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: \[ \begin{aligned} \frac{1}{x} + \frac{1}{y} + \frac{1}{z} &= 5 \quad \text{(1)} \\ \frac{1}{x} + \frac{1}{y} &= 2 \quad \text{(2)} \\ \frac{1}{x} - \frac{1}{z} &= -2 \quad \text{(3)} \end{aligned} \] Let's introduce substitution variables to simplify the equations: \[ a = \frac{1}{x}, \quad b = \frac{1}{y}, \quad c = \frac{1}{z} \] Substituting these into the original system, we get: \[ \begin{aligned} a + b + c &= 5 \quad \text{(1')} \\ a + b &= 2 \quad \text{(2')} \\ a - c &= -2 \quad \text{(3')} \end{aligned} \] **Step 1:** Subtract equation (2') from equation (1') to find \( c \): \[ (a + b + c) - (a + b) = 5 - 2 \\ c = 3 \] **Step 2:** Substitute \( c = 3 \) into equation (3') to find \( a \): \[ a - 3 = -2 \\ a = 1 \] **Step 3:** Substitute \( a = 1 \) into equation (2') to find \( b \): \[ 1 + b = 2 \\ b = 1 \] **Step 4:** Now, convert back to the original variables using \( a = \frac{1}{x}, b = \frac{1}{y}, \) and \( c = \frac{1}{z} \): \[ \frac{1}{x} = 1 \quad \Rightarrow \quad x = 1 \\ \frac{1}{y} = 1 \quad \Rightarrow \quad y = 1 \\ \frac{1}{z} = 3 \quad \Rightarrow \quad z = \frac{1}{3} \] **Solution:** \[ x = 1, \quad y = 1, \quad z = \frac{1}{3} \]