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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

$$x=9y=4z=1$$$$\mathrm{sorry}\mathrm{I}\mathrm{saw}\mathrm{this}\mathrm{at}\mathrm{first}\mathrm{glance}...$$

Commented by A8;15: last updated on 19/Sep/20

thanks sir

Answered by bemath last updated on 19/Sep/20

$$u+v^2+w^2=8\to u=8-(v^2+w^2)$$$$u^2+v+w^2=12\to u^2=12-v-w^2$$$$u^2+v^2+w=14\to u^2=14-v^2-w$$$$\iff 12-v-w^2=14-v^2-w$$$$\iff v^2-v=2+w^2-w$$$$\iff v^2-w^2-(v-w)=2$$$$\iff (v-w)\{(v+w)-1\}=2$$$$v-w=\frac{2}{v+w-1}$$

Commented by MJS_new last updated on 19/Sep/20

$$\mathrm{yes}\mathrm{but}\mathrm{try}\mathrm{to}\mathrm{solve}\mathrm{it}\mathrm{to}\mathrm{the}\mathrm{end}.{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{getting}$$$$\mathrm{complicated}$$$$\mathrm{if}\mathrm{we}\mathrm{have}\mathrm{nice}\mathrm{numbers}\mathrm{as}\mathrm{here}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{possible}$$$$\mathrm{to}\mathrm{find}\mathrm{them}\mathrm{but}\mathrm{trt}$$$$\sqrt{x}+y+z=7$$$$x+\sqrt{y}+z=11$$$$x+y+\sqrt{z}=13$$

Commented by A8;15: last updated on 19/Sep/20

thank you sir !

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