3 equations for 3 variables -- simple right?

Algebra Level 2

Let x, y and z be positive real numbers.

Given 3 equations:

1) x 2 + y + z = 6 x^{2} + y + z = 6

2) x + y 2 + z = 8 x + y^{2} + z = 8

3) x + y + z 2 = 12 x + y + z^{2} = 12

Compute x 7 + y 4 + z x^{7} + y^{4} + z


The answer is 20.

Denton Young by Denton Young , 47, USA

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1 solution

Denton Young
Jul 8, 2015

By inspection, x = 1, y = 2 and z = 3

So: 1 + 16 + 3 = 20

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

How do you know that x = 1 , y = 2 , z = 3 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 ) (y^{2} - y) - ( x 2 − x ) (x^{2} - x) = 2

Similarly, ( z 2 − z ) (z^{2} - z) - ( y 2 − y ) (y^{2} - y) = 4 and ( z 2 − z ) (z^{2} - z) - ( x 2 − x ) (x^{2} - x) = 6

A few easy manipulations later, uniqueness of solution is proved.

Denton Young - 5 years, 11 months ago

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