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Solve-for-real-numbers-the-following-system-of-equations-x-2-yz-3-y-2-xz-1-z-2-xy-1-




Question Number 152892 by mathdanisur last updated on 02/Sep/21
Solve for real numbers the following  system of equations:   { ((x^2  - yz = 3)),((y^2  - xz = 1)),((z^2  - xy = - 1)) :}
$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{the}\:\mathrm{following} \\ $$$$\mathrm{system}\:\mathrm{of}\:\mathrm{equations}: \\ $$$$\begin{cases}{\mathrm{x}^{\mathrm{2}} \:-\:\mathrm{yz}\:=\:\mathrm{3}}\\{\mathrm{y}^{\mathrm{2}} \:-\:\mathrm{xz}\:=\:\mathrm{1}}\\{\mathrm{z}^{\mathrm{2}} \:-\:\mathrm{xy}\:=\:-\:\mathrm{1}}\end{cases} \\ $$
Commented by MJS_new last updated on 03/Sep/21
simply put y=px∧z=qx and it′s easy to solve
$$\mathrm{simply}\:\mathrm{put}\:{y}={px}\wedge{z}={qx}\:\mathrm{and}\:\mathrm{it}'\mathrm{s}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{solve} \\ $$
Commented by mathdanisur last updated on 03/Sep/21
Yes, x=±(6/3) ; y=∓(2/3) ; z=±(1/3)
$$\mathrm{Yes},\:\mathrm{x}=\pm\frac{\mathrm{6}}{\mathrm{3}}\:;\:\mathrm{y}=\mp\frac{\mathrm{2}}{\mathrm{3}}\:;\:\mathrm{z}=\pm\frac{\mathrm{1}}{\mathrm{3}} \\ $$

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