Precalculus, Chapter 1, Review Exercises, Section Review Exercises, Problem 38

Determine the slope and $y$-intercept of the line $3x + 4y = 12$. Graph the line, labeling any intercepts.

We write the equation $3x + 4y = 12$ in slope intercept form to find the slope. We get


$
\begin{equation}
\begin{aligned}

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

\end{aligned}
\end{equation}
$



The slope is $\displaystyle \frac{-3}{4}$. To find the $y$-intercept, we let $x = 0$ and solve for $y$. So we have


$
\begin{equation}
\begin{aligned}

y =& \frac{-3}{4} (0) + 3
\qquad x = 0
\\
y =& 3

\end{aligned}
\end{equation}
$


The $y$-intercept is $3$.