Let x, y and z be positive real numbers.
Given 3 equations:
1) x 2 + y + z = 6
2) x + y 2 + z = 8
3) x + y + z 2 = 1 2
Compute x 7 + y 4 + z
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How do you know that x = 1 , y = 2 , z = 3 is the only possible solution. 3 equations with 3 variables does not guarantee that there is only one solution.
Subtracting (1) from (2), ( y 2 − y ) - ( x 2 − x ) = 2
Similarly, ( z 2 − z ) - ( y 2 − y ) = 4 and ( z 2 − z ) - ( x 2 − x ) = 6
A few easy manipulations later, uniqueness of solution is proved.
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By inspection, x = 1, y = 2 and z = 3
So: 1 + 16 + 3 = 20