+1734 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?
+130466 yz / (y + z) = 2 xz / (x + z) = 1
yz = 2y + 2z xz = x + z
yz - 2y = 2z xz - x = z
y ( z -2) = 2z x ( z -1) = z
y = 2z / ( z - 2) x = z/ (z -1)
xy / (x + y) = 1
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 ( z -2) + 2z ( z -1)
2z^2 = 3z^2 - 4z
z^2 - 4z = 0
z ( z -4) = 0
z = 0 ( reject ) and z = 4
