Can you find the neat solution?
If and are real numbers that satisfy the system of equations above, find the sum of all distinct value(s) of .
The answer is 9.
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1 solution
Adding the three equations yields x 2 − 6 y + 1 7 + y 2 − 1 0 z + 4 1 + z 2 − 2 x − 2 3 = 0
⇒ x 2 + y 2 + z 2 − 2 x − 6 y − 1 0 z + 3 5 = 0
This factorises to give ( x − 1 ) 2 + ( y − 3 ) 2 + ( z − 5 ) 2 = 0
Squares are always non-negative so the expression on the RHS above is always more than or equal to 0, whatever the values of x, y and z are. In our case, the RHS equals zero so this is only true when x − 1 = 0 , y − 3 = 0 and z − 5 = 0 . Therefore the only solution is x = 1, y = 3 and z = 5 so the answer is 1 + 3 + 5 = 9 .