Solve the following equations:
$2 x^{2} - x y + y^{2} = 2 y, 2 x^{2} + 4 x y = 5 y$ .

(0, 0)

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(1, 2)

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(2, 3)

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(0, 1)

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Solution

The correct options are
B $(0, 0)$
C $(1, 2)$
Given, $2 x^{2} + 4 x y = 5 y$ .......(i)
Also given $2 x^{2} - x y + y^{2} = 2 y$

Substituting $y$ on the R.H.S from (i)

$2 x^{2} - x y + y^{2} = 2 (\frac{2 x^{2} + 4 x y}{5}) \Rightarrow 10 x^{2} - 5 x y + 5 y^{2} = 4 x^{2} + 8 x y \Rightarrow 6 x^{2} - 5 x y + 5 y^{2} = 8 x y \Rightarrow 6 x^{2} - 5 x y + 2 x y = 8 x y - 5 y^{2} + 2 x y \Rightarrow 6 x^{2} - 3 x y = 10 x y - 5 y^{2} \Rightarrow 3 x (2 x - y) = 5 y (2 x - y) \Rightarrow 3 x (2 x - y) - 5 y (2 x - y) = 0 \Rightarrow (3 x - 5 y) (2 x - y) = 0 \Rightarrow 3 x = 5 y, 2 x = y$

Substituting $y$ in (i), we get

(1). $3 x = 5 y$

$\Rightarrow y = \frac{3 x}{5}$
Therefore, $2 x^{2} + 4 x . \frac{3 x}{5} = 5. \frac{3 x}{5}$
$\Rightarrow 22 x^{2} = 15 x \Rightarrow 22 x^{2} - 15 x = 0 \Rightarrow x (22 x - 15) = 0 \Rightarrow x = 0, \frac{15}{22}$
$y = \frac{3 x}{5} x = 0 \Rightarrow y = 0 x = \frac{15}{22} \Rightarrow y = \frac{9}{22}$

(2). $y = 2 x$

$2 x^{2} + 4 x (2 x) = 5 (2 x) \Rightarrow 10 x^{2} - 10 x = 0 \Rightarrow 10 x (x - 1) = 0 \Rightarrow x = 0, 1 y = 2 x x = 0 \Rightarrow y = 0 x = 1 \Rightarrow y = 2$

So, the values of $x$ are $0, 1, \frac{15}{22}$ and the corresponding values of $y$ are $0, 2, \frac{9}{22}$ .

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