To solve for \( k \) in the equation
\[ \frac{2}{3}(9a + 18b) = 6a + kb, \]
let's start by simplifying the left side of the equation.
Distributing \( \frac{2}{3} \) to both terms inside the parentheses:
\[ \frac{2}{3}(9a) + \frac{2}{3}(18b) = \frac{2 \cdot 9a}{3} + \frac{2 \cdot 18b}{3}. \]
Calculating each term:
\[ \frac{2 \cdot 9a}{3} = 6a, \] \[ \frac{2 \cdot 18b}{3} = 12b. \]
So the left side simplifies to:
\[ 6a + 12b. \]
Now we equate the left side to the right side of the original equation:
\[ 6a + 12b = 6a + kb. \]
To find \( k \), we can equate the coefficients of \( b \) from both sides:
\[ 12b = kb. \]
Since \( b \neq 0 \) (we assume \( b \) is not equal to zero), we can divide both sides by \( b \):
\[ 12 = k. \]
Thus, the value of \( k \) that makes the equation true is
\[ \boxed{12}. \]
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