1. **State the problem:** We want to understand the system of inequalities: $$y - x < 2$$ $$2y + 3x \leq 12$$ $$x \geq 0$$ and describe the solution region they define. 2. **Rewrite each inequality for clarity:** - From $$y - x < 2$$, add $$x$$ to both sides to get $$y < x + 2$$ This means the region below the line $$y = x + 2$$, not including the line since it's strict inequality. - From $$2y + 3x \leq 12$$, isolate $$y$$: $$2y \leq 12 - 3x$$ $$y \leq \frac{12 - 3x}{2} = 6 - \frac{3}{2}x$$ This is the region below or on the line $$y = 6 - \frac{3}{2}x$$. - The third inequality $$x \geq 0$$ represents all points on or to the right of the $$y$$-axis. 3. **Interpret the combined solution region:** - The feasible region satisfies all three: - Below the line $$y = x + 2$$ (strictly less than) - Below or on the line $$y = 6 - \frac{3}{2}x$$ - To the right of or on the $$y$$-axis This means the solution set is the intersection of these three regions in the coordinate plane. 4. **Final answer:** The solution region is all points $$ (x,y) $$ such that $$x \geq 0, \quad y < x + 2, \quad y \leq 6 - \frac{3}{2}x.$$