Intermediate Algebra, Chapter 3, Test, Section Test, Problem 14

Determine an equation of the line "through $(-7,2)$ and parallel to $3x + 5y = 6$", and write it in the following form:

a.) Slope-intercept form

We write the equation $3x + 5y = 6$ in slope-intercept form


$
\begin{equation}
\begin{aligned}

3x + 5y =& 6
&& \text{Given equation}
\\
5y =& -3x + 6
&& \text{Subtract each side by $3x$}
\\
y =& - \frac{3}{5}x + \frac{6}{5}
&& \text{Slope-intercept form}

\end{aligned}
\end{equation}
$


The slope is $\displaystyle - \frac{3}{5}$.

Using Point Slope Form


$
\begin{equation}
\begin{aligned}

y - y_1 =& m(x - x_1)
&& \text{Point Slope Form}
\\
\\
y - 2 =& - \frac{3}{5} [x - (-7)]
&& \text{Substitute } x = -7, y = 2 \text{ and } m = - \frac{3}{5}
\\
\\
y - 2 =& - \frac{3}{5}x - \frac{21}{5}
&& \text{Distributive Property}
\\
\\
5y - 10 =& -3x - 21
&& \text{Multiply each side by $5$}
\\
\\
5y =& -3x - 11
&& \text{Add each side by $10$}
\\
\\
y =& - \frac{3}{5}x - \frac{11}{5}
&& \text{Divide each side by $5$}


\end{aligned}
\end{equation}
$



b.) Standard Form


$
\begin{equation}
\begin{aligned}

& y = - \frac{3}{5}x - \frac{11}{5}
&& \text{Slope Intercept Form}
\\
\\
& \frac{3}{5}x + y = - \frac{11}{5}
&& \text{Standard Form}
\\
& \text{or}
&&
\\
& 3x + 5x = -11
&&

\end{aligned}
\end{equation}
$