+859 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?
+129881 Rewrite as rhe second and third equations
xz = x + z → xz - x = z → x(z -1) = z → x = z / (z -1)
yz = 2 (y+ z) → yz = 2y + 2z → yz - 2y = 2z → y ( z -2) = 2z → y = (2z) / (z -2)
Rewrite the first equation. sub for x and y
[ z / ( z -1)] [(2z) / (z -2)] = z/ (z -1) + (2z) / ( 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 because it makes the second equation false )
z = 4
