Solve the following equations:
$4 x^{2} + 5 y = 6 + 20 x y - 25 y^{2} + 2 x, 7 x - 11 y = 17$

4, 1

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\frac{107}{13}, \frac{48}{13}

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5, 2

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12, 5

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Solution

The correct options are
A $4, 1$
C $\frac{107}{13}, \frac{48}{13}$
Given equations are ${4 x}^{2} + 5 y = 6 + 20 x y - 25 y^{2} + 2 x$ .......(i)

and $7 x - 11 y = 17$
$\Rightarrow 7 x = 17 + 11 y$
$\Rightarrow x = \frac{17 + 11 y}{7}$

Substituting $x$ in (i), we get

${4 x}^{2} + 5 y = 6 + 20 x y - 25 y^{2} + 2 x$ .......(ii)
$\Rightarrow 4 {(\frac{17 + 11 y}{7})}^{2} + 5 y = 6 + 20 (\frac{17 + 11 y}{7}) y - 25 y^{2} + 2 (\frac{17 + 11 y}{7}) \Rightarrow \frac{1156 + 484 y^{2} + 1469 y + 245 y}{49} = \frac{42 + 340 y + 220 y^{2} - 175 y^{2} + 34 + 22 y}{7} \Rightarrow 484 y^{2} + 1741 y + 1156 = 7 (45 y^{2} 362 y + 76) \Rightarrow 484 y^{2} + 1741 y + 1156 = 315 y^{2} + 2534 y + 532 \Rightarrow 169 y^{2} - 793 y + 624 = 0 \Rightarrow 13 y^{2} - 61 y + 48 = 0 \Rightarrow 13 y^{2} - 13 y - 48 y + 48 = 0 \Rightarrow 13 y (y - 1) - 48 (y - 1) = 0 \Rightarrow (13 y - 48) (y - 1) = 0 \Rightarrow y = \frac{48}{13}, 1 \Rightarrow x = \frac{17 + 11 y}{7}$
For $y = 1 \Rightarrow x = \frac{17 + 11}{7} = 4$
For $y = \frac{48}{13} \Rightarrow x = \frac{17 + 11 (\frac{48}{13})}{7} = \frac{107}{13}$

So, the values of $x$ are $4, \frac{107}{13}$ and values of $y$ are $1, \frac{48}{13}$ .

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