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The number of ordered pairs $(x, y)$ where $x, y ϵ$ R satisfying $x^{2} + y^{2} - x y = 4 (x + y - 4)$ is

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Solution

$x^{2} + y^{2} - x y = 4 (x + y - 4)$
$2 x^{2} + 2 y^{2} - 2 x y = 8 (x + y - 4)$
$(x^{2} - 2 x y + y^{2}) + (x^{2} - 8 x + 16) + (y^{2} - 8 y + 16) = 0$
$(x - y)^{2} + (x - 4)^{2} + (y - 4)^{2} = 0$
$\Rightarrow x - y = 0, x - 4 = 0, y - 4 = 0$
Hence there is only one solution $(4, 4)$

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