Answer
$\text{a) Slope-Intercept Form: }
y=4x-29
\\\\\text{b) Standard Form: }
4x-y=29$
Work Step by Step
$\bf{\text{Solution Outline:}}$
Find the slope of the given equation, $
4x-y=3
.$ Then use this slope and the given point , $(
7,-1
)$ to find the equation of the needed line. Give the equation in the Slope-Intercept Form and in the Standard Form.
$\bf{\text{Solution Details:}}$
In the form $y=mx+b$ or the Slope-Intercept Form (where $m$ is the slope), the given equation is equivalent to
\begin{array}{l}\require{cancel}
4x-y=3
\\\\
-y=-4x+3
\\\\
-1(-y)=-1(-4x+3)
\\\\
y=4x-3
.\end{array}
Hence, the slope of the given line is $
m=4
.$
Since parallel lines have the same slope, then the needed line has the following characteristics:
\begin{array}{l}\require{cancel}
m=4
\\\text{Through: }
(7,-1)
.\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=-1
,\\x_1=7
,\\m=4
,\end{array}
is
\begin{array}{l}\require{cancel}
y-y_1=m(x-x_1)
\\\\
y-(-1)=4(x-7)
\\\\
y+1=4(x-7)
.\end{array}
Using the properties of equality, in the form $y=mx+b$ or the Slope-Intercept Form, the equation above is equivalent to
\begin{array}{l}\require{cancel}
y+1=4(x-7)
\\\\
y+1=4(x)+4(-7)
\\\\
y+1=4x-28
\\\\
y=4x-28-1
\\\\
y=4x-29
.\end{array}
Using the properties of equality, in the form $ax+by=c$ or the Standard Form, the equation above is equivalent to
\begin{array}{l}\require{cancel}
y=4x-29
\\\\
-4x+y=-29
\\\\
-1(-4x+y)=-1(-29)
\\\\
4x-y=29
.\end{array}
The equation of the line is
\begin{array}{l}\require{cancel}
\text{a) Slope-Intercept Form: }
y=4x-29
\\\\\text{b) Standard Form: }
4x-y=29
.\end{array}
