1. The problem asks for the three inequalities that define the unshaded triangular region bounded by three lines. 2. The first boundary is a solid decreasing line through points (0,12) and (6,0). The slope is calculated as: $$m = \frac{0 - 12}{6 - 0} = \frac{-12}{6} = -2$$ The equation of the line using point-slope form with point (0,12) is: $$y - 12 = -2(x - 0) \implies y = -2x + 12$$ Since the region is to the right of this line and the line is solid, the inequality is: $$y \leq -2x + 12$$ 3. The second boundary is a dashed increasing line through points (0,3) and (8,11). The slope is: $$m = \frac{11 - 3}{8 - 0} = \frac{8}{8} = 1$$ The equation of the line is: $$y - 3 = 1(x - 0) \implies y = x + 3$$ Since the region is above this dashed line, the inequality is: $$y > x + 3$$ 4. The third boundary is a solid vertical line at: $$x = 4$$ The region is to the left of this line, so the inequality is: $$x \leq 4$$ 5. Therefore, the three inequalities defining the unshaded triangular region are: $$\boxed{\begin{cases} y \leq -2x + 12 \\ y > x + 3 \\ x \leq 4 \end{cases}}$$