#### Answer

$\text{a) Slope-Intercept Form: }
y=3x+11
\\\text{b) Standard Form: }
3x-y=-11$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
Find the slope of the line defined by the given equation, $
3x-y=4
.$ Then use the Point-Slope Form of linear equations with the given point, $
(-2,5)
$ 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=4
\\\\
-y=-3x+4
\\\\
-1(-y)=-1(-3x+4)
\\\\
y=3x-4
.\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 same slopes, then the needed line has the following characteristics:
\begin{array}{l}\require{cancel}
m=3
\\\text{Through: }
(-2,5)
.\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=
-2
,\\m=
3
,\end{array}
is
\begin{array}{l}\require{cancel}
y-y_1=m(x-x_1)
\\\\
y-5=3(x-(-2))
\\\\
y-5=3(x+2)
.\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=3(x+2)
\\\\
y-5=3(x)+3(2)
\\\\
y-5=3x+6
\\\\
y=3x+6+5
\\\\
y=3x+11
.\end{array}
In the form $ax+by=c$ or the Standard Form, the equation above is equivalent to
\begin{array}{l}\require{cancel}
y=3x+11
\\\\
-3x+y=11
\\\\
-1(-3x+y)=-1(11)
\\\\
3x-y=-11
.\end{array}
Hence, $
\text{a) Slope-Intercept Form: }
y=3x+11
\\\text{b) Standard Form: }
3x-y=-11
.$