1. **State the problem:** We need to determine which points are solutions to the system defined by the shaded triangular region bounded by two lines and the x-axis. 2. **Write the system of inequalities:** - The solid line passes through points (-3,4) and (4,-3). Calculate its slope: $$m=\frac{-3-4}{4-(-3)}=\frac{-7}{7}=-1$$ Equation form: $$y=m x + b$$ Using point (-3,4): $$4 = -1 \times (-3) + b \Rightarrow 4 = 3 + b \Rightarrow b=1$$ So, solid line equation: $$y = -x + 1$$ - The dotted line passes through (-3,-2) and (0,3). Calculate slope: $$m=\frac{3 - (-2)}{0 - (-3)}=\frac{5}{3}$$ Using point (0,3): $$3 = \frac{5}{3} \times 0 + b \Rightarrow b=3$$ So, dotted line equation: $$y = \frac{5}{3} x + 3$$ - The x-axis is $$y=0$$ 3. **Write the system of inequalities describing the shaded region:** - The shaded region is below the solid line: $$y \leq -x + 1$$ - Above the dotted line: $$y \geq \frac{5}{3} x + 3$$ - Above the x-axis: $$y \geq 0$$ 4. **Check each labeled point:** - A(3,2): $$2 \leq -3 + 1 = -2 \text{? No}$$ Not a solution. - B(2,1): $$1 \leq -2 + 1 = -1 \text{? No}$$ Not a solution. - C(-1.5,3.5): $$3.5 \leq 1.5 + 1 = 2.5 \text{? No}$$ Not a solution. - D(0,3): $$3 \leq 0 + 1 = 1 \text{? No}$$ Not a solution. - E(-1,1): $$1 \leq 1 + 1 = 2 \text{Yes}$$ $$1 \geq \frac{5}{3} \times (-1) + 3 = -\frac{5}{3} + 3 = \frac{4}{3} \approx 1.33 \text{? No}$$ Not a solution. - F(-2,0): $$0 \leq 2 + 1 = 3 \text{Yes}$$ $$0 \geq \frac{5}{3} \times (-2) + 3 = -\frac{10}{3} + 3 = -\frac{1}{3} \text{Yes}$$ $$0 \geq 0 \text{Yes}$$ F is a solution. - G(1,1): $$1 \leq -1 + 1 = 0 \text{No}$$ Not a solution. - H(-2,-2): $$-2 \leq 2 + 1 = 3 \text{Yes}$$ $$-2 \geq -\frac{10}{3} + 3 = -\frac{1}{3} \text{No}$$ Not a solution. **Final answer:** Only point F(-2,0) satisfies the system. --- **Summary:** The system is: $$\begin{cases} y \leq -x + 1 \\ y \geq \frac{5}{3} x + 3 \\ y \geq 0 \end{cases}$$ Only point F(-2,0) is a solution to this system.