Intermediate Algebra, Chapter 3, 3.3, Section 3.3, Problem 30

(a) Write the equation $3x + 4y = 12$ in slope-intercept form.


$
\begin{equation}
\begin{aligned}

3x + 4y =& 12
&& \text{Given equation}
\\
4y =& -3x + 12
&& \text{Subtract each side by $3x$}
\\
y =& - \frac{3}{4}x + 3
&& \text{Divide each side by $4$}

\end{aligned}
\end{equation}
$



(b) Give the slope of the line.

Here the slope is $\displaystyle - \frac{3}{4}$. It can be interpreted as $\displaystyle m = \frac{\text{rise}}{\text{run}} = \frac{-3}{4} \text{ or } \frac{3}{-4}$

(c) Give the $y$-intercept.

The $y$-intercept is $3$.

(d) Graph the line.

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