The number of ordered pairs (x,y) of real numbers that satisfy the simultaneous equations $x + y^{2} = x^{2} + y = 12$ is

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Solution

The correct option is D
4

$x + y^{2} = x^{2} + y = 12 x + y^{2} = x^{2} + y x - y = x^{2} - y^{2} \Rightarrow x = y, x + y = 1 When x = y x^{2} + x = 12 x^{2} + x - 12 = 0 x^{2} + 4 x - 3 x - 12 = 0 (x + 4) (x - 3) = 0 x = - 4, 3 (3, 3) (- 4, - 4) When y = 1 - x x + (1 + x)^{2} = 12 x^{2} - x - 11 = 0 x = \frac{1 \pm \sqrt{1 + 44}}{2}$ for two values of x there are two values of y. So four pairs.

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