1. **State the problem:** Graph the inequality $$-x + 3y > -12$$. 2. **Rewrite the inequality in slope-intercept form:** Start by isolating $y$: $$-x + 3y > -12$$ Add $x$ to both sides: $$3y > x - 12$$ Divide both sides by 3: $$y > \frac{\cancel{3}x}{\cancel{3}} - \frac{12}{3}$$ which simplifies to: $$y > \frac{1}{3}x - 4$$ 3. **Interpret the inequality:** The boundary line is $$y = \frac{1}{3}x - 4$$. Since the inequality is strict ($>$), the line is dashed. 4. **Graph the boundary line:** - The y-intercept is at $(0, -4)$. - The slope is $\frac{1}{3}$, meaning for every 3 units right, go 1 unit up. 5. **Shade the solution region:** Because the inequality is $$y > \frac{1}{3}x - 4$$, shade the region above the line. 6. **Summary:** - Draw the dashed line $$y = \frac{1}{3}x - 4$$. - Shade above this line to represent all points satisfying the inequality. This completes the graphing of the inequality.