To solve the equation
\[ \frac{9}{z} + \frac{9}{z - 2} = 12, \]
let's first find a common denominator for the fractions. The common denominator is \( z(z - 2) \). We can rewrite the equation:
\[ \frac{9(z - 2) + 9z}{z(z - 2)} = 12. \]
This simplifies to:
\[ \frac{9z - 18 + 9z}{z(z - 2)} = 12, \]
which further simplifies to:
\[ \frac{18z - 18}{z(z - 2)} = 12. \]
Now cross-multiply to eliminate the fraction:
\[ 18z - 18 = 12z(z - 2). \]
Distributing on the right side gives us:
\[ 18z - 18 = 12z^2 - 24z. \]
Rearranging this gives:
\[ 12z^2 - 24z - 18z + 18 = 0, \]
which simplifies to:
\[ 12z^2 - 42z + 18 = 0. \]
Dividing the entire equation by 6 to simplify:
\[ 2z^2 - 7z + 3 = 0. \]
Now, we can use the quadratic formula, \( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2 \), \( b = -7 \), and \( c = 3 \):
Calculating the discriminant:
\[ b^2 - 4ac = (-7)^2 - 4 \cdot 2 \cdot 3 = 49 - 24 = 25. \]
Now substituting in the quadratic formula:
\[ z = \frac{-(-7) \pm \sqrt{25}}{2 \cdot 2} = \frac{7 \pm 5}{4}. \]
This leads to two possible solutions for \( z \):
\[ z = \frac{12}{4} = 3 \quad \text{and} \quad z = \frac{2}{4} = \frac{1}{2}. \]
Thus, the solutions to the equation are
\[ z = 3 \quad \text{and} \quad z = \frac{1}{2}. \]
The final answer is:
\( z = 3 \) and \( z = \frac{1}{2} \).
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