#### Answer

neither

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To determine if the given pair of lines, $
2y=3x+12
$ and $
3y=2x-5
,$ are parallel, perpendicular, or neither, find and compare the slopes of each line. If the slopes are the same, the lines are parallel. If the product of the slopes is $-1,$ the lines are perpendicular.
$\bf{\text{Solution Details:}}$
In the form $y=mx+b$ or the Slope-Intercept Form (where $m$ is the slope), the first equation is equivalent to
\begin{array}{l}\require{cancel}
2y=3x+12
\\\\
\dfrac{2y}{2}=\dfrac{3x}{2}+\dfrac{12}{2}
\\\\
y=\dfrac{3}{2}x+6
.\end{array}
Hence the slope of the line defined by the first equation is $
m_1=\dfrac{3}{2}
.$
In the form $y=mx+b$ or the Slope-Intercept Form (where $m$ is the slope), the second equation is equivalent to
\begin{array}{l}\require{cancel}
3y=2x-5
\\\\
\dfrac{3y}{3}=\dfrac{2x}{3}-\dfrac{5}{3}
\\\\
y=\dfrac{2}{3}x-\dfrac{5}{3}
.\end{array}
Hence the slope of the line defined by second equation is $
m_2=\dfrac{2}{3}
.$
Since neither the slopes are equal nor the product of $m_1$ and $m_2$ is equal to $-1,$ then the given lines are $\text{
neither parallel nor perpendicular
}.$