The system of equations

(xy) / (x+y) = 1, (xz) / (x+z) = 2, (yz) / (y+z) = 3 has exactly one solution. What is $z$ in this solution?

z = - 12

Step-by-step explanation:

Rewritting the equations, we have:

x + y = xy (eq1)

2x + 2z = xz (eq2)

3y + 3z = yz (eq3)

From the first equation:

x = y / (y-1) (eq4)

From the third equation:

y = 3z / (z - 3) (eq5)

Using the value of y from (eq5) in (eq4), we have:

x = [3z / (z - 3) ] / [3z / (z - 3) - 1]

x = [3z / (z - 3) ] / [ (3z - z + 3) / (z - 3) ]

x = 3z / (2z + 3) (eq6)

Using the value of x from (eq6) in (eq2), we have:

6z / (2z + 3) + 2z = (3z / (2z + 3)) * z

(6z + 4z^2 + 6z) / (2z + 3) = 3z^2 / (2z + 3)

12z + 4z^2 = 3z^2

z^2 = - 12z

z = - 12