Solve in R^3 { ((x^2 +2xy+y^2 −4z^2 =0)),((3x−2y+z=3)) :} x^2 +y^2 +3z^2 −4xy+yz+x+2y−3z+7=0


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Question Number 107065 by mathocean1 last updated on 08/Aug/20
$Solve in R^3 { ((x^2 +2xy+y^2 −4z^2 =0)),((3x−2y+z=3)) :} x^2 +y^2 +3z^2 −4xy+yz+x+2y−3z+7=0$
$S o l v e i n R^{3}$ ${\begin{cases} x^{2} + 2 x y + y^{2} - 4 z^{2} = 0 \\ 3 x - 2 y + z = 3 \end{cases}$ $x^{2} + y^{2} + 3 z^{2} - 4 x y + y z + x + 2 y - 3 z + 7 = 0$
	Answered by Her_Majesty last updated on 09/Aug/20

	$I I \Rightarrow z = - 3 x + 2 y + 3$ $I + I I I \Rightarrow y = \frac{7 x^{2} - 24 x + 8}{7 x - 11}$ $\Rightarrow z = \frac{- 7 x^{2} + 6 x - 17}{7 x - 11}$ $\Rightarrow$ $I o r I I I$ $\frac{(23 x + 26) (28 x^{2} - 47 x + 42)}{{(7 x - 11)}^{2}} = 0$ $\Rightarrow$ $x = - \frac{26}{23}, y = - \frac{268}{115}, z = \frac{199}{115}$
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