graph the inequality on the axes below. 3x + 4y < 12

Explanation:

Step1: Rewrite in slope - intercept form

First, solve $3x + 4y<12$ for $y$. Subtract $3x$ from both sides: $4y<-3x + 12$. Then divide by 4: $y<-\frac{3}{4}x+3$.

Step2: Find the boundary line

The boundary line of the inequality $y<-\frac{3}{4}x + 3$ is the equation $y=-\frac{3}{4}x+3$. Since the inequality is strict ($<$), the boundary line is dashed.

Step3: Find the $x$ - intercept

Set $y = 0$ in $y=-\frac{3}{4}x+3$. Then $0=-\frac{3}{4}x+3$. Add $\frac{3}{4}x$ to both sides: $\frac{3}{4}x=3$. Multiply both sides by $\frac{4}{3}$ to get $x = 4$. So the $x$ - intercept is $(4,0)$.

Step4: Find the $y$ - intercept

Set $x = 0$ in $y=-\frac{3}{4}x+3$. Then $y=3$. So the $y$ - intercept is $(0,3)$.

Step5: Determine the shaded region

Choose a test - point not on the line, say $(0,0)$. Substitute $x = 0$ and $y = 0$ into the original inequality $3x+4y<12$. We get $3(0)+4(0)=0<12$, which is true. So we shade the region that contains the point $(0,0)$.

To graph:

1. Plot the $x$ - intercept $(4,0)$ and the $y$ - intercept $(0,3)$.
2. Draw a dashed line through these two points.
3. Shade the region below the dashed line (the region that contains the origin $(0,0)$).

Answer:

Graph a dashed line through the points $(4,0)$ and $(0,3)$ and shade the region below the line.