solve: x+3y=1 xy=y^2 −3


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Question Number 171836 by Mikenice last updated on 21/Jun/22

$s o l v e :$ $x + 3 y = 1$ $x y = y^{2} - 3$
	Answered by Rasheed.Sindhi last updated on 21/Jun/22
	$x+3y=1⇒x=1−3y xy=y^2 −3⇒y(1−3y)=y^2 −3 ⇒y−3y^2 =y^2 −3 4y^2 −y−3=0 (y−1)(4y+3)=0 y=1 ∣ y=−(3/4) •x=1−3y=1−3(1)=−2 • x=1−3(−(3/4))=((13)/4) (x,y)={(((13)/4),−(3/4)),(−2,1)}$
	$x + 3 y = 1 \Rightarrow x = 1 - 3 y$ $x y = y^{2} - 3 \Rightarrow y (1 - 3 y) = y^{2} - 3$ $\Rightarrow y - 3 y^{2} = y^{2} - 3$ $4 y^{2} - y - 3 = 0$ $(y - 1) (4 y + 3) = 0$ $y = 1 ∣ y = - \frac{3}{4}$ $∙ x = 1 - 3 y = 1 - 3 (1) = - 2$ $∙ x = 1 - 3 (- \frac{3}{4}) = \frac{13}{4}$ $(x, y) = {(\frac{13}{4}, - \frac{3}{4}), (- 2, 1)}$
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