The system of equations
\[\frac{xy}{x + y} = 1, \quad \frac{xz}{x + z} = 2, \quad \frac{yz}{y + z} = 1\]
has exactly one solution. What is $z$ in this solution?
+397 Clear the fractions
xy = x + y .................(1)
xz = 2x + 2z .............(2)
yz = y + z .................(3)
Subtract eq(1) from eq(3),
yz - yx = z - x,
y(z - x) - (z - x) = 0,
(z - x)(y - 1) = 0.
Implies that z = x, or y = 1.
If y = 1 then, in eq(1), x = x + 1, so reject that.
If z = x, then in eq(2), z^2 = 4z,
z(z - 4) = 0,
z = 0 or z = 4.
If z = 0 then x = y = 0, so reject that.
Finally, z = 4, x = 4, y = 4/3.
