Problem 31 *

Find all real numbers x, y and z which satisfy the simultaneous equations x^2 - 4y + 7 = 0, y^2 - 6z + 14 = 0, z^2 - 2x - 7 = 0.

[originally posted by und on TSR]

Adding these we have:

(x^2 - 2x) + (y^2 - 4y) + (z^2 - 6z) + 14 = 0

Writing:

x^2 - 2x = (x - 1)^2 - 1

y^2 - 4y = (y - 2)^2 - 4

z^2 - 6z = (z - 3)^2 - 9

So:

(x - 1)^2 - 1 + (y - 2)^2 - 4 + (z - 3)^2 - 9 + 14 = 0

hence:

(x - 1)^2 + (y - 2)^2 + (z - 3)^2 = 0

since these terms are necessarily non-negative we must have:

x - 1 = y - 2 = z - 3 = 0

that is:

x = 1, y = 2, z = 3.