Intermediate Algebra, Chapter 3, 3.3, Section 3.3, Problem 34

(a) Write the equation $x + 3y = -9$ in slope-intercept form.


$\begin{equation} \begin{aligned} x + 3y =& -9 && \text{Given equation} \\ \\ 3y =& -x - 9 && \text{Subtract each side by $x$} \\ \\ y =& \frac{-1}{3}x - \frac{9}{3} && \text{Divide each side by $3$} \\ \\ y =& \frac{-1}{3}x - 3 && \end{aligned} \end{equation}$



(b) Give the slope of the line.

Here the slope is $\displaystyle \frac{-1}{3}$. It can be interpreted as $\displaystyle m = \frac{\text{rise}}{\text{run}} = \frac{-1}{3} \text{ or } \frac{1}{-3}$

(c) Give the $y$-intercept.

The $y$-intercept is $-3$.

(d) Graph the line.

From $(0,-3)$, move 1 unit down and 3 units to the right and plot a second point at $(3,-4)$. Draw a line through the two points.