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xy-yz-8-yz-xz-9-zx-xy-5-




Question Number 93039 by i jagooll last updated on 10/May/20
 { ((xy+yz = 8)),((yz+xz = 9)),((zx+xy = 5)) :}
$$\begin{cases}{{xy}+{yz}\:=\:\mathrm{8}}\\{{yz}+{xz}\:=\:\mathrm{9}}\\{{zx}+{xy}\:=\:\mathrm{5}}\end{cases} \\ $$
Commented by i jagooll last updated on 10/May/20
2xy+2xz+2yz = 22  xy+xz+yz = 11  ⇒ xz = 3 , yz = 6 , xy = 2   (xyz)^2  = 36 ⇒ xyz = ± 6  (1) xyz=6  { ((x=1)),((y= 2)),((z= 3)) :}  (2) xyz = −6  { ((x=−1)),((y=−2)),((z=−3)) :}
$$\mathrm{2xy}+\mathrm{2xz}+\mathrm{2yz}\:=\:\mathrm{22} \\ $$$$\mathrm{xy}+\mathrm{xz}+\mathrm{yz}\:=\:\mathrm{11} \\ $$$$\Rightarrow\:\mathrm{xz}\:=\:\mathrm{3}\:,\:\mathrm{yz}\:=\:\mathrm{6}\:,\:\mathrm{xy}\:=\:\mathrm{2}\: \\ $$$$\left(\mathrm{xyz}\right)^{\mathrm{2}} \:=\:\mathrm{36}\:\Rightarrow\:\mathrm{xyz}\:=\:\pm\:\mathrm{6} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{xyz}=\mathrm{6}\:\begin{cases}{\mathrm{x}=\mathrm{1}}\\{\mathrm{y}=\:\mathrm{2}}\\{\mathrm{z}=\:\mathrm{3}}\end{cases} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{xyz}\:=\:−\mathrm{6}\:\begin{cases}{\mathrm{x}=−\mathrm{1}}\\{\mathrm{y}=−\mathrm{2}}\\{\mathrm{z}=−\mathrm{3}}\end{cases} \\ $$

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