrt{{x}}+\sqrt{{x}−\mathrm{48}}}\:=\:\mathrm{2}\sqrt{\mathrm{6}} \\ $$$$\Rightarrow\:\:\:\sqrt{{x}}+\sqrt{{x}−\mathrm{48}}\:\:=\:\mathrm{4}\sqrt{\mathrm{6}}\:\:\:\:\:\:…\left({v}\right) \\ $$$${Adding}\:\:\left({iv}\right),\:\left({v}\right) \\ $$$$\:\:\:\:\:\:\mathrm{2}\sqrt{{x}}\:=\:\mathrm{6}\sqrt{\mathrm{6}}\:\:\:\:{or}\:\:\: \\ $$$$\:\:\:\:\:{x}=\mathrm{54},\:\:{z}=\mathrm{6}\:,\:{y}=\mathrm{24}. \\ $$$$\left({One}\:{set}\:{of}\:{possible}\:{solution}\right)! \\ $$
Answered by 1549442205 last updated on 24/Jun/20
