Solve the following equations:
$3 x^{2} + x y + y^{2} = 15$ ,
$31 x y - 3 x^{2} - 5 y^{2} = 45$ .

x = \pm 2; y = \pm 1

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x = \pm 1; y = \pm 3

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x = \pm 2; y = \pm 2

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x == \pm 3; y = \pm 4

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Solution

The correct options are
B $x = \pm 1; y = \pm 3$
D $x = \pm 2; y = \pm 1$
$x^{2} + x y + y^{2} = 15$ ........(i)
$31 x y - 3 x^{2} - 5 y^{2} = 45$

Put $y = v x$

Thus $3 x^{2} + v x^{2} + v^{2} x^{2} = 15$ .......(ii)
$31 v x^{2} - 3 x^{2} - 5 v^{2} x^{2} = 45$ ......(iii)

Dividing (ii) by (iii), we get
$\frac{3 x^{2} + v x^{2} + v^{2} x^{2}}{31 v x^{2} - 3 x^{2} - 5 v^{2} x^{2}} = \frac{15}{45} \Rightarrow \frac{3 + v + v^{2}}{31 v - 3 - 5 v^{2}} = \frac{1}{3} \Rightarrow 9 + 3 v + 3 v^{2} = 31 v - 3 - 5 v^{2} \Rightarrow 8 v^{2} - 28 v + 12 = 0 \Rightarrow 2 v^{2} - 7 v + 3 = 0 \Rightarrow 2 v^{2} - 6 v - v + 3 = 0 \Rightarrow 2 v (v - 3) - 1 (v - 3) = 0 \Rightarrow (2 x - 1) (v - 3) = 0 \Rightarrow v = \frac{1}{2}, 3 \Rightarrow y = \frac{x}{2}, 3 x$

On substituting $y$ in (i), we get
$3 x^{2} + x y + y^{2} = 15$

(i) Put $y = \frac{x}{2}$

Therefore, $3 x^{2} + x (\frac{x}{2}) + {(\frac{x}{2})}^{2} = 15$
$\Rightarrow \frac{15 x^{2}}{4} = 15$
$\Rightarrow x = \pm 2$
Thus $y = \frac{(\pm 2)}{2} = \pm 1$

(ii) Put $y = 3 x$

Therefore, $3 x^{2} + x (3 x) + {(3 x)}^{2} = 15$
$\Rightarrow 15 x^{2} = 15 \Rightarrow x = \pm 1$
Thus $y = 3 (\pm 1) = \pm 3$

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