Question

which is the graph of the linear inequality $2x - 3y < 12$?

Explanation:

Step1: Rewrite the inequality in slope - intercept form

We start with the inequality \(2x - 3y<12\).
Subtract \(2x\) from both sides: \(- 3y<-2x + 12\).
Divide each term by \(-3\). Remember that when we divide an inequality by a negative number, the direction of the inequality sign changes. So we get \(y>\frac{2}{3}x - 4\).

Step2: Analyze the boundary line

The equation of the boundary line is \(y=\frac{2}{3}x - 4\). Since the inequality is \(y>\frac{2}{3}x - 4\) (not \(y\geq\frac{2}{3}x - 4\)), the boundary line should be a dashed line (to indicate that the points on the line are not included in the solution set).

Step3: Determine the region to shade

We test the point \((0,0)\) in the inequality \(y>\frac{2}{3}x - 4\).
Substitute \(x = 0\) and \(y = 0\) into the inequality: \(0>\frac{2}{3}(0)-4\), which simplifies to \(0>- 4\). This is a true statement. So the region that contains the point \((0,0)\) (the region above the line \(y=\frac{2}{3}x - 4\)) should be shaded.

Now let's analyze the graphs:
- The first graph has a solid line, which is incorrect because the inequality is strict (\(<\) or \(>\)), so the boundary line should be dashed.
- The second graph: The boundary line is dashed (good), and we check the shading. Let's see if the region above the line (since \(y>\frac{2}{3}x - 4\)) is shaded. The line \(y=\frac{2}{3}x - 4\) has a y - intercept of \(- 4\) and a slope of \(\frac{2}{3}\). The shading in the second graph is below the line? Wait, no. Wait, when we have \(y>\frac{2}{3}x - 4\), we shade above the line. Wait, let's re - check the test point. For the line \(y=\frac{2}{3}x - 4\), when \(x = 0\), \(y=-4\). The point \((0,0)\) is above the line \(y=\frac{2}{3}x - 4\) (since \(0>-4\)). Now let's look at the third graph: The boundary line is dashed, and the shading is above the line (the region that includes \((0,0)\) since when we look at the grid, the top - left part (where \((0,0)\) is) is shaded? Wait, no, let's look at the four graphs again.

Wait, let's re - express the steps:

1. Boundary line: \(y=\frac{2}{3}x - 4\). The slope is \(\frac{2}{3}\) (rise 2, run 3) and y - intercept is \(-4\).
2. Since the inequality is \(y>\frac{2}{3}x - 4\), the boundary line is dashed (because the inequality is strict, \(>\) not \(\geq\)).
3. We test the point \((0,0)\): \(0>\frac{2}{3}(0)-4\) is \(0 > - 4\), which is true. So we shade the region that contains \((0,0)\), which is above the line \(y=\frac{2}{3}x - 4\).

Now let's analyze each graph:

- First graph: Solid line (incorrect, should be dashed), so eliminate.
- Second graph: Dashed line. Let's see the shading. The line has a y - intercept of \(-4\) and slope \(\frac{2}{3}\). The shaded region is below the line? Wait, no, the second graph's shaded region is the area with the grid (the non - white area) is below the line? Wait, maybe I got the direction wrong. Wait, let's solve for \(y\) again. \(2x-3y < 12\) => \(-3y< - 2x + 12\) => \(y>\frac{2}{3}x - 4\). So \(y\) is greater than the line, so the region above the line.

Third graph: Dashed line, and the shaded region is above the line (the area with the grid is above the line? Wait, the third graph's shaded area is the left - most area (including \((0,0)\) which is above the line \(y=\frac{2}{3}x - 4\) when \(x = 0\), \(y=-4\), so \((0,0)\) is above the line). Wait, the fourth graph: solid line (incorrect). The second graph: dashed line, and the shaded area is below the line? Wait, no, let's take a point in the shaded area of the second graph. Let's say the shaded area is the area with the grid (the non - white area). If we take a point in that area, say \((0,-5)\), plug into \(y>\frac{2}{3}x - 4\): \(-5>\frac{2}{3}(0)-4\) => \(-5 > - 4\), which is false. So the second graph's shaded region is incorrect.

