1. **State the problem:** We need to graph the system of linear inequalities: $$y < 3x + 2$$ $$y \geq 3x - 4$$ and find the solution region where both inequalities are true. 2. **Understand the inequalities:** - The first inequality $y < 3x + 2$ means the solution is all points below the line $y = 3x + 2$ (not including the line). - The second inequality $y \geq 3x - 4$ means the solution is all points on or above the line $y = 3x - 4$. 3. **Graph the boundary lines:** - For $y = 3x + 2$, plot the line with slope 3 and y-intercept 2. This line is dashed because the inequality is strict ($<$). - For $y = 3x - 4$, plot the line with slope 3 and y-intercept -4. This line is solid because the inequality includes equality ($\geq$). 4. **Shade the regions:** - Shade below the line $y = 3x + 2$. - Shade above the line $y = 3x - 4$. 5. **Find the solution region:** - The solution to the system is the overlap of the two shaded regions. 6. **Check for intersection:** - Since both lines have the same slope 3, they are parallel and never intersect. - The region between these two parallel lines is the solution. **Final answer:** The solution region is the set of points $y$ such that $$3x - 4 \leq y < 3x + 2$$ which is the area between the two parallel lines, including the lower line but not the upper line.