The system of equations \frac{xy}{x + y} = 1, \quad \frac{xz}{x + z} = 1, \quad \frac{yz}{y + z} = 1 has exactly one solution. What is $z$ in this solution?
xy / (x + y) =1 xz / (x + z) = 1
xy = x + y xz = x + z
xy - y = x xz - z = x
y ( x -1) = x z ( x -1) = x
y = x / (x -1) z = x / ( x -1)
So y = z
So
yz / (y + z) = 1
z*z / ( z + z) = 1
z^2 = 2z
z^2 - 2z = 0
z ( z - 2) = 0
z= 0 reject or z = 2
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