We are asked to find the equation of a line that is parallel to the line $2y = 3(x-2)$ and passes through the point $(2, 3)$.

1. Problem Description

We are asked to find the equation of a line that is parallel to the line $2y = 3(x-2)$ and passes through the point $(2, 3)$.

2. Solution Steps

First, we need to find the slope of the given line $2y = 3(x-2)$. We can rewrite this equation in slope-intercept form ($y = mx + b$), where $m$ is the slope and $b$ is the y-intercept.

$2y = 3(x-2)$

$2y = 3x - 6$

$y = \frac{3}{2}x - 3$

So, the slope of the given line is $\frac{3}{2}$.

Since we want to find a line parallel to the given line, the parallel line will have the same slope. Thus, the slope of the parallel line is also $\frac{3}{2}$.

Now we know the slope of the parallel line is $\frac{3}{2}$, and it passes through the point $(2, 3)$. We can use the point-slope form of a line, which is:

$y - y_1 = m(x - x_1)$

where $m$ is the slope and $(x_1, y_1)$ is a point on the line. In this case, $m = \frac{3}{2}$ and $(x_1, y_1) = (2, 3)$.

Plugging in these values, we get:

$y - 3 = \frac{3}{2}(x - 2)$

$y - 3 = \frac{3}{2}x - 3$

$y = \frac{3}{2}x - 3 + 3$

$y = \frac{3}{2}x$

3. Final Answer

The equation of the line is $y = \frac{3}{2}x$. The correct answer is C.