Solve the following equations:
$3 x^{2} + 165 = 16 x y$ ,
$7 x y + 3 y^{2} = 132$ .

x = \pm 4; y = \pm 2

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x = \pm 5; y = \pm 3

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x = \pm 3; y = \pm 4

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

x = \pm 1; y = \pm 2

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Solution

The correct options are
B $x = \pm 5; y = \pm 3$
C $x = \pm 3; y = \pm 4$
Given equations are $3 x^{2} + 165 = 16 x y$
$\Rightarrow 3 x^{2} - 16 x y = - 165$ .......(i)
and $7 x y + 3 y^{2} = 132$

Put $y = v x$
$3 x^{2} - 16 v x^{2} = - 165$ ........(ii)
$7 v x^{2} + 3 v^{2} x^{2} = 132$ .........(iii)

Dividing (ii) by (iii), we get

$\frac{3 x^{2} - 16 v x^{2}}{7 v x^{2} + 3 v^{2} x^{2}} = - \frac{165}{132} \Rightarrow \frac{3 - 16 v}{7 v + 3 v^{2}} = \frac{- 5}{4} \Rightarrow 12 - 64 v = - 35 v - 15 v^{2} \Rightarrow 15 v^{2} - 29 v + 12 = 0 \Rightarrow 15 v^{2} - 20 v - 9 v + 12 = 0 \Rightarrow 5 v (3 v - 4) - 3 (3 v - 4) = 0 \Rightarrow (5 v - 3) (3 v - 4) = 0 \Rightarrow v = \frac{3}{5}, \frac{4}{3} \Rightarrow y = \frac{3 x}{5}, \frac{4 x}{3}$

Substituting $y$ in (i), we have
(i) Put $y = \frac{3 x}{5}$

Therefore, $3 x^{2} - 16 x (\frac{3 x}{5}) = - 165$
$\Rightarrow \frac{- 33 x^{2}}{5} = - 165 \Rightarrow x^{2} = 25 \Rightarrow x = \pm 5$
Thus $y = \frac{3 (\pm 5)}{5} = \pm 3$

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

Therefore, $3 x^{2} - 16 x (\frac{4 x}{3}) = - 165$
$\Rightarrow \frac{- 55 x^{2}}{3} = - 165 \Rightarrow x = \pm 3$
Thus $y = \frac{4 (\pm 3)}{3} = \pm 4$

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