Question

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?

Answer 1

-12

Explanation:

Taking the reciprocal of the first equation, we get

\[\frac{x + y}{xy} = 1.\]We can split the fraction, and the equation becomes

\[\frac{1}{x} + \frac{1}{y} = 1.\]Similarly, the second and third equations can be rewritten as

\[\frac{1}{x} + \frac{1}{z} = \frac{1}{2} \quad \text{and} \quad \frac{1}{y} + \frac{1}{z} = \frac{1}{3},\]respectively.

Substituting $a = \frac{1}{x}$, $b = \frac{1}{y}$, and $c = \frac{1}{z}$, we obtain the linear system of equations

\begin{align*}

a + b &= 1, \\

a + c &= \frac{1}{2}, \\

b + c &= \frac{1}{3}.

\end{align*}

Adding all three equations, we get

\[2a + 2b + 2c = \frac{11}{6},\]so

\[a + b + c = \frac{11}{12}.\]Then

\[a = (a + b + c) - (b + c) = \frac{11}{12} - \frac{1}{3} = \frac{7}{12},\]which means $x = \frac{1}{a} = \frac{12}{7}.$

Similarly,

\[b = (a + b + c) - (a + c) = \frac{11}{12} - \frac{1}{2} = \frac{5}{12},\]so $y = \frac{1}{b} = \frac{12}{5}$, and

\[c = (a + b + c) - (a + b) = \frac{11}{12} - 1 = -\frac{1}{12},\]so $z = \frac{1}{c} = -12$.

Therefore, the solution is $(x,y,z) = \left( \frac{12}{7}, \frac{12}{5}, -12 \right)$. In particular, $z = \boxed{-12}$.

Answer 2

z = -12

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.