Question Number 114375 by A8;15: last updated on 18/Sep/20
Commented by A8;15: last updated on 18/Sep/20
please help
Commented by MJS_new last updated on 18/Sep/20
$${x}=\mathrm{9}\:{y}=\mathrm{4}\:{z}=\mathrm{1} \\ $$$$\mathrm{sorry}\:\mathrm{I}\:\mathrm{saw}\:\mathrm{this}\:\mathrm{at}\:\mathrm{first}\:\mathrm{glance}... \\ $$
Commented by A8;15: last updated on 19/Sep/20
thanks sir
Answered by bemath last updated on 19/Sep/20
$${u}+{v}^{\mathrm{2}} +{w}^{\mathrm{2}} =\mathrm{8}\rightarrow{u}=\mathrm{8}−\left({v}^{\mathrm{2}} +{w}^{\mathrm{2}} \right) \\ $$$${u}^{\mathrm{2}} +{v}+{w}^{\mathrm{2}} =\mathrm{12}\rightarrow{u}^{\mathrm{2}} =\mathrm{12}−{v}−{w}^{\mathrm{2}} \\ $$$${u}^{\mathrm{2}} +{v}^{\mathrm{2}} +{w}=\mathrm{14}\rightarrow{u}^{\mathrm{2}} =\mathrm{14}−{v}^{\mathrm{2}} −{w} \\ $$$$\Leftrightarrow\:\mathrm{12}−{v}−{w}^{\mathrm{2}} =\mathrm{14}−{v}^{\mathrm{2}} −{w} \\ $$$$\Leftrightarrow{v}^{\mathrm{2}} −{v}\:=\mathrm{2}+{w}^{\mathrm{2}} −{w} \\ $$$$\Leftrightarrow{v}^{\mathrm{2}} −{w}^{\mathrm{2}} −\left({v}−{w}\right)=\mathrm{2} \\ $$$$\Leftrightarrow\:\left({v}−{w}\right)\left\{\left({v}+{w}\right)−\mathrm{1}\right\}=\mathrm{2} \\ $$$${v}−{w}\:=\:\frac{\mathrm{2}}{{v}+{w}−\mathrm{1}}\: \\ $$
Commented by MJS_new last updated on 19/Sep/20
$$\mathrm{yes}\:\mathrm{but}\:\mathrm{try}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{it}\:\mathrm{to}\:\mathrm{the}\:\mathrm{end}.\:\mathrm{it}'\mathrm{s}\:\mathrm{getting} \\ $$$$\mathrm{complicated} \\ $$$$\mathrm{if}\:\mathrm{we}\:\mathrm{have}\:\mathrm{nice}\:\mathrm{numbers}\:\mathrm{as}\:\mathrm{here}\:\mathrm{it}'\mathrm{s}\:\mathrm{possible} \\ $$$$\mathrm{to}\:\mathrm{find}\:\mathrm{them}\:\mathrm{but}\:\mathrm{trt} \\ $$$$\sqrt{{x}}+{y}+{z}=\mathrm{7} \\ $$$${x}+\sqrt{{y}}+{z}=\mathrm{11} \\ $$$${x}+{y}+\sqrt{{z}}=\mathrm{13} \\ $$
Commented by A8;15: last updated on 19/Sep/20
thank you sir !