which shows the graph of the solution set of 6x + 4y < 12?

Explanation:

Step1: Rewrite the inequality in slope - intercept form

First, solve $6x + 4y<12$ for $y$. Subtract $6x$ from both sides: $4y<-6x + 12$. Then divide by 4: $y<-\frac{3}{2}x+3$.

Step2: Analyze the boundary line

The boundary line of the inequality $y<-\frac{3}{2}x + 3$ is the equation $y=-\frac{3}{2}x+3$. Since the inequality is strict ($<$), the boundary line is dashed.

Step3: Find the y - intercept and slope

For the line $y =-\frac{3}{2}x+3$, the y - intercept is 3 (the point $(0,3)$) and the slope is $-\frac{3}{2}$. This means for every 2 units we move to the right along the x - axis, we move 3 units down along the y - axis.

Step4: Test a point

We can test the point $(0,0)$. Substitute $x = 0$ and $y = 0$ into the original inequality $6x+4y<12$. We get $6(0)+4(0)=0<12$, which is true. So the region that contains the origin $(0,0)$ is part of the solution set.

The graph with a dashed line having a y - intercept of 3, a slope of $-\frac{3}{2}$, and the region below the line and containing the origin is the correct graph.

Answer:

The graph with a dashed line passing through $(0,3)$ and $(2,0)$ and the region below the line (including the origin) is the solution - set graph.