\( \begin{array}{l}\frac{1}{2} x+\frac{1}{3} y=3 \\ \text { Which of the following equations is equivalent to the given equation? } \\ 3 x+2 y=18 \\ 3 x+2 y=3 \\ 2 x+3 y=18\end{array} \)

To solve for an equivalent equation, let's first eliminate the fractions in the original equation \(\frac{1}{2} x + \frac{1}{3} y = 3\). We can do this by multiplying every term by 6 (the least common multiple of 2 and 3). Doing so gives us: \[ 6 \left(\frac{1}{2} x\right) + 6 \left(\frac{1}{3} y\right) = 6 \times 3 \] This simplifies to: \[ 3x + 2y = 18 \] Now, looking at the options provided, we can see that \(3x + 2y = 18\) is indeed equivalent to the original equation. So, the answer is: **Answer:** \(3x + 2y = 18\)