Solve the following equations:
$5 y^{2} - 7 x^{2} = 17$ ,
$5 x y - 6 x^{2} = 6$ .

x = \pm 2; y = \pm 3

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x = \pm 3; y = \pm 4

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x = \pm 1; y = \pm 2

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x = \pm 4; y = \pm 1

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Solution

The correct options are
A $x = \pm 2; y = \pm 3$
D $x = \pm 3; y = \pm 4$
Given equations are ${5 y}^{2} - {7 x}^{2} = 17$ .....(i)
and $5 x y - 6 x^{2} = 6$

Put $y = v x$
$5 v^{2} x^{2} - {7 x}^{2} = 17$ ......(ii)
$5 v x^{2} - {6 x}^{2} = 6$ ......(iii)

Dividing (ii) by (iii), we get
$\frac{5 v^{2} x^{2} - {7 x}^{2}}{5 v x^{2} - {6 x}^{2}} = \frac{17}{6} \Rightarrow \frac{5 v^{2} - 7}{5 v - 6} = \frac{17}{6} \Rightarrow 30 v^{2} - 42 = 85 v - 102 \Rightarrow 30 v^{2} - 85 v + 60 = 0 \Rightarrow 6 v^{2} - 17 v + 12 = 0 \Rightarrow 6 v^{2} - 9 v - 8 v + 12 = 0 \Rightarrow 3 v (2 v - 3) - 4 (2 v - 3) = 0 \Rightarrow (3 v - 4) (2 v - 3) = 0 \Rightarrow v = \frac{4}{3}, \frac{3}{2} \Rightarrow y = \frac{4 x}{3}, \frac{3 x}{2}$

Substituting in (i), we get
${5 y}^{2} - {7 x}^{2} = 17$

(i) Put $y = \frac{4 x}{3}$

Therefore, $5 {(\frac{4 x}{3})}^{2} - 7 x^{2} = 17$
$\Rightarrow \frac{17 x^{2}}{9} = 17 \Rightarrow x = \pm 3$
Thus $y = \frac{4 (\pm 3)}{3} = \pm 4$

(ii) Put $y = \frac{3 x}{2}$

Therefore, $5 {(\frac{3 x}{2})}^{2} - 7 x^{2} = 17$
$\frac{17 x^{2}}{4} = 17 \Rightarrow x = \pm 2$
Thus $y = \frac{3 (\pm 2)}{2} = \pm 3$

Suggest Corrections

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