+231 The system of equations
\(\frac{xy}{x + y} = 1, \quad \frac{xz}{x + z} = 2, \quad \frac{yz}{y + z} = 3\)
has exactly one solution. What is \(z\) in this solution?
+23252 xy/(x + y) = 1 ---> xy = x + y ---> xy - x = y ---> x(y - 1) = y ---> x = y/(y - 1)
xz/(x + z) = 2 ---> x = 2z/(x - z)
yz/(y + z) = 3 ---> z = 3y/(y - 3)
Substituting the third equation into the second: x = [ 2(3y/(y - 3)) ] / [ 3y/(y - 3) - 2 ]
---> x = (6y) / (y + 6)
Combining this equation and the first equation: y/(y - 1) = (6y)/(y + 6)
---> y(y + 6) = 6y(y - 1)
---> y2 + 6y = 6y2 - 6y
---> 0 = 5y2 - 12y
---> 0 = y(5y - 12)
Either y = 0 or y = 12/5
If y = 12/5 ---> x = y/(y - 1) ---> x = 12/7
If x = 12/7 and y = 12/5 ---> z = 3y/(y - 3) ---> z = -12
