olympiad

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

2 Answers

x = (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) [ tâ‰ -1]

Similarly, making y & z the subject,you get the solution sets : (-2-t , -2 , t) [ tâ‰ -1 ] & (-2-t , t , -2) [ tâ‰ -1 ]

Also,you have the trivial solution.

341

Hari Shankar ·2009-07-19 07:01:02

Simon's Favourite Factoring Technique again!

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)

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