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 → xy - x = y → x ( y - 1) = y → x = y / (y - 1)
xz = x + z
yz = y + z → yz - z = y → z (y - 1) = y → z = y/(y - 1)
x = z
So
z*z = z + z
z^2 = 2z
z^2 - 2z = 0
z ( z - 2) = 0
z = 0 reject
z = 2
+1341 
