{ ((2x^2 +3y^2 −6xy=12)),((x^2 −y^2 =4)) :}


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Question Number 211548 by MrGaster last updated on 12/Sep/24
${ ((2x^2 +3y^2 −6xy=12)),((x^2 −y^2 =4)) :}$
${\begin{cases} 2 x^{2} + 3 y^{2} - 6 x y = 12 \\ x^{2} - y^{2} = 4 \end{cases}$
	Answered by Frix last updated on 12/Sep/24

	$(2) \Rightarrow y^{2} = x^{2} - 4$ $(1) \Rightarrow y = \frac{5 x^{2} - 24}{6 x}$ $Inserting in (1) or (2) gives$ $x^{4} + \frac{96 x^{2}}{11} - \frac{576}{11} = 0$ $x^{2} = - \frac{48}{11} \pm \frac{24 \sqrt{15}}{11}$ $. . .$
	Answered by Rasheed.Sindhi last updated on 13/Sep/24
	${ ((2x^2 +3y^2 −6xy=12...(i))),((x^2 −y^2 =4....(ii))) :} 2x^2 +3y^2 −6xy=3×4 2x^2 +3y^2 −6xy=3×(x^2 −y^2 ) x^2 +6xy−6y^2 =0............A x=((−6y±(√(36y^2 +24y^2 )) )/2) x=((−6y±2y(√(15)) )/2)=(−3±(√(15)) )y x^2 =(24∓6(√(15)) )y^2 (ii)⇒(24∓6(√(15)) )y^2 −y^2 =4 (23∓6(√(15)) )y^2 =4 y^2 =(4/(23∓6(√(15)) ))∙((23±6(√(15)) )/(23±6(√(15)) )) y^2 =((92±24(√(15)) )/(529−36(15))) =−((92)/(11))∓((24(√(15)) )/(11)) A⇒6y^2 −6xy−x^2 =0 y=((6x±(√(36x^2 +24x^2 )) )/(12)) y=((6x±2x(√(15)) )/(12))=(((3±(√(15)))/6))x y^2 =((24±6(√(15)) )/(36))x^2 (ii)⇒x^2 −((24±6(√(15)) )/(36))x^2 =4 (1−((24±6(√(15)) )/(36)))x^2 =4 (((12∓6(√(15)) )/(36)))x^2 =4 x^2 =((144)/(12∓6(√(15)) ))∙((12±6(√(15)) )/(12±6(√(15)) )) =((1728±864(√(15)))/(144−36(15)))=((−1728∓864(√(15)))/(396)) =−((48)/(11))∓((24(√(15)))/(11)) (x^2 ,y^2 )=(−((48)/(11))±((24(√(15)))/(11)) , −((92)/(11))±((24(√(15)) )/(11))) (x,y)={((√(−((48)/(11))+((24(√(15)))/(11)))) ,(√(−((92)/(11))+((24(√(15)) )/(11)))) ), (−(√(−((48)/(11))+((24(√(15)))/(11)))) ,−(√(−((92)/(11))+((24(√(15)) )/(11)))) ), ((√(−((48)/(11))−((24(√(15)))/(11)))) ,(√(−((92)/(11))−((24(√(15)) )/(11)))) ), (−(√(−((48)/(11))−((24(√(15)))/(11)))) ,−(√(−((92)/(11))−((24(√(15)) )/(11)))) )}$
	${\begin{cases} 2 x^{2} + 3 y^{2} - 6 x y = 12 . . . (i) \\ x^{2} - y^{2} = 4 . . . . (i i) \end{cases}$ $2 x^{2} + 3 y^{2} - 6 x y = 3 \times 4$ $2 x^{2} + 3 y^{2} - 6 x y = 3 \times (x^{2} - y^{2})$ $x^{2} + 6 x y - 6 y^{2} = 0 . . . . . . . . . . . . A$ $x = \frac{- 6 y \pm \sqrt{36 y^{2} + 24 y^{2}}}{2}$ $x = \frac{- 6 y \pm 2 y \sqrt{15}}{2} = (- 3 \pm \sqrt{15}) y$ $x^{2} = (24 \mp 6 \sqrt{15}) y^{2}$ $(i i) \Rightarrow (24 \mp 6 \sqrt{15}) y^{2} - y^{2} = 4$ $(23 \mp 6 \sqrt{15}) y^{2} = 4$ $y^{2} = \frac{4}{23 \mp 6 \sqrt{15}} \cdot \frac{23 \pm 6 \sqrt{15}}{23 \pm 6 \sqrt{15}}$ $y^{2} = \frac{92 \pm 24 \sqrt{15}}{529 - 36 (15)}$ $= - \frac{92}{11} \mp \frac{24 \sqrt{15}}{11}$ $A \Rightarrow 6 y^{2} - 6 x y - x^{2} = 0$ $y = \frac{6 x \pm \sqrt{36 x^{2} + 24 x^{2}}}{12}$ $y = \frac{6 x \pm 2 x \sqrt{15}}{12} = (\frac{3 \pm \sqrt{15}}{6}) x$ $y^{2} = \frac{24 \pm 6 \sqrt{15}}{36} x^{2}$ $(i i) \Rightarrow x^{2} - \frac{24 \pm 6 \sqrt{15}}{36} x^{2} = 4$ $(1 - \frac{24 \pm 6 \sqrt{15}}{36}) x^{2} = 4$ $(\frac{12 \mp 6 \sqrt{15}}{36}) x^{2} = 4$ $x^{2} = \frac{144}{12 \mp 6 \sqrt{15}} \cdot \frac{12 \pm 6 \sqrt{15}}{12 \pm 6 \sqrt{15}}$ $= \frac{1728 \pm 864 \sqrt{15}}{144 - 36 (15)} = \frac{- 1728 \mp 864 \sqrt{15}}{396}$ $= - \frac{48}{11} \mp \frac{24 \sqrt{15}}{11}$ $(x^{2}, y^{2}) = (- \frac{48}{11} \pm \frac{24 \sqrt{15}}{11}, - \frac{92}{11} \pm \frac{24 \sqrt{15}}{11})$ $(x, y) = {(\sqrt{- \frac{48}{11} + \frac{24 \sqrt{15}}{11}}, \sqrt{- \frac{92}{11} + \frac{24 \sqrt{15}}{11}}),$ $(- \sqrt{- \frac{48}{11} + \frac{24 \sqrt{15}}{11}}, - \sqrt{- \frac{92}{11} + \frac{24 \sqrt{15}}{11}}),$ $(\sqrt{- \frac{48}{11} - \frac{24 \sqrt{15}}{11}}, \sqrt{- \frac{92}{11} - \frac{24 \sqrt{15}}{11}}),$ $(- \sqrt{- \frac{48}{11} - \frac{24 \sqrt{15}}{11}}, - \sqrt{- \frac{92}{11} - \frac{24 \sqrt{15}}{11}})}$
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