1. The problem is to graph the linear equations: $$y = -\frac{1}{3}x - 3$$ and $$y = -2x + 2$$ 2. These are linear equations in slope-intercept form $y = mx + b$, where $m$ is the slope and $b$ is the y-intercept. 3. For the first equation $y = -\frac{1}{3}x - 3$: - The slope $m = -\frac{1}{3}$ means the line falls 1 unit vertically for every 3 units it moves horizontally to the right. - The y-intercept $b = -3$ means the line crosses the y-axis at $(0, -3)$. 4. For the second equation $y = -2x + 2$: - The slope $m = -2$ means the line falls 2 units vertically for every 1 unit it moves horizontally to the right. - The y-intercept $b = 2$ means the line crosses the y-axis at $(0, 2)$. 5. To graph each line, plot the y-intercept and use the slope to find another point: - For $y = -\frac{1}{3}x - 3$, start at $(0, -3)$, then move right 3 units and down 1 unit to $(3, -4)$. - For $y = -2x + 2$, start at $(0, 2)$, then move right 1 unit and down 2 units to $(1, 0)$. 6. Draw straight lines through these points to complete the graphs. Final answer: The graphs are lines with slopes $-\frac{1}{3}$ and $-2$ crossing the y-axis at $-3$ and $2$ respectively.