Question-160302

Question Number 160302 by cortano last updated on 27/Nov/21

Answered by Rasheed.Sindhi last updated on 27/Nov/21

{\begin{cases} 8 x^{2} - 8 y^{2} = 24 \\ 3 x^{2} + 3 x y + 6 y^{2} = 24 \end{cases} \Rightarrow 5 x^{2} - 3 x y - 14 y^{2} = 0

\Rightarrow 5 x^{2} - 10 x y + 7 x y - 14 y^{2} = 0

5 x (x - 2 y) + 7 y (x - 2 y) = 0

(x - 2 y) (5 x + 7 y) = 0

x = 2 y ∣ x = - \frac{7 y}{5}

\underset{―}{x = 2 y}

x^{2} - y^{2} = 3 \Rightarrow 4 y^{2} - y^{2} = 3 \Rightarrow y = \pm 1

\Rightarrow x = 2 (\pm 1) = \pm 2

(x, y) = (1, 2), (- 1, - 2)

x = - \frac{7 y}{5}

{(- \frac{7 y}{5})}^{2} - y^{2} = 3 \Rightarrow 49 y^{2} - 25 y^{2} = 75

\Rightarrow 24 y^{2} = 75 \Rightarrow y^{2} = \frac{25}{8} \Rightarrow y = \pm \frac{5}{2 \sqrt{2}}

\Rightarrow x = - \frac{7}{5} (\pm \frac{5}{2 \sqrt{2}}) = \mp \frac{7}{2 \sqrt{2}}

(x, y) = (\frac{7}{2 \sqrt{2}}, - \frac{5}{2 \sqrt{2}}), (- \frac{7}{2 \sqrt{2}}, \frac{5}{2 \sqrt{2}})

(x, y) = (1, 2), (- 1, - 2), (\frac{7}{2 \sqrt{2}}, - \frac{5}{2 \sqrt{2}}), (- \frac{7}{2 \sqrt{2}}, \frac{5}{2 \sqrt{2}})

Commented by cortano last updated on 27/Nov/21

yes nice

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