1. Let's analyze the system of inequalities for each case and find the feasible regions. 2. For A: - Inequalities: $$2x + 3y \leq 12$$ $$x - y \leq 2$$ $$x \geq 0, y \geq 0$$ - These define a polygon in the first quadrant bounded by the lines and inequalities. 3. For B: - Inequalities: $$2x + 3y \leq 12$$ $$x - y \geq 2$$ $$x \geq 0, y \geq 0$$ - The feasible region lies where $x-y \geq 2$ instead of $\leq 2$. 4. For C: - Equalities and inequalities: $$2x + 3y = 12$$ $$x - y \leq 2$$ $$x \geq 0, y \geq 0$$ - The feasible region is restricted to the line $2x + 3y=12$ and the other inequalities. 5. For D: - Inequalities: $$3x + 2y \leq 12$$ $$x - y \geq 2$$ $$x \geq 0, y \geq 0$$ - Different linear combination for the first inequality and $x - y \geq 2$. 6. For E: - Inequalities: $$3x + 2y \leq 12$$ $$x - y \leq 2$$ $$x \geq 0, y \geq 0$$ - Similar to D but with $x - y \leq 2$. 7. Each system describes a feasible region bounded by the lines and coordinate axes. Summary: - For each case, the intersections of the lines and axes define polygon vertices. - Inequalities restrict regions either above/below or left/right of these lines. Final note: Solving numerically or graphically gives exact feasible region shapes depending on inequality directions and constants.