We can solve for $x,$ $y,$ $z$ separately.
First, $\frac{xy}{x + y} = 1$ simplifies to $xy = x + y.$ Adding 1 to both sides gives $xy + 1 = x + y + 1,$ which factors as $(x + 1)(y + 1) = 2.$ Since 2 is not the product of two integers that differ by 1, we have no solution for $x$ and $y.$
Next, $\frac{xz}{x + z} = 2$ simplifies to $xz = 2x + 2z.$ Then $xz - 2x - 2z = 0,$ which factors as $(x - 2)(z - 2) = 4.$ Thus, the solutions for $x$ and $z$ are the pairs
\[(x,z) = (2,6), \quad (x,z) = (3,-2), \quad (x,z) = (1,1).\]Since not all of these satisfy the third equation, $\frac{yz}{y + z} = 1,$ the unique system solution is $(x,y,z) = (1,1,3),$ so $z = \boxed{3}.$
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?
1 answer