`x + y le 10, 4x + 3y le 24, x ge , x ge 0, y ge 0`

Explanation

To solve the given system of linear inequalities step by step, we will follow these steps:

### Step 1: Write down the inequalities
The inequalities we need to consider are:
1. \( x + y \leq 10 \)
2. \( 4x + 3y \leq 24 \)
3. \( x \geq 0 \)
4. \( y \geq 0 \)

### Step 2: Find the boundary lines
To graph the inequalities, we first convert them into equations to find the boundary lines.

1. For \( x + y = 10 \):
- If \( x = 0 \), then \( y = 10 \) (Point: \( (0, 10) \))
- If \( y = 0 \), then \( x = 10 \) (Point: \( (10, 0) \))

2. For \( 4x + 3y = 24 \):
- If \( x = 0 \), then \( 3y = 24 \) → \( y = 8 \) (Point: \( (0, 8) \))
- If \( y = 0 \), then \( 4x = 24 \) → \( x = 6 \) (Point: \( (6, 0) \))

### Step 3: Graph the boundary lines
Now we will plot the points on a graph:
- Plot \( (0, 10) \) and \( (10, 0) \) for the line \( x + y = 10 \).
- Plot \( (0, 8) \) and \( (6, 0) \) for the line \( 4x + 3y = 24 \).

### Step 4: Determine the shaded regions
Next, we need to determine which side of each line to shade:
1. For \( x + y \leq 10 \): Shade below the line.
2. For \( 4x + 3y \leq 24 \): Shade below the line.
3. Since \( x \geq 0 \) and \( y \geq 0 \): We only consider the first quadrant.

### Step 5: Identify the feasible region
The feasible region is where all shaded areas overlap. This will be a polygon formed by the intersection of the lines and the axes.

### Step 6: Identify the vertices of the feasible region
The vertices of the feasible region can be found by identifying the intersection points:
1. The points of intersection are:
- \( (0, 0) \) (origin)
- \( (0, 8) \) from \( 4x + 3y = 24 \)
- \( (6, 0) \) from \( 4x + 3y = 24 \)
- The intersection of the two lines can be found by solving the equations:
- \( x + y = 10 \)
- \( 4x + 3y = 24 \)

Solving these simultaneously:
- From \( x + y = 10 \), we get \( y = 10 - x \).
- Substitute \( y \) in the second equation:
\[ 4x + 3(10 - x) = 24 \]
\[ 4x + 30 - 3x = 24 \]
\[ x + 30 = 24 \Rightarrow x = -6 \text{ (not valid since } x \geq 0\text{)} \]
Therefore, the intersection does not yield a valid point in the first quadrant.

### Step 7: Conclusion
The feasible region is bounded by the points \( (0, 0) \), \( (0, 8) \), and \( (6, 0) \).