1. **Stating the problem:** We are given a piecewise linear graph with two line segments. The first segment has an open circle at $(-7,5)$ and passes through $(-10,-9)$. The second segment has an open circle at $(2,6)$ and extends to $(10,-4)$. We want to find the equations of these two line segments. 2. **Formula used:** The equation of a line through two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by the point-slope form: $$y - y_1 = m(x - x_1)$$ where the slope $m$ is $$m = \frac{y_2 - y_1}{x_2 - x_1}$$ 3. **First line segment:** Points are $(-10,-9)$ and $(-7,5)$ (open circle at $(-7,5)$ means the segment does not include this point). Calculate slope: $$m = \frac{5 - (-9)}{-7 - (-10)} = \frac{14}{3}$$ Equation using point $(-10,-9)$: $$y - (-9) = \frac{14}{3}(x - (-10))$$ Simplify: $$y + 9 = \frac{14}{3}(x + 10)$$ $$y = \frac{14}{3}x + \frac{14}{3} \times 10 - 9$$ $$y = \frac{14}{3}x + \frac{140}{3} - 9$$ Convert 9 to thirds: $$9 = \frac{27}{3}$$ So: $$y = \frac{14}{3}x + \frac{140}{3} - \frac{27}{3} = \frac{14}{3}x + \frac{113}{3}$$ 4. **Second line segment:** Points are $(2,6)$ (open circle) and $(10,-4)$. Calculate slope: $$m = \frac{-4 - 6}{10 - 2} = \frac{-10}{8} = -\frac{5}{4}$$ Equation using point $(2,6)$: $$y - 6 = -\frac{5}{4}(x - 2)$$ Simplify: $$y - 6 = -\frac{5}{4}x + \frac{5}{2}$$ $$y = -\frac{5}{4}x + \frac{5}{2} + 6$$ Convert 6 to halves: $$6 = \frac{12}{2}$$ So: $$y = -\frac{5}{4}x + \frac{5}{2} + \frac{12}{2} = -\frac{5}{4}x + \frac{17}{2}$$ 5. **Summary:** - First segment equation: $$y = \frac{14}{3}x + \frac{113}{3}$$ for $x$ in $[-10,-7)$ (open circle at $-7$) - Second segment equation: $$y = -\frac{5}{4}x + \frac{17}{2}$$ for $x$ in $(2,10]$ (open circle at $2$) These equations describe the two linear pieces of the graph as given.