Find all positive real numbers x,y,z such that 2 x &#x2212;<!-- − --> 2 y +

Janiyah Huffman

Janiyah Huffman

Answered question

2022-05-20

Find all positive real numbers x,y,z such that
2 x 2 y + 1 z = 1 2014 , 2 y 2 z + 1 x = 1 2014 , 2 z 2 x + 1 y = 1 2014

Answer & Explanation

Michaela Alvarado

Michaela Alvarado

Beginner2022-05-21Added 11 answers

Step 1
Consider the system of equations
x y + z = 1 , y z + x = 1 , z x + y = 1
These equations are related by cyclic permutations of (x, y, z), but they are satisified by (1,1,0) (and its cyclic permutations) when x, y and z are not all equal.
There are also solutions where x = y = z = ± 5 1 2 , but these are not the only solutions.
Carly Roy

Carly Roy

Beginner2022-05-22Added 4 answers

Step 1
However, in this case you can make a simple argument that starts with an observation based on cyclic invariance, namely that you may as well assume x y , z (i.e., cycle through (x, y, z), (y, z, x) and (z, x, y) and pick the one that starts with the largest of the three numbers).
If x y , z , then 2 x 2 y 0 while 2 z 2 x 0 , so that
1 2014 = 2 x 2 y + 1 z 1 z z 2014
and
1 2014 = 2 z 2 x + 1 y 1 y 2014 y
so we now have x z 2014 y . But this now tells us 2 y 2 z 0 , so that
1 2014 = 2 y 2 z + 1 x 1 x 2014 x
so we now have 2014 x z 2014 y , from which we see x = z = 2014 y . The final equality, 2014 = y , comes by sustituting x = z = 2014 into any of the three equations.
Note, the implication 1 2014 1 z z 2014 requires the assumption z > 0 .

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