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When graphing a linear inequality, the first and crucial step is to rewrite the inequality as an equation. This equation represents the boundary line of the inequality's solution region. In our exercise, the inequality is given as (2x + 3y > 12). To create the boundary equation, we simply replace the inequality symbol with an equals sign, yielding (2x + 3y = 12).

It's important to note that this new equation will not include the inequality symbol, and it only serves to determine where the boundary of the solution lies on a graph. In further steps, we will consider the original inequality to establish the correct side of the boundary where the solution set is located.

Once we have the boundary equation (2x + 3y = 12), we can plot the boundary line on a coordinate plane. We need two points to draw a straight line, and the easiest ones to obtain are generally the x- and y-intercepts. To determine the x-intercept, we set (y = 0) in our equation, solving for (x), which gives us (6, 0). Similarly, for the y-intercept, we set (x = 0), solving for (y), resulting in (0, 4).

With these points, we can draw the boundary line. However, since our original inequality is strict (using '>' rather than '>='), the boundary line should be dashed to indicate that points on the line are not included in the solution set. The test point method, which we will discuss in the next step, will help us determine which side of this line to shade.

The test point method is used to determine which side of the boundary line represents the solution region for an inequality. After plotting our dashed boundary line, we select a test point that is not on the line—usually (0, 0) is the most convenient if it's not on the boundary itself. We then substitute the coordinates of this test point into the original inequality.

If the inequality is satisfied with these values, the region that includes the test point is part of the solution set. If not satisfied, the opposite region is the solution set. In our case, substituting (0, 0) into (2x + 3y > 12) gives us (0 > 12), which is false. Hence, we do not shade the side that includes (0, 0). Instead, we shade the opposite side.

Shading is the final step in graphically representing a linear inequality and visually distinguishing the solution region from the rest of the graph. We've already established that the region containing our test point (0, 0) is not the solution region because the inequality wasn't true for this point.

Consequently, we need to shade the area on the opposite side of the boundary line. In this exercise, since (0, 0) does not satisfy the inequality, we shade the region above the boundary line, indicating that any point in this area (not on the line itself due to the dashed nature of the line) will satisfy the initial inequality (2x + 3y > 12). This visually communicates the set of all possible solutions to anyone reading the graph.