Feasible Region 2Ca4B3

1. **State the problem:** We need to graph the feasible region for the system of inequalities: $$x + y \leq 5$$ $$x - y \geq 6$$ 2. **Rewrite inequalities as equations to find boundary lines:** - Line 1: $$x + y = 5$$ - Line 2: $$x - y = 6$$ 3. **Find intersection point of the boundary lines:** From line 2, express $y$: $$x - y = 6 \implies y = x - 6$$ Substitute into line 1: $$x + (x - 6) = 5$$ $$2x - 6 = 5$$ $$2x = 11$$ $$x = \frac{11}{2}$$ Then, $$y = \frac{11}{2} - 6 = \frac{11}{2} - \frac{12}{2} = -\frac{1}{2}$$ So the lines intersect at $$\left( \frac{11}{2}, -\frac{1}{2} \right)$$. 4. **Determine the half-planes for each inequality:** - For $$x + y \leq 5$$, the feasible region is the half-plane **below or on** the line $x + y = 5$. - For $$x - y \geq 6$$, rewrite as $$y \leq x - 6$$, so the feasible region is the half-plane **below or on** the line $y = x - 6$. 5. **Feasible region:** The overlap of these two half-planes is the region below both lines. 6. **Shape and position:** The feasible region is unbounded downward to the right, bounded above by the two lines intersecting at $$\left( \frac{11}{2}, -\frac{1}{2} \right)$$. **Final answer:** The feasible region is the intersection of the half-planes below the lines $x + y = 5$ and $y = x - 6$, bounded above by their intersection point at $$\left( \frac{11}{2}, -\frac{1}{2} \right)$$, and extending unbounded downward to the right.