Solve the following system of equations: $\;$ $\dfrac{xy}{x + y} = 1, \;\; \dfrac{xz}{x + z} = 2, \;\; \dfrac{yz}{y + z} = 3$
Given system of equations:
$\dfrac{xy}{x + y} = 1$ $\;\;\; \cdots \; (1)$, $\;$ $\dfrac{xz}{x +z} = 2$ $\;\;\; \cdots \; (2)$ $\;$ and $\;$ $\dfrac{yz}{y + z} = 3$ $\;\;\; \cdots \; (3)$
The given equations can be respectively simplified as
$\dfrac{x + y}{xy} = 1$ $\;\;\;$ i.e. $\;$ $\dfrac{1}{x} + \dfrac{1}{y} = 1$ $\;\;\; \cdots \; (1a)$
$\dfrac{x + z}{xz} = \dfrac{1}{2}$ $\;\;\;$ i.e. $\;$ $\dfrac{1}{x} + \dfrac{1}{z} = \dfrac{1}{2}$ $\;\;\; \cdots \; (2a)$
$\dfrac{y + z}{yz} = \dfrac{1}{3}$ $\;\;\;$ i.e. $\;$ $\dfrac{1}{y} + \dfrac{1}{z} = \dfrac{1}{3}$ $\;\;\; \cdots \; (3a)$
Subtracting equations $(1a)$ and $(2a)$ gives
$\dfrac{1}{y} - \dfrac{1}{z} = \dfrac{1}{2}$ $\;\;\; \cdots \; (4)$
Adding equations $(3a)$ and $(4)$ gives
$\dfrac{2}{y} = \dfrac{1}{2} + \dfrac{1}{3} = \dfrac{5}{6}$ $\implies$ $y = \dfrac{12}{5}$
Substituting $\;$ $y = \dfrac{12}{5}$ $\;$ in equation $(4)$ gives
$\dfrac{1}{12/5} - \dfrac{1}{z} = \dfrac{1}{2}$
i.e. $\;$ $\dfrac{1}{z} = \dfrac{5}{12} - \dfrac{1}{2} = \dfrac{-1}{12}$ $\implies$ $z = -12$
Substituting $\;$ $z = -12$ $\;$ in equation $(2a)$ gives
$\dfrac{1}{x} - \dfrac{1}{12} = \dfrac{1}{2}$
i.e. $\;$ $\dfrac{1}{x} = \dfrac{1}{12} + \dfrac{1}{2} = \dfrac{7}{12}$ $\implies$ $x = \dfrac{12}{7}$
$\therefore \;$ The solution to the given system of equations is $\;\;$ $\left(x, y, z\right) = \left\{\left(\dfrac{12}{7}, \dfrac{12}{5}, -12\right) \right\}$