9. \( \begin{aligned} \frac{1}{x}+\frac{1}{y}+\frac{1}{z} & =5 \\ \frac{1}{x}+\frac{1}{y} & =2 \\ \frac{1}{x}-\frac{1}{z} & =-2\end{aligned} \)

To solve the system of equations, we can first express the variables in terms of their reciprocals. Let's define \( a = \frac{1}{x} \), \( b = \frac{1}{y} \), and \( c = \frac{1}{z} \). Rewriting the equations in terms of \( a, b, c \) gives us: 1. \( a + b + c = 5 \) 2. \( a + b = 2 \) 3. \( a - c = -2 \) From the second equation, we can express \( c \): \[ c = 5 - (a + b) \] Substituting \( a + b = 2 \) into the expression gives: \[ c = 5 - 2 = 3 \] Now substituting \( c = 3 \) into the third equation: \[ a - 3 = -2 \] This simplifies to: \[ a = 1 \] Now that we have \( a = 1 \) and \( b = 2 - a = 2 - 1 = 1 \), we know: \[ 1 = \frac{1}{x} \Rightarrow x = 1 \\ 1 = \frac{1}{y} \Rightarrow y = 1 \\ 3 = \frac{1}{z} \Rightarrow z = \frac{1}{3} \] Thus, the solutions are: \[ x = 1, \quad y = 1, \quad z = \frac{1}{3} \] So the final answer is: \((x, y, z) = \left(1, 1, \frac{1}{3}\right)\)