Solve the given linear inequality and specify the bounded region.
`2x + y ge 2, x - y le 1, x + 2y le 8, x ge 0, y ge 0`.

Explanation

To solve the given linear inequalities and specify the bounded region, we will follow these steps:

### Step 1: Convert Inequalities to Equalities
We start by converting each inequality into an equality to find the boundary lines:
1. \(2x + y = 2\)
2. \(x - y = 1\)
3. \(x + 2y = 8\)
4. \(x = 0\) (y-axis)
5. \(y = 0\) (x-axis)

### Step 2: Find Intercepts of Each Line
Next, we find the intercepts for each line to plot them on the graph.

1. **For \(2x + y = 2\)**:
- When \(x = 0\), \(y = 2\) (point: (0, 2))
- When \(y = 0\), \(x = 1\) (point: (1, 0))

2. **For \(x - y = 1\)**:
- When \(x = 0\), \(y = -1\) (point: (0, -1))
- When \(y = 0\), \(x = 1\) (point: (1, 0))

3. **For \(x + 2y = 8\)**:
- When \(x = 0\), \(y = 4\) (point: (0, 4))
- When \(y = 0\), \(x = 8\) (point: (8, 0))

### Step 3: Plot the Lines
Now, we will plot the lines on a graph using the points found above.

1. **Line for \(2x + y = 2\)**: Plot points (0, 2) and (1, 0) and draw the line.
2. **Line for \(x - y = 1\)**: Plot points (0, -1) and (1, 0) and draw the line.
3. **Line for \(x + 2y = 8\)**: Plot points (0, 4) and (4, 2) and draw the line.
4. **Lines for \(x = 0\) and \(y = 0\)**: These are the axes.

### Step 4: Determine the Feasible Region
Next, we need to determine which side of each line satisfies the inequality.

1. **For \(2x + y \geq 2\)**: Test the origin (0, 0):
- \(2(0) + 0 \geq 2\) → False. So, the solution lies on the right side of the line.

2. **For \(x - y \leq 1\)**: Test the origin (0, 0):
- \(0 - 0 \leq 1\) → True. So, the solution lies below the line.

3. **For \(x + 2y \leq 8\)**: Test the origin (0, 0):
- \(0 + 2(0) \leq 8\) → True. So, the solution lies below the line.

4. **For \(x \geq 0\) and \(y \geq 0\)**: This restricts us to the first quadrant.

### Step 5: Identify the Bounded Region
The bounded region is where all the shaded areas from the inequalities overlap. This will be a polygon formed by the intersection points of the lines.

### Step 6: Find Intersection Points
To find the vertices of the bounded region, we need to solve the equations pairwise:

1. **Intersection of \(2x + y = 2\) and \(x - y = 1\)**:
- Solve the system:
\[ 2x + y = 2 \quad (1) \]
\[ x - y = 1 \quad (2) \]
- From (2), \(y = x - 1\). Substitute into (1):
\[ 2x + (x - 1) = 2 \implies 3x - 1 = 2 \implies 3x = 3 \implies x = 1 \]
\[ y = 1 - 1 = 0 \implies (1, 0) \]

2. **Intersection of \(x - y = 1\) and \(x + 2y = 8\)**:
- Solve:
\[ x - y = 1 \quad (3) \]
\[ x + 2y = 8 \quad (4) \]
- From (3), \(x = y + 1\). Substitute into (4):
\[ (y + 1) + 2y = 8 \implies 3y + 1 = 8 \implies 3y = 7 \implies y = \frac{7}{3} \]
\[ x = \frac{7}{3} + 1 = \frac{10}{3} \implies \left(\frac{10}{3}, \frac{7}{3}\right) \]

3. **Intersection of \(2x + y = 2\) and \(x + 2y = 8\)**:
- Solve:
\[ 2x + y = 2 \quad (5) \]
\[ x + 2y = 8 \quad (6) \]
- From (5), \(y = 2 - 2x\). Substitute into (6):
\[ x + 2(2 - 2x) = 8 \implies x + 4 - 4x = 8 \implies -3x = 4 \implies x = -\frac{4}{3} \text{ (not in the first quadrant)} \]

4. **Intersection of \(x + 2y = 8\) and \(y = 0\)**:
- Substitute \(y = 0\) into \(x + 2(0) = 8\):
\[ x = 8 \implies (8, 0) \]

### Conclusion
The vertices of the bounded region are:
1. (1, 0)
2. (0, 4)
3. (0, 2)
4. \(\left(\frac{10}{3}, \frac{7}{3}\right)\)

The bounded region is the area enclosed by these points in the first quadrant.