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?

z = -12

Step-by-step explanation:

The given system of equations is:

xy/(x + y) = 1 (1)

xz/(x + z) = 2(2)

yz/(y + z) = 3(3)

From (1): x + y = xy

=> y = xy - x

y = x(y - 1)

x = y/(y - 1)(4)

From (2): 2(x + z) = xz

=> 2x + 2z = xz

2x = xz - 2z

2x = z(x - 2)

z = 2x/(x - 2) (5)

From (3): 3(y + z) = yz

=> 3y + 3z = yz

3y = yz - 3z

3y = z(y - 3)

z = 3y/(y - 3)(6)

Comparing (5) and (6)

2x/(x - 2) = 3y/(y - 3)

2x(y - 3) = 3y(x - 2)

2xy - 6x = 3xy - 6y

6(y - x) = xy (7)

But from (1): xy = x + y

Using this in (7), we have

6(y - x) = x + y

6y - y - 6x - x = 0

5y - 7x = 0

5y = 7x

x = 5y/7(8)

Using this in (4)

5y/7 = y/(y - 1)

1/(y - 1) = 5/7

(y - 1) = 7/5

y = 1 + 7/5

y = 12/5(9)

Using this in (8)

x = 5(12/5)/7 = 12/7 (10)

Using (10) in (5)

z = 2x/(x - 2)

z = 2(12/7) ÷ (12/7 - 2)

= 24/7 ÷ -2/7

= 24/7 × (-7/2)

= -24/2 = -12

z = -12.

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step-by-step explanation: