graph the inequality on the axes below.\n3x + 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: Graph the boundary line

The boundary line is $y =-\frac{3}{4}x + 3$. Since the inequality is $y<-\frac{3}{4}x+3$ (not $\leq$), the boundary line is dashed. Find the $x$ - intercept by setting $y = 0$: $0=-\frac{3}{4}x+3$, then $\frac{3}{4}x=3$, and $x = 4$. The $x$ - intercept is $(4,0)$. Find the $y$ - intercept by setting $x = 0$: $y=3$. The $y$ - intercept is $(0,3)$. Plot these two points and draw a dashed line through them.

Step3: Determine the shaded region

Choose a test point not on the line, say $(0,0)$. Substitute into the original inequality: $3(0)+4(0)<12$, which is $0<12$ (a true statement). So, shade the region that contains the point $(0,0)$.

Answer:

Graph a dashed line with $x$ - intercept $(4,0)$ and $y$ - intercept $(0,3)$, and shade the region below the line.