Question Number 223965 by Rojarani last updated on 11/Aug/25
Answered by Frix last updated on 11/Aug/25
$${z}={x}+{y}−\mathrm{7} \\ $$$$\mathrm{Insert}\:\mathrm{in}\:\left(\mathrm{2}\right)\:\Rightarrow\:{y}=\frac{\mathrm{7}{x}−\mathrm{43}}{{x}−\mathrm{7}} \\ $$$$\mathrm{Insert}\:\mathrm{in}\:\left(\mathrm{3}\right)\:\Rightarrow\:{x}^{\mathrm{2}} −\mathrm{19}{x}+\mathrm{90}=\mathrm{0} \\ $$$$\Rightarrow \\ $$$${x}=\mathrm{9}\wedge{y}=\mathrm{10}\wedge{z}=\mathrm{12} \\ $$$$\vee \\ $$$${x}=\mathrm{10}\wedge{y}=\mathrm{9}\wedge{z}=\mathrm{12} \\ $$
Answered by mr W last updated on 11/Aug/25
$${x}+{y}=\mathrm{7}+{z} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{2}{xy} \\ $$$$\mathrm{37}+{z}^{\mathrm{2}} =\left(\mathrm{7}+{z}\right)^{\mathrm{2}} −\mathrm{2}{xy} \\ $$$$\mathrm{37}=\mathrm{49}+\mathrm{14}{z}−\mathrm{2}{xy} \\ $$$$\Rightarrow{xy}=\mathrm{6}+\mathrm{7}{z} \\ $$$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} =\left({x}+{y}\right)^{\mathrm{3}} −\mathrm{3}{xy}\left({x}+{y}\right) \\ $$$$\mathrm{1}+{z}^{\mathrm{3}} =\left(\mathrm{7}+{z}\right)^{\mathrm{3}} −\mathrm{3}\left(\mathrm{6}+\mathrm{7}{z}\right)\left(\mathrm{7}+{z}\right) \\ $$$$\Rightarrow{z}=\mathrm{12} \\ $$$$\Rightarrow{x}+{y}=\mathrm{19},\:{xy}=\mathrm{90} \\ $$$$\Rightarrow\left({x},\:{y}\right)=\left(\mathrm{9},\:\mathrm{10}\right) \\ $$
Commented by Rojarani last updated on 12/Aug/25
$${Sir},\:{excellent}. \\ $$