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 $3$.