Let's start by graphing the first inequality's boundary line. Since the inequality is strict, the line will be dashed to indicate that the points on the line are not part of the solution set.
In order to determine which region to shade, we will use the test point (0,0). If we obtain a true statement, we will shade the region that contains the point. Otherwsie, we will shade the region that does not contain the point.
Since the values made the inequality false, we will shade the region that does not contain test point.
We will follow the same procedure to graph the boundary line of the second inequality. Since the second inequality is non-strict, the boundary line will be solid. We will use the same test point to determine which region to shade for the second inequality.
Notice that both lines have a slope of 3. At the same time, the lines are not identical. For example, let's look at the values of y for x=0.
Equation (I)
Equation (II)
y=3x+3
y=3x−5
y=3(0)+3
y=3(0)−5
y=3
y=-5
Since the lines have the same slope and are not identical, we know that the lines are parallel and will never intersect.
Let's take another look at the graph from Part A.
We can see that the shaded regions do not overlap. Additionally, because the boundary lines are parallel, as we found in Part B, the shaded regions will remain an equal distance apart for as long as the lines continue. Therefore, they will never overlap.
The solution set of a system of linear inequalities can be interpreted as the intersection of shaded regions of graphs of all the inequalities in the system. Since the graphs do not, and will never, overlap, there are no solutions to this system.
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