find all ordered triples (x,y,z)of real numbers which satisfy the following system of equations:
xy = z -x - y
xz = y - x - z
yz = x - y - z
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2 Answers
11x = (z-y)/(1+y) = (y-z)/(1+z)
(1+y) = - (1+z)
y+z = -2
y = -2-z
x = [(-2-z)-z]/(1+z) = -2
From this you get the solution set : (-2 , -2-t , t) [ t≠-1]
Similarly, making y & z the subject,you get the solution sets : (-2-t , -2 , t) [ t≠-1 ] & (-2-t , t , -2) [ t≠-1 ]
Also,you have the trivial solution.
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The three equations can be written as
(1+x) (1+y) = (1+z)..................................1
(1+y) (1+z) = (1+x).................................2
(1+z) (1+x) = (1+y).................................3
Multiplying the three equations we get
A2 = A where A = (1+x) (1+y) (1+z)
Hence A = 0 or A = 1
If A = 0, then we have x=y=z= -1
If A = 1, then we have
first we see that each of x,y and z can be 0 or -2
Obviously x = y = z = 0 is one solution
The other solutions are permutations of (0,-2,-2)