You can draw the area bounded by the unequality [tex]y\ \textgreater \ \frac{3}{4} x-2[/tex] in following way:
1. Draw a dashed line [tex]y=\frac{3}{4} x-2[/tex] with a slope [tex] \frac{3}{4} [/tex] (a dashed, because all points from this line, for example when x=0, y=-2 do not belong to the area);
2. Define which part (under the line or over the line) is suitable by substituting values. For example, when x=0 and y=0 you obtain that [tex]0\ \textgreater \ \frac{3}{4} \cdot 0-2[/tex] is a right unequality, that means that the point (0,0) is a solution and belongs to the area;
3. Shade that part of the plane to which point (0,0) belongs, you receive the area over the line (see image).
From the done steps you can see that only fourth statement is true.
