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 A $4$
$x + y^{2} = x^{2} + y = 12$
$x + y^{2} = x^{2} + y$
$x - y = x^{2} - y^{2} \Rightarrow x = y, x + y = 1$
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 value of $x$ there are two value of $y$ .
So four pair.

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