Intermediate Algebra, Chapter 3, 3.3, Section 3.3, Problem 68

Determine an equation of the line that satisfies the condition "through $(4,1)$; parallel to $2x + 5y = 10$".

(a) Write the equation in slope intercept form.

We find the slope of the line $2x + 5y = 10$ and write it in slope intercept form


$
\begin{equation}
\begin{aligned}

2x + 5y =& 10
&& \text{Given equation}
\\
\\
5y =& -2x + 10
&& \text{Subtract each side by $2x$}
\\
\\
y =& - \frac{2}{5}x + \frac{10}{5}
&& \text{Divide each side by $5$}
\\
\\
y =& - \frac{2}{5}x + 2
&&

\end{aligned}
\end{equation}
$



The slope is $\displaystyle - \frac{2}{5}$. Using the Point Slope Form, with point $(4,1)$, we have


$
\begin{equation}
\begin{aligned}


y - y_1 =& m(x - x_1)
&& \text{Point Slope Form}
\\
\\
y - 1 =& - \frac{2}{5} (x-4)
&& \text{Substitute } x = 4, y = 1 \text{ and } m = - \frac{2}{5}
\\
\\
y - 1 =& - \frac{2}{5}x + \frac{8}{5}
&& \text{Distributive Property}
\\
\\
5y - 5 =& -2x + 8
&& \text{Multiply each side by $5$}
\\
\\
5y =& -2x + 13
&& \text{Add each side by $5$}
\\
\\
y =& - \frac{2}{5}x + \frac{13}{5}
&& \text{Slope Intercept Form}

\end{aligned}
\end{equation}
$


(b) Write the equation in standard form.


$
\begin{equation}
\begin{aligned}

y =& - \frac{2}{5}x + \frac{13}{5}
&& \text{Slope Intercept Form}
\\
\\
5y =& -2x + 13
&& \text{Multiply each side by $5$}
\\
\\
2x + 5y =& 13
&& \text{Standard Form}

\end{aligned}
\end{equation}
$