Question

which is the graph of the linear inequality $2y > x - 2$?

Explanation:

Step1: Rewrite the inequality

First, we rewrite the inequality \(2y > x - 2\) in slope - intercept form (\(y=mx + b\)) by dividing both sides by 2. We get \(y>\frac{1}{2}x - 1\).

Step2: Analyze the boundary line

The boundary line of the inequality \(y>\frac{1}{2}x - 1\) is the line \(y = \frac{1}{2}x-1\). Since the inequality is \(y>\frac{1}{2}x - 1\) (not \(y\geq\frac{1}{2}x - 1\)), the boundary line should be a dashed line (because the points on the line do not satisfy the inequality).

Step3: Determine the shaded region

To find the direction of the shading, we can use a test point. A common test point is \((0,0)\). Substitute \(x = 0\) and \(y = 0\) into the inequality \(y>\frac{1}{2}x-1\):
\(0>\frac{1}{2}(0)-1\), which simplifies to \(0>- 1\). This is a true statement. So, the region that contains the point \((0,0)\) should be shaded.

Now let's analyze the four graphs:
- The first two graphs have a solid boundary line, which is incorrect because our inequality is strict (\(>\)), so we can eliminate the first two graphs.
- For the third and fourth graphs, the boundary line is dashed (correct for \(>\)). Now we check the shading. The line \(y=\frac{1}{2}x - 1\) has a \(y\) - intercept of \(- 1\) and a slope of \(\frac{1}{2}\). The point \((0,0)\) is in the upper - left region relative to the line \(y=\frac{1}{2}x - 1\). The fourth graph's shaded region includes \((0,0)\) (since when we look at the fourth graph, the shaded area is below the dashed line? Wait, no. Wait, let's re - check the slope - intercept form. The line \(y=\frac{1}{2}x-1\): when \(x = 0\), \(y=-1\); when \(y = 0\), \(x = 2\). So the line passes through \((0,-1)\) and \((2,0)\). The inequality \(y>\frac{1}{2}x - 1\) means we shade above the line. Wait, no: if \(y>\frac{1}{2}x-1\), for a given \(x\), we take all \(y\) values greater than \(\frac{1}{2}x - 1\). So the shading is above the line \(y=\frac{1}{2}x - 1\).

Wait, let's re - do the test point. For the line \(y=\frac{1}{2}x-1\), when \(x = 0\), \(y=-1\). The point \((0,0)\) has a \(y\) - value of \(0\), which is greater than \(-1\), so \((0,0)\) is above the line \(y=\frac{1}{2}x - 1\). So we need to shade above the dashed line \(y=\frac{1}{2}x - 1\).

Looking at the third graph: the dashed line, and the shading is above? Wait, no, the third graph's shaded area seems to be below? Wait, maybe I made a mistake. Wait, let's look at the four graphs again.

Wait, the fourth graph: the dashed line, and the shaded area is below? No, wait, the \(y\) - axis: the fourth graph, the shaded area is from the bottom (lower \(y\) - values) up? Wait, no, let's re - express the inequality.

Wait, \(2y>x - 2\) => \(y>\frac{1}{2}x-1\). So the boundary line is \(y=\frac{1}{2}x - 1\) (dashed). The region where \(y>\frac{1}{2}x - 1\) is above the line.

Now, let's check the four graphs:

- Graph 1: Solid line, so eliminate (since inequality is strict).
- Graph 2: Solid line, eliminate.
- Graph 3: Dashed line, but the shading seems to be below the line? Wait, no, maybe the grid. Wait, the line \(y=\frac{1}{2}x - 1\) passes through \((0,-1)\) and \((2,0)\). Let's take a point in the shaded region of graph 4. Let's take \((0, - 2)\): does \(-2>\frac{1}{2}(0)-1\)? \(-2>-1\)? No. Take \((0,0)\): \(0>\frac{1}{2}(0)-1\) => \(0 > - 1\), which is true. So \((0,0)\) should be in the shaded region.

