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Question Number 109377 by bemath last updated on 23/Aug/20

	Answered by john santu last updated on 23/Aug/20
	$((≈ JS ≈)/(★■★)) { ((2x^2 +5y^2 +4xy = 12)),((3x^2 +3y^2 +3xy = 12)) :} ⇔ x^2 −2y^2 −xy = 0 ⇒(x−2y)(x+y) = 0 { ((x=2y)),((x=−y)) :} (•) x=2y ⇒4y^2 +y^2 +2y^2 =4 y = ± (2/3)⇒ x= ±(4/3) (••)x=−y ⇒y^2 +y^2 −y^2 =4 y = ±2 ⇒x = ∓ 2 the solution set is {(±(4/3),±(2/3)),(∓2, ±2 )}$
	$\frac{\approx J S \approx}{★ ◼ ★} {\begin{cases} 2 x^{2} + 5 y^{2} + 4 x y = 12 \\ 3 x^{2} + 3 y^{2} + 3 x y = 12 \end{cases}$ $\Leftrightarrow x^{2} - 2 y^{2} - x y = 0$ $\Rightarrow (x - 2 y) (x + y) = 0$ ${\begin{cases} x = 2 y \\ x = - y \end{cases}$ $(∙) x = 2 y \Rightarrow 4 y^{2} + y^{2} + 2 y^{2} = 4$ $y = \pm \frac{2}{3} \Rightarrow x = \pm \frac{4}{3}$ $(∙ ∙) x = - y \Rightarrow y^{2} + y^{2} - y^{2} = 4$ $y = \pm 2 \Rightarrow x = \mp 2$ $t h e s o l u t i o n s e t i s {(\pm \frac{4}{3}, \pm \frac{2}{3}), (\mp 2, \pm 2)}$
	Commented by bobhans last updated on 23/Aug/20

	$\frac{j o o s s - - - - -}{}$
	Commented by bemath last updated on 23/Aug/20

	$c o o l l$
	Commented by mnjuly1970 last updated on 23/Aug/20

	Commented by mnjuly1970 last updated on 23/Aug/20

	$p l e a s e s o l v e :$
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