Respuesta :
Answer:
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.
Answer:
-12
Step-by-step 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, $z = \boxed{-12}$.
