Find real values of $x, y, z$ satisfying the equations
$x + y = 2, x y - z^{2} = 1.$

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Solution

We have $x + y = 2, x y = 1 + z^{2}$
Eliminating $y$ , we get
$x (2 - x) = 1 + z^{2}$
$∴ x^{2} - 2 x + (1 + z^{2}) = 0$ . For real $x$
$△ = 4 - 4 (1 + z^{2}) \geq 0$ or $- 4 z^{2} \geq 0$
$∴ z = 0$
Hence $x + y = 2, x y = 1$
$∴ x, y$ are roots of $t^{2} - 2 t + 1 = 0$
$∴ x = 1, y = 1, z = 0$

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