Question

Use the equation to answer the question.
2/3(9a + 18b) = 6a +kb


Which value of
makes the equation true?

Responses

Answer 1

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

\[
\boxed{12}.
\]