Find the values of x and y x^2 − 2xy − y^2 = 14 ............ equation (i) 2x^2 + 3xy + y^2 = − 2 ............ equation (ii)


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Question Number 13166 by tawa tawa last updated on 15/May/17

$Find the values of x and y$ $x^{2} - 2 xy - y^{2} = 14 . . . . . . . . . . . . equation (i)$ ${2 x}^{2} + 3 xy + y^{2} = - 2 . . . . . . . . . . . . equation (ii)$
	Answered by RasheedSindhi last updated on 16/May/17
	$x^2 − 2xy − y^2 = 14 ..... (i) 2x^2 + 3xy + y^2 = − 2 ...(ii) (i)×1: x^2 − 2xy − y^2 = 14 (ii)×7: 14x^2 + 21xy + 7y^2 = −14 (i)×1 + (ii)×7: 15x^2 +19xy+6y^2 =0 (3x+2y)(5x+3y)=0 x=−(2/3)y ∣ x=−(3/5)y x=−(2/3)y ∧ x^2 − 2xy − y^2 = 14 ⇒(−(2/3)y)^2 −2(−(2/3)y)y−y^2 =14 4y^2 +12y^2 −9y^2 =126 7y^2 =126⇒y=±3(√2) x=−(2/3)(±3(√2))=∓2(√2) x=−(3/5)y ∧ x^2 − 2xy − y^2 = 14 ⇒(−(3/5)y)^2 −2(−(3/5)y)y−y^2 =14 9y^2 +30y^2 −25y^2 =350 14y^2 =350⇒y=±5 x=−(3/5)(±5)=∓3 {(±2(√2),∓3(√2)),(±3,∓5)}$
	$x^{2} - 2 xy - y^{2} = 14 . . . . . (i)$ ${2 x}^{2} + 3 xy + y^{2} = - 2 . . . (ii)$ $(i) \times 1 : x^{2} - 2 xy - y^{2} = 14$ $(ii) \times 7 : {14 x}^{2} + 21 xy + {7 y}^{2} = - 14$ $(i) \times 1 + (ii) \times 7 :$ ${15 x}^{2} + 19 xy + {6 y}^{2} = 0$ $(3 x + 2 y) (5 x + 3 y) = 0$ $x = - \frac{2}{3} y ∣ x = - \frac{3}{5} y$ $x = - \frac{2}{3} y \land x^{2} - 2 xy - y^{2} = 14$ $\Rightarrow {(- \frac{2}{3} y)}^{2} - 2 (- \frac{2}{3} y) y - y^{2} = 14$ ${4 y}^{2} + {12 y}^{2} - {9 y}^{2} = 126$ ${7 y}^{2} = 126 \Rightarrow y = \pm 3 \sqrt{2}$ $x = - \frac{2}{3} (\pm 3 \sqrt{2}) = \mp 2 \sqrt{2}$ $x = - \frac{3}{5} y \land x^{2} - 2 xy - y^{2} = 14$ $\Rightarrow {(- \frac{3}{5} y)}^{2} - 2 (- \frac{3}{5} y) y - y^{2} = 14$ ${9 y}^{2} + {30 y}^{2} - {25 y}^{2} = 350$ ${14 y}^{2} = 350 \Rightarrow y = \pm 5$ $x = - \frac{3}{5} (\pm 5) = \mp 3$ ${(\pm 2 \sqrt{2}, \mp 3 \sqrt{2}), (\pm 3, \mp 5)}$
	Commented by tawa tawa last updated on 16/May/17

	$God bless you sir$
	Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 16/May/17

	${(\frac{x}{y})}^{2} - 2 \frac{x}{y} - 1 = \frac{14}{y^{2}} . . . . . . . . (A)$ $2 {(\frac{x}{y})}^{2} + 3 \frac{x}{y} + 1 = \frac{- 2}{y^{2}} . . . . . . . (B) \Rightarrow (\frac{x}{y} = t) \Rightarrow$ $15 t^{2} + 19 t + 6 = 0 \Rightarrow t = \frac{- 19 \pm \sqrt{19^{2} - 24 \times 15}}{2 \times 15}$ $t = \frac{- 19 \pm 1}{30} = - \frac{2}{3}, - \frac{3}{5}$ $1) \frac{x}{y} = - \frac{2}{3} \Rightarrow x^{2} - 2 x (- \frac{3}{2} x) - {(- \frac{3}{2} x)}^{2} = 14$ $\Rightarrow x^{2} + 3 x^{2} - \frac{9}{4} x^{2} = 14 \Rightarrow x = \pm 2 \sqrt{2}, y = \mp 3 \sqrt{2}$ $2) \frac{x}{y} = - \frac{3}{5} \Rightarrow x^{2} - 2 x (- \frac{5}{3} x) - {(- \frac{5}{3} x)}^{2} = 14$ $\Rightarrow \frac{9 + 30 - 25}{9} x^{2} = 14 \Rightarrow x = \pm 3, y = \mp 5$ $\Rightarrow (x, y) = (\pm 2 \sqrt{2}, \mp 3 \sqrt{2}), (\pm 3, \mp 5) . ◼$
	Commented by tawa tawa last updated on 16/May/17

	$God bless you sir .$ $please sir . How did you get . {15 t}^{2} + 19 t + 6 = 0$
	Commented by tawa tawa last updated on 16/May/17

	$have seen it sir . you continue . .$
	Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 16/May/17

	$A + 7 B = 15 t^{2} + 19 t + 6 = 0$
	Commented by tawa tawa last updated on 16/May/17

	$I understand now sir .$ $God bless you sir$
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