Intermediate Algebra, Chapter 3, 3.2, Section 3.2, Problem 46

Determine the slope of of the line $x + 3y = -6$ and sketch the graph.

The intercepts can be used as the two different points needed to find the slope. So

$x$-intercepts


$
\begin{equation}
\begin{aligned}

x + 3y =& -6
&& \text{Given equation}
\\
x + 3(0) =& -6
&& \text{To find the $x$-intercepts, we let $y=0$ and solve for $x$}
\\
x =& -6
&&

\end{aligned}
\end{equation}
$


The $x$-intercept is $-6$.


$y$-intercepts


$
\begin{equation}
\begin{aligned}

x + 3y =& -6
&& \text{Given equation}
\\
0 + 3y =& -6
&& \text{To find the $y$-intercepts, we let $x=0$ and solve for $y$}
\\
y =& -2
&& \text{Divide each side by $3$}

\end{aligned}
\end{equation}
$


The $y$-intercept is $-2$.

Thus, the points are $(-6,0)$ and $(0,-2)$.

Using the two points in slope formula


$
\begin{equation}
\begin{aligned}

m = \frac{y_2 - y_1}{x_2 - x_1} =& \frac{-2-0}{0- (-6)}
&& \text{Substitute } (x_1, y_1) = (-6,0) \text{ and } (x_2, y_2) = (0,-2)
\\
\\
=& \frac{-2}{6}
&& \text{Simplify}
\\
\\
=& - \frac{1}{3}


\end{aligned}
\end{equation}
$