Question Number 153605 by bramlexs22 last updated on 09/Sep/21
$$\:\:\begin{cases}{\sqrt{\mathrm{x}}\:+\sqrt{\mathrm{y}+\mathrm{z}}\:=\mathrm{5}}\\{\sqrt{\mathrm{y}}+\sqrt{\mathrm{z}+\mathrm{x}}\:=\:\mathrm{7}}\\{\sqrt{\mathrm{z}}+\sqrt{\mathrm{x}+\mathrm{y}}\:=\:\mathrm{7}}\end{cases} \\ $$
Commented by Rasheed.Sindhi last updated on 08/Sep/21
$$\:\:\begin{cases}{\sqrt{\mathrm{x}}\:+\sqrt{\mathrm{y}+\mathrm{z}}\:=\mathrm{5}}\\{\sqrt{\mathrm{y}}+\sqrt{\mathrm{z}+\mathrm{x}}\:=\:\mathrm{7}}\\{\sqrt{{z}}+\sqrt{\mathrm{x}+\mathrm{y}}\:=\:\mathrm{7}}\end{cases} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:? \\ $$
Commented by bramlexs22 last updated on 09/Sep/21
$$\mathrm{yes} \\ $$
Answered by mindispower last updated on 09/Sep/21
$$\left(\mathrm{2}\right)^{\mathrm{2}} −\left(\mathrm{3}\right)^{\mathrm{2}} \Rightarrow\left(\sqrt{{y}\left({z}+{x}\right)}−\sqrt{{z}\left({x}+{y}\right)}\right)=\mathrm{0} \\ $$$$\Rightarrow{y}\left({z}+{x}\right)={z}\left({x}+{y}\right)\mathrm{9} \\ $$$${x}\left({y}−{z}\right)=\mathrm{0} \\ $$$${x}=\mathrm{0}\Rightarrow{y}+{z}=\mathrm{25},\sqrt{{y}}+\sqrt{{z}}=\mathrm{7}\Rightarrow\mathrm{2}\sqrt{{yz}}=\mathrm{24} \\ $$$$\Rightarrow{yz}=\mathrm{144} \\ $$$${T}^{\mathrm{2}} −\mathrm{25}{T}+\mathrm{144}=\mathrm{0}\Rightarrow{T}\in\left\{\mathrm{16},\mathrm{9}\right\} \\ $$$$\left(\mathrm{0},\mathrm{16},\mathrm{9}\right)\:{withe}\:{all}\:{permutation}\: \\ $$$${y}={z}\Rightarrow\sqrt{{x}}+\sqrt{\mathrm{2}{y}}=\mathrm{5} \\ $$$$\sqrt{{y}}+\sqrt{{x}+{y}}=\mathrm{7} \\ $$$$\left(\mathrm{1}\right)^{\mathrm{2}} \sqrt{{y}}+\sqrt{{x}+{y}}=\mathrm{7}\Rightarrow,{x}+{y}=\mathrm{49}+{y}−\mathrm{14}\sqrt{{y}} \\ $$$${x}=\mathrm{7}\left(\mathrm{7}−\mathrm{2}\sqrt{{y}}\right)\Rightarrow\sqrt{{y}}=\frac{\mathrm{1}}{\mathrm{14}}\left(\mathrm{49}−{x}\right) \\ $$$$\Rightarrow\sqrt{{x}}+\frac{\mathrm{7}}{\:\sqrt{\mathrm{2}}}−\frac{{x}}{\:\mathrm{7}\sqrt{\mathrm{2}}}=\mathrm{5}…{solve}\:{and}\:{find}\:{all}\:{solution} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$