Step 1: Graph Interpretation

The graph above shows the lines corresponding to the inequalities:

• $y = -2x + 6$ (blue line)
• $y = -x + 3$ (green line)
• $y = 2$ (red line)

The shaded gray region represents the intersection of these inequalities, which we label as region $S$.

Step 2: Calculate the Area of Region S

To find the area of region $S$, we first need to identify the vertices of the polygon that forms this region. Let's determine these vertices by solving the system of equations at their intersection points.

Finding Intersection Points:

1. Intersection of $y = -2x + 6$ and $y = -x + 3$: $-2x + 6 = -x + 3 \implies x = 3, \quad y = 0$ Vertex: $(3, 0)$

2. Intersection of $y = -x + 3$ and $y = 2$: $2 = -x + 3 \implies x = 1, \quad y = 2$ Vertex: $(1, 2)$

3. Intersection of $y = -2x + 6$ and $y = 2$: $2 = -2x + 6 \implies x = 2, \quad y = 2$ Vertex: $(2, 2)$

The vertices of the region $S$ are $(1, 2)$, $(2, 2)$, and $(3, 0)$.

Calculating the Area:

We can calculate the area of this triangle using the formula:

$\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$ Where the vertices are $(x_1, y_1) = (1, 2)$, $(x_2, y_2) = (2, 2)$, $(x_3, y_3) = (3, 0)$.

Let's calculate this next.### Step 2: Area of Region S

The area of the required region $S$ is $1$ square unit.

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Do you want any further details or have any questions? Here are some related problems you might find interesting:

1. How would the area change if the inequality $y \leq 2$ was replaced by $y \leq 3$?
2. Can you determine the coordinates of a point that lies within the region $S$?
3. What would be the area if one of the inequalities was reversed?
4. How would the graph change if the inequality $2x + y - 6 \geq 0$ was replaced by $2x + y - 6 \leq 0$?
5. What is the significance of the intersection points in the context of linear programming?

Tip: When dealing with systems of inequalities, graphing can help you visually identify the feasible region where all conditions are met.