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Question Number 161159 by daus last updated on 13/Dec/21

	Answered by Rasheed.Sindhi last updated on 13/Dec/21

	$x = \frac{2 + y^{2}}{y}; y \neq 0$ ${(\frac{2 + y^{2}}{y})}^{2} - 5 (\frac{2 + y^{2}}{y}) y + 6 y^{2} = 0$ ${(2 + y^{2})}^{2} - 10 y^{2} - 5 y^{4} + 6 y^{4} = 0$ $4 + 4 y^{2} + y^{4} - 10 y^{2} - 5 y^{4} + 6 y^{4} = 0$ $y^{4} - 3 y^{2} + 2 = 0$ $(y^{2} - 1) (y^{2} - 2) = 0$ $y = \pm 1, \pm \sqrt{2}$ $x = \frac{2 + y^{2}}{y} = \frac{2 + {(\pm 1)}^{2}}{\pm 1} = \pm 3$ $x = \frac{2 + {(\pm \sqrt{2})}^{2}}{\pm \sqrt{2}} \frac{4}{\pm \sqrt{2}} . \frac{\pm \sqrt{2}}{\pm \sqrt{2}} = \pm 2 \sqrt{2}$ $(x, y) = (\pm 3, \pm 1), (\pm 2 \sqrt{2}, \pm \sqrt{2})$
	Commented by daus last updated on 13/Dec/21

	$t h a n k s$
	Answered by Rasheed.Sindhi last updated on 13/Dec/21
	$x^2 −5xy+6y^2 =0_((i)) ∧ xy−y^2 =2_((ii)) ; x,y=? (i):x^2 −5xy+6y^2 =0 (x−2y)(x−3y)=0 x=2y ∣ x=3y xy−y^2 ^((ii): ) =2⇒ { ((2y^2 −y^2 =2⇒y=±(√2) ...(iii))),((3y^2 −y^2 =2⇒y=±1...(iv))) :} (iii)⇒x=2y=2(±(√2) )=±2(√2) (iv)⇒x=3y=3(±1)=±3 (x,y)=(±2(√2) ,±(√2) ),(±3,±1)$
	$\underset{(i)}{\underset{⏟}{x^{2} - 5 x y + 6 y^{2} = 0}} \land \underset{(i i)}{\underset{⏟}{x y - y^{2} = 2}}; x, y = ?$ $(i) : x^{2} - 5 x y + 6 y^{2} = 0$ $(x - 2 y) (x - 3 y) = 0$ $x = 2 y ∣ x = 3 y$ $\overset{(i i) :}{x y - y^{2}} = 2 \Rightarrow {\begin{cases} 2 y^{2} - y^{2} = 2 \Rightarrow y = \pm \sqrt{2} . . . (i i i) \\ 3 y^{2} - y^{2} = 2 \Rightarrow y = \pm 1 . . . (i v) \end{cases}$ $(i i i) \Rightarrow x = 2 y = 2 (\pm \sqrt{2}) = \pm 2 \sqrt{2}$ $(i v) \Rightarrow x = 3 y = 3 (\pm 1) = \pm 3$ $(x, y) = (\pm 2 \sqrt{2}, \pm \sqrt{2}), (\pm 3, \pm 1)$
	Commented by daus last updated on 13/Dec/21

	$t h a n k s$
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