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Add the three equations together we have:
x 2 − 9 x + y 2 + 7 y + z 2 + 1 1 z = − 6 2 . 2 5
( x − 4 . 5 ) 2 − 2 0 . 2 5 + ( y + 3 . 5 ) 2 − 1 2 . 2 5 + ( z + 5 . 5 ) 2 − 3 0 . 2 5 = − 6 2 . 2 5
( x − 4 . 5 ) 2 + ( y + 3 . 5 ) 2 + ( z + 5 . 5 ) 2 − 6 2 . 2 5 = − 6 2 . 2 5
( x − 4 . 5 ) 2 + ( y + 3 . 5 ) 2 + ( z + 5 . 5 ) 2 = 0
Since the three squared quantities are always positive, the above is true only when:
( x − 4 . 5 ) 2 = 0 , ( y + 3 . 5 ) 2 = 0 and ( z + 5 . 5 ) 2 = 0
⇒ x = 4 . 5 , y = − 3 . 5 and z = − 5 . 5
⇒ ∣ x ∣ + ∣ y ∣ + ∣ z ∣ = 4 . 5 + 3 . 5 + 5 . 5 = 1 3 . 5