1. **Problem:** Write equations for the lines shown on the graph. 2. **Step 1: Find the equation of the first line passing through points (-6,0) and (0,6).** - Use the slope formula: $$m=\frac{y_2-y_1}{x_2-x_1}=\frac{6-0}{0-(-6)}=\frac{6}{6}=1$$ - Use point-slope form with point (0,6): $$y - 6 = 1(x - 0)$$ - Simplify to slope-intercept form: $$y = x + 6$$ 3. **Step 2: Find the equation of the second line passing through points (0,6) and (6,-8).** - Calculate slope: $$m=\frac{-8-6}{6-0}=\frac{-14}{6}=-\frac{7}{3}$$ - Use point-slope form with point (0,6): $$y - 6 = -\frac{7}{3}(x - 0)$$ - Simplify: $$y = -\frac{7}{3}x + 6$$ 4. **Step 3: Describe how to find the solution to the system by looking at the graph.** - The solution is the point where the two lines intersect. - From the graph, the intersection is approximately at (1,5). 5. **Step 4: Describe how to find the solution using the equations.** - Set the two equations equal to each other since at the intersection their y-values are equal: $$x + 6 = -\frac{7}{3}x + 6$$ - Subtract 6 from both sides: $$x + \cancel{6} - \cancel{6} = -\frac{7}{3}x + \cancel{6} - \cancel{6}$$ $$x = -\frac{7}{3}x$$ - Add $\frac{7}{3}x$ to both sides: $$x + \frac{7}{3}x = 0$$ - Combine like terms: $$\frac{3}{3}x + \frac{7}{3}x = \frac{10}{3}x = 0$$ - Divide both sides by $\frac{10}{3}$: $$\cancel{\frac{10}{3}}x = \frac{0}{\cancel{\frac{10}{3}}}$$ $$x = 0$$ - Substitute $x=0$ into one equation, e.g., $y = x + 6$: $$y = 0 + 6 = 6$$ - The algebraic solution is $(0,6)$, which differs from the approximate graph intersection (1,5) due to rough graph reading. 6. **Step 5: For the second problem, find two equations whose solution is (5, -19).** - Substitute $x=5$, $y=-19$ into each equation: - A: $y = -3x - 6 \Rightarrow -19 = -3(5) - 6 = -15 - 6 = -21$ (No) - B: $y = 2x - 23 \Rightarrow -19 = 2(5) - 23 = 10 - 23 = -13$ (No) - C: $y = -7x + 16 \Rightarrow -19 = -7(5) + 16 = -35 + 16 = -19$ (Yes) - D: $y = x - 17 \Rightarrow -19 = 5 - 17 = -12$ (No) - E: $y = -2x - 9 \Rightarrow -19 = -2(5) - 9 = -10 - 9 = -19$ (Yes) - So equations C and E form a system with solution (5, -19).