Beginning Algebra With Applications, Chapter 5, 5.2, Section 5.2, Problem 106

Graph $3x+4y =12$ by using $x$- and $y$-intercepts

$x$-intercept:


$
\begin{equation}
\begin{aligned}

3x+4y =& 12
&& \text{Given equation}
\\
3x + 4(0) =& 12
&& \text{To find the $x$-intercept, let } y = 0
\\
3x =& 12
&& \text{Divide by } 3
\\
x =& 4
&&

\end{aligned}
\end{equation}
$


The $x$-intercept is $(4,0)$

$y$-intercept:


$
\begin{equation}
\begin{aligned}

3x+4y =& 12
&& \text{Given equation}
\\
3(0) + 4y =& 12
&& \text{To find the $y$-intercept, let } x=0
\\
4y =& 12
&& \text{Divide by } 4
\\
y =& 3
&&

\end{aligned}
\end{equation}
$



The $y$-intercept is $(0,3)$

Graph the ordered pairs $(4,0)$ and $(0,3)$. Draw a straight line through the points.