Third graph: Dashed line. Take a point in the shaded area, say \((0,0)\): \(0>\frac{2}{3}(0)-4\) is true. The line is dashed, and the shaded region is above the line.

Wait, maybe I made a mistake in the first analysis. Let's re - check the four graphs:

Top - left: Solid line, shaded above the line (but solid line is wrong).

Top - right: Dashed line, shaded below the line (test point \((0,0)\) is not in the shaded area, since \((0,0)\) is above the line, and the shaded area is below, so incorrect).

Bottom - left: Dashed line, shaded above the line (test point \((0,0)\) is in the shaded area, and the line is dashed).

Bottom - right: Solid line, shaded below the line (incorrect, solid line and wrong shading).

Wait, no, the bottom - left graph: the line is dashed, and the shaded area is the left - hand side (including \((0,0)\)), which is above the line \(y=\frac{2}{3}x - 4\). The bottom - right graph has a solid line, so it's out. The top - left has a solid line, out. The top - right has a dashed line but shaded below, out. So the bottom - left graph (third graph) is the correct one? Wait, no, let's check the equation of the line again.

Wait, the line \(2x-3y = 12\) can be rewritten as \(y=\frac{2}{3}x - 4\). When \(x = 0\), \(y=-4\); when \(y = 0\), \(2x=12\) => \(x = 6\)? Wait, wait, I made a mistake earlier! When \(y = 0\), \(2x-3(0)=12\) => \(x = 6\), not \(x = 5\). Wait, the graphs have the line passing through \((0,-4)\) and \((6,0)\)? But the graphs show the line passing through \((0,-4)\) and \((3, - 2)\) or something? Wait, maybe the x - intercept is at \(x = 6\), but the graphs show the line passing through \((0,-4)\) and \((3, - 2)\) (since slope \(\frac{2}{3}\), from \((0,-4)\), rise 2, run 3: \((0 + 3,-4+2)=(3,-2)\), then \((3 + 3,-2 + 2)=(6,0)\)). But the graphs in the problem have the line passing through \((0,-4)\) and \((3,-2)\) and \((6,0)\)? Wait, the graphs show the line with x - intercept at \(x = 5\)? Maybe the graphs are drawn with a different scale, but the key points are:

- Boundary line: dashed (because of \(<\) in \(2x - 3y<12\), so the inequality is strict, so dashed line).
- Shading: above the line (because \(y>\frac{2}{3}x - 4\)).

So the correct graph should have a dashed line and shading above the line. Looking at the four graphs:

- Top - left: solid line (wrong).
- Top - right: dashed line, shading below (wrong, because \(y>\frac{2}{3}x - 4\) means above).
- Bottom - left: dashed line, shading above (correct).
- Bottom - right: solid line (wrong).

Wait, but maybe I messed up the direction of the inequality. Let's re - solve \(2x-3y<12\) for \(y\):

\(2x-3y<12\)

\(-3y< - 2x + 12\)

Divide by \(-3\) (reverse inequality):

\(y>\frac{2}{3}x - 4\)

Yes, so \(y\) is greater than the line, so shade above the line. The line is dashed. So the graph with dashed line and shading above the line is the correct one. Among the four graphs, the bottom - left graph (third graph) has a dashed line and shading above the line (the shaded area includes \((0,0)\) which satisfies \(y>\frac{2}{3}x - 4\)).

Answer:

The graph in the bottom - left (third graph) is the correct graph of \(2x - 3y<12\). (Assuming the four graphs are labeled as top - left, top - right, bottom - left, bottom - right in order. If we consider the bottom - left graph as the third one in the 2x2 grid, that is the correct one.)