#### Answer

$\text{a) Slope-Intercept Form: }
y=-\dfrac{1}{3}x+\dfrac{23}{3}
\\\text{b) Standard Form: }
x+3y=23$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
Find the slope of the line defined by the given equation, $
3x-y=7
.$ Then use the Point-Slope Form of linear equations with the given point, $
(-1,3)
$ to find the equation of the needed line. Finally, Express the answer in both the Slope-Intercept and Standard forms.
$\bf{\text{Solution Details:}}$
In the form $y=mx+b,$ the given equation is equivalent to
\begin{array}{l}\require{cancel}
3x-y=7
\\\\
-y=-3x+7
\\\\
-1(-y)=-1(-3x+7)
\\\\
y=3x-7
.\end{array}
Using $y=mx+b$ or the Slope-Intercept Form, where $m$ is the slope, then the slope of the line with the equation above is
\begin{array}{l}\require{cancel}
m=3
.\end{array}
Since parallel lines have negative reciprocal slopes, then the needed line has the following characteristics:
\begin{array}{l}\require{cancel}
\text{Through: }
(8,5)
\\
m=-\dfrac{1}{3}
.\end{array}
Using $y-y_1=m(x-x_1)$ or the Point-Slope Form of linear equations, the equation of the line with the given conditions,
\begin{array}{l}\require{cancel}
y_1=5
,\\x_1=
8
,\\m=
-\dfrac{1}{3}
,\end{array}
is
\begin{array}{l}\require{cancel}
y-y_1=m(x-x_1)
\\\\
y-5=-\dfrac{1}{3}(x-8)
.\end{array}
In the form $y=mx+b$ or the Slope-Intercept Form, the equation above is equivalent to
\begin{array}{l}\require{cancel}
y-5=-\dfrac{1}{3}(x)-\dfrac{1}{3}(-8)
\\\\
y-5=-\dfrac{1}{3}x+\dfrac{8}{3}
\\\\
y=-\dfrac{1}{3}x+\dfrac{8}{3}+5
\\\\
y=-\dfrac{1}{3}x+\dfrac{8}{3}+\dfrac{15}{3}
\\\\
y=-\dfrac{1}{3}x+\dfrac{23}{3}
.\end{array}
In the form $ax+by=c$ or the Standard Form, the equation above is equivalent to
\begin{array}{l}\require{cancel}
3(y)=3\left( -\dfrac{1}{3}x+\dfrac{23}{3} \right)
\\\\
3y=-x+23
\\\\
x+3y=23
.\end{array}
Hence, $
\text{a) Slope-Intercept Form: }
y=-\dfrac{1}{3}x+\dfrac{23}{3}
\\\text{b) Standard Form: }
x+3y=23
.$