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?

Guest Jun 13, 2023

#1**+1 **

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.

Tiggsy Jun 13, 2023