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

ou would have 12 of each

linear pair postulates