{ ((x^2 −xy+y^2 =7)),(((x+3)(y−2)=(√(xy+3))+(√(xy−2)))) :} Find x,y


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Question Number 143584 by Huy last updated on 16/Jun/21
${ ((x^2 −xy+y^2 =7)),(((x+3)(y−2)=(√(xy+3))+(√(xy−2)))) :} Find x,y$
${\begin{cases} x^{2} - x y + y^{2} = 7 \\ (x + 3) (y - 2) = \sqrt{x y + 3} + \sqrt{x y - 2} \end{cases}$ $F i n d x, y$
	Answered by MJS_new last updated on 16/Jun/21
	$assuming both (√(xy+3)) and (√(xy−2)) are ∈N the only possibility is xy=6 ⇒ y=(6/x) inserting and transforming we get { ((x^4 −13x^2 +36=0)),((x^2 +((5x)/2)−9=0)) :} ⇒ x=2 ⇒ y=3$
	$assuming both \sqrt{x y + 3} and \sqrt{x y - 2} are \in N$ $the only possibility is x y = 6 \Rightarrow y = \frac{6}{x}$ $inserting and transforming we get$ ${\begin{cases} x^{4} - 13 x^{2} + 36 = 0 \\ x^{2} + \frac{5 x}{2} - 9 = 0 \end{cases}$ $\Rightarrow x = 2 \Rightarrow y = 3$
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