Question

graph the inequality.
$-4x + 3y < 6$

Explanation:

Step1: Rewrite in slope-intercept form

Isolate $y$ to solve for the boundary line.

$$\begin{align*} -4x + 3y &< 6 \\ 3y &< 4x + 6 \\ y &< \frac{4}{3}x + 2 \end{align*}$$

Step2: Identify boundary line

The boundary is $y=\frac{4}{3}x+2$, dashed (since $<$ is used, no equality).

• Y-intercept: $(0, 2)$
• Slope: $\frac{4}{3}$ (rise 4, run 3)

Step3: Determine shaded region

Test the origin $(0,0)$: $0 < \frac{4}{3}(0)+2 \implies 0 < 2$, which is true. Shade the region containing the origin (below the dashed line).

Answer:

1. Draw a dashed line for $y=\frac{4}{3}x+2$: start at the y-intercept $(0,2)$, then use the slope to plot another point (e.g., $(3,6)$) and connect them with a dashed line.
2. Shade the entire region below this dashed line (the area that includes the point $(0,0)$).