Looking at the fourth graph: the shaded region includes \((0,0)\)? Wait, the fourth graph's shaded area is the lower part? No, maybe I messed up the direction. Wait, let's solve for \(y\): \(y>\frac{1}{2}x - 1\). So for a given \(x\), \(y\) must be greater than \(\frac{1}{2}x - 1\). So when \(x = 0\), \(y>-1\); when \(x = 2\), \(y>0\). So the region is above the line.

Now, the third graph: the dashed line, and the shaded area is above? Wait, the third graph's shaded area is the upper part (higher \(y\) - values), and the fourth graph's shaded area is the lower part? Wait, no, the third graph has a shaded area that is the upper grid (from \(y = - 1\) up), and the fourth graph's shaded area is the lower grid (from \(y=-5\) up to the line). Wait, I think I made a mistake in the direction.

Wait, let's rewrite the inequality as \(y>\frac{1}{2}x - 1\). So the slope is \(\frac{1}{2}\) (positive), \(y\) - intercept is \(-1\). So the line goes from \((0,-1)\) (down on the \(y\) - axis) to \((2,0)\) (to the right). So the region above the line (where \(y\) is greater than \(\frac{1}{2}x - 1\)) is the region that is above this line. So for example, at \(x = 0\), \(y>-1\) (so above \(y=-1\)), at \(x = 2\), \(y>0\) (above \(y = 0\)).

Now, looking at the four graphs:

- Graph 1: Solid line, wrong (since \(>\) means dashed).
- Graph 2: Solid line, wrong.
- Graph 3: Dashed line, and the shaded area is above the line (the upper part of the grid), which includes \((0,0)\) (since \(0>-1\)).
- Graph 4: Dashed line, but the shaded area is below the line (the lower part of the grid), which would not include \((0,0)\) (since \(0\) is above \(-1\), not below). Wait, no, maybe the graphs are drawn with the shaded area in a different way. Wait, maybe I got the inequality direction wrong. Wait, \(2y>x - 2\) => \(y>\frac{1}{2}x-1\). So if we have a dashed line, and we shade above the line. So the correct graph should have a dashed line \(y=\frac{1}{2}x - 1\) and shading above it.

Looking at the third graph: the dashed line, and the shaded area is above (the upper part of the grid), which is correct. Wait, but let's check the \(y\) - intercept. The line in the third graph: when \(x = 0\), the line is at \(y=-1\) (dashed), and the shaded area is above (higher \(y\) - values), which includes \((0,0)\) (since \(0>-1\)). The fourth graph's shaded area is below the dashed line, which would not include \((0,0)\) (since \(0\) is above \(-1\)). So the correct graph is the third one? Wait, no, maybe I made a mistake. Wait, let's check the original inequality again.

Wait, \(2y>x - 2\). Let's solve for \(y\): \(y>\frac{x - 2}{2}=\frac{1}{2}x-1\). So the boundary line is \(y=\frac{1}{2}x - 1\) (dashed, because the inequality is strict). To find the shaded region, we can use the test point \((0,0)\): \(0>\frac{1}{2}(0)-1\) => \(0>-1\), which is true. So \((0,0)\) is in the solution region. So the solution region is the set of all points above the line \(y=\frac{1}{2}x - 1\) (since \(y\) must be greater than \(\frac{1}{2}x - 1\)).

Now, looking at the four graphs:

- Graph 1: Solid line (incorrect, should be dashed), so eliminate.
- Graph 2: Solid line (incorrect), eliminate.
- Graph 3: Dashed line, and the shaded area is above the line (includes \((0,0)\)), correct?
- Graph 4: Dashed line, but the shaded area is below the line (does not include \((0,0)\) since \(0\) is above \(-1\)), incorrect.

Wait, but maybe the graphs are labeled differently. Wait, the third graph: the dashed line, and the shaded area is the upper part (the part with higher \(y\) - values), which is correct. So the correct graph is the third one (the one in the bottom - left, with the dashed line and shading above the line).

Answer:

The graph in the bottom - left (third graph) with the dashed line \(y=\frac{1}{2}x - 1\) and shading above the line. (Assuming the third graph is the one with the dashed line and the shaded area above the line. If we consider the labels, if the four graphs are labeled as top - left, top - right, bottom - left, bottom - right, then the bottom - left graph is the correct one.)