1. **State the problem:** Graph the system of inequalities: $$x + 3y > 6$$ $$2x + y \leq -3$$ 2. **Rewrite inequalities as equations for boundary lines:** - For $$x + 3y > 6$$, the boundary line is $$x + 3y = 6$$ (dashed because inequality is strict). - For $$2x + y \leq -3$$, the boundary line is $$2x + y = -3$$ (solid because inequality includes equality). 3. **Find intercepts for $$x + 3y = 6$$:** - When $$x=0$$, $$3y=6 \Rightarrow y=2$$. - When $$y=0$$, $$x=6$$. 4. **Find intercepts for $$2x + y = -3$$:** - When $$x=0$$, $$y=-3$$. - When $$y=0$$, $$2x=-3 \Rightarrow x=-\frac{3}{2}$$. 5. **Determine shading for $$x + 3y > 6$$:** - Test point $$ (0,0) $$: $$0 + 3(0) = 0 \not> 6$$, so shade opposite side of line from origin. 6. **Determine shading for $$2x + y \leq -3$$:** - Test point $$ (0,0) $$: $$0 + 0 = 0 \not\leq -3$$, so shade opposite side of line from origin. 7. **Solution region:** - The overlapping shaded region is where both inequalities hold. **Final answer:** Graph the dashed line $$x + 3y = 6$$ with shading above it (away from origin). Graph the solid line $$2x + y = -3$$ with shading below it (away from origin). The solution is the intersection of these shaded regions.