Question Number 171756 by Mikenice last updated on 20/Jun/22
$${solve}\:{for}\:{real}\:{numbers}: \\ $$$$\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{y}^{\mathrm{2}} +{z}^{\mathrm{2}} =\mathrm{7} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} =\sqrt{\mathrm{2}}\:{z}\left({x}+{y}\right) \\ $$
Commented by mr W last updated on 20/Jun/22
$${you}\:{cant}\:{solve}\:{for}\:\mathrm{3}\:{variables}\:{from} \\ $$$${only}\:\mathrm{2}\:{equations}. \\ $$
Commented by Mikenice last updated on 20/Jun/22
$${it}\:{is}\:{possible},{although}\:{i}\:{don}'{t}\:{know}\:{the}\:{answer}\:{but}\:{it}\:{is}\:{possible}\:{for}\:{getting}\:{three}\:{variable}\:{in}\:{two}\:{equation}. \\ $$
Commented by mr W last updated on 21/Jun/22
$${x}={y}=\pm\mathrm{1} \\ $$$${z}=\pm\sqrt{\mathrm{2}} \\ $$
Commented by Mikenice last updated on 21/Jun/22
$${please}\:{sir}\:{show}\:{the}\:{solution} \\ $$