+769 The system of equations
\frac{xy}{x + y} = 1, \quad \frac{xz}{x + z} = 1, \quad \frac{yz}{y + z} = 2
has exactly one solution. What is $z$ in this solution?
+129895 xz = x + z yz = 2 ( y + z)
xz - x = z yz = 2y + 2z
x ( z -1) = z yz - 2y = 2z
x = z / ( z -1) y ( z - 2) =2z
y = (2z) / (z -2)
xy = x + y
z/ (z-1)* (2z) ( z -2) = z/(z -1) + (2z) (z -2)
[2z^2] / [ (z-1)(z -2)] = [ z (z -2) + 2z ( z -1)] / [ (z -1) ( z -2) ]
2z^2 = z^2 - 2z + 2z^2 - 2z
2z^2 = 3z^2 - 4z
z^2 - 4z = 0
z ( z - 4) = 0
Reject z = 0
z = 4
