Olympiad Corner 3
⎩ ⎪ ⎨ ⎪ ⎧ x + x y + x y z = 1 2 y + y z + x y z = 2 1 z + z x + x y z = 3 0
Given that x , y and z are real numbers that satisfy the system of equations above, find the sum of all possible values of x + y + z .
Give your answer to 3 decimal places.
The answer is 15.707.
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2 solutions
I would like to know where I got it wrong. In your equations of x, y and z, I put k = 1 to get x, y, and z as 4/7, 7/10 and 5/2 giving me a perfect k = xyz = 1. But, it does not satisfy your original equations. That means that these three equations have an inherent fault which results in an invalid answer. Where is the fault?
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Sir, from the equations we find that the values of x,y,z depend upon the value of xyz so evaluating the equation by a preset value of xyz would provide solutions which might be incompitable with the original equation.
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'Preset' is a misnomer here. I solved for all possible xyz to get 1, 10, 27 and 28 as solutions. Then why is 1 singled out for giving wrong results.
Can you suggest some more problem of the same type ?
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x+xy+xyz=12
y+yz+xyz=21
z+zx+xyz=30
Now multiply the first equation by z and subtract from the third equation, multiply the second equation by x and subtract from the first and similarly multiply the third equation by y and subtract from the second equation to get,
(13z - 30)/z = (22x - 12)/x = (31y - 21)/y = xyz = k
Now,
x = (12)/(22-k),
y = (21)/(31-k),
z = (30)/(13-k).
Put these values of x,y,z in the first equation to get,
[k^(3)] - [65 (k^(2))] + 1306k - 7560 = 0
We get the solutions k = 10,27,28.
For each of these we get,
(x,y,z) = (1,1,10)
Whose sum =15.707