Beginning Algebra With Applications, Chapter 5, 5.3, Section 5.3, Problem 48

Graph $\displaystyle x+3y = 6$ by using the slope and $y$-intercept.

$y$-intercept:


$
\begin{equation}
\begin{aligned}

x+3y =& 6
&& \text{Given equation}
\\
0 +3y =& 6
&& \text{To find the $y$-intercept, let } x = 0
\\
3y =& 6
&& \text{Simplify}
\\
y =& 2
&&

\end{aligned}
\end{equation}
$


The $y$-intercept is $(0,2)$

Writing the equation in slope form, $y = mx+b$


$
\begin{equation}
\begin{aligned}

3y =& 6-x
\\
\\
y =& \frac{6-x}{3}
\\
\\
y =& \frac{-1}{3}x + 2

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

m =& \frac{\text{change in } y}{\text{change in } x}
\\
\\
m =& \frac{-1}{3}

\end{aligned}
\end{equation}
$


Beginning at the $y$-intercept, move to the right 3 units and then down 1 unit.







$(3, 1)$ are the coordinates of a second point on the graph.

Draw a line through $(0,2)$ and $(3, 1)$