x , y ∈ R , x^( 2) + xy = 12 y^( 2) + 2xy = 7 −−−−− x , y =?


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Question Number 197475 by mnjuly1970 last updated on 19/Sep/23

$x, y \in R,$ $x^{2} + x y = 12$ $y^{2} + 2 x y = 7$ $- - - - - x, y = ?$
	Answered by Sutrisno last updated on 19/Sep/23

	$x^{2} + x y = 12 \to y = \frac{12}{x} - x$ ${(\frac{12}{x} - x)}^{2} + 2 x (\frac{12}{x} - x) = 7$ $\frac{144}{x^{2}} - 24 + x^{2} + 24 - 2 x^{2} = 7$ $x^{4} - 7 x^{2} - 144 = 0$ $(x + 3) (x - 3) (x^{2} + 16) = 0$ $x = - 3 a n d y = - 1$ $x = 3 a n d y = 1$
	Commented by mnjuly1970 last updated on 19/Sep/23

	$t h x a l o t s i r$
	Answered by Frix last updated on 19/Sep/23

	$y = p x$ $x^{2} (1 + p) = 12$ $x^{2} (p^{2} + 2 p) = 7$ $7 (p + 1) = 12 (p^{2} + 2 p)$ $p^{2} + \frac{17}{12} p - \frac{7}{12} = 0$ $p = - \frac{7}{4} \lor p = \frac{1}{3}$ $x^{2} = \frac{12}{p + 1}$ $x^{2} = - 16 \lor x^{2} = 9$ $x \in R \Rightarrow x = \pm 3 \Rightarrow$ $x = - 3 \land y = - 1 \lor x = 3 \land y = 1$
	Answered by Rasheed.Sindhi last updated on 21/Sep/23
	$7x^2 +7xy=84 ∧ 12y^2 +24xy=84 7x^2 −17xy−12y^2 =0 (x−3y)(7x+4y)=0 x=3y ∨ x=−(4/7)y y^( 2) + 2xy = 7 ⇒ (a) y^2 +2(3y)y=7⇒7y^2 =7⇒y=±1⇒x=±3 { ((x=3,y=1)),((x=−3,y=−1)) :} (b) y^2 +2(((−4y)/7))y=7⇒7y^2 −8y^2 =49⇒y=±7i∉R$
	$7 x^{2} + 7 x y = 84 \land 12 y^{2} + 24 x y = 84$ $7 x^{2} - 17 x y - 12 y^{2} = 0$ $(x - 3 y) (7 x + 4 y) = 0$ $x = 3 y \lor x = - \frac{4}{7} y$ $y^{2} + 2 x y = 7$ $\Rightarrow$ $(a) y^{2} + 2 (3 y) y = 7 \Rightarrow 7 y^{2} = 7 \Rightarrow y = \pm 1 \Rightarrow x = \pm 3$ ${\begin{cases} x = 3, y = 1 \\ x = - 3, y = - 1 \end{cases}$ $(b) y^{2} + 2 (\frac{- 4 y}{7}) y = 7 \Rightarrow 7 y^{2} - 8 y^{2} = 49 \Rightarrow y = \pm 7 i \notin R$
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