#### Answer

perpendicular

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To determine if the given pair of lines, $
5x-y=8
$ and $
5y=-x+3
,$ 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}
5x-y=8
\\\\
-y=-5x+8
\\\\
-1(-y)=-1(-5x+8)
\\\\
y=5x-8
.\end{array}
Hence the slope of the line defined by the first equation is $
m_1=5
.$
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}
5y=-x+3
\\\\
y=-\dfrac{1}{5}x+\dfrac{3}{5}
.\end{array}
Hence the slope of the line defined by second equation is $
m_2=-\dfrac{1}{5}
.$
Since the product of $m_1$ and $m_2$ is equal to $-1,$ then the given lines are $\text{
perpendicular
}.$