`x + 2y le 3, 3x + 4y ge 12, x ge 0, y ge 1`

Explanation

To solve the given linear inequalities step by step, we will analyze each inequality and find the feasible region graphically.

### Given Inequalities:
1. \( x + 2y \leq 3 \)
2. \( 3x + 4y \geq 12 \)
3. \( x \geq 0 \)
4. \( y \geq 1 \)

### Step 1: Graph the first inequality \( x + 2y \leq 3 \)

1. **Convert to equality**: \( x + 2y = 3 \)
2. **Find intercepts**:
- Set \( x = 0 \): \( 0 + 2y = 3 \) → \( y = \frac{3}{2} \) (Point: \( (0, \frac{3}{2}) \))
- Set \( y = 0 \): \( x + 0 = 3 \) → \( x = 3 \) (Point: \( (3, 0) \))
3. **Plot the line**: Draw a line through points \( (0, \frac{3}{2}) \) and \( (3, 0) \).
4. **Shade the region**: Since it is \( \leq \), shade below the line.

### Step 2: Graph the second inequality \( 3x + 4y \geq 12 \)

1. **Convert to equality**: \( 3x + 4y = 12 \)
2. **Find intercepts**:
- Set \( x = 0 \): \( 3(0) + 4y = 12 \) → \( y = 3 \) (Point: \( (0, 3) \))
- Set \( y = 0 \): \( 3x + 0 = 12 \) → \( x = 4 \) (Point: \( (4, 0) \))
3. **Plot the line**: Draw a line through points \( (0, 3) \) and \( (4, 0) \).
4. **Shade the region**: Since it is \( \geq \), shade above the line.

### Step 3: Graph the constraints \( x \geq 0 \) and \( y \geq 1 \)

1. **For \( x \geq 0 \)**: This is the right half-plane to the right of the y-axis.
2. **For \( y \geq 1 \)**: This is the region above the horizontal line \( y = 1 \).

### Step 4: Identify the feasible region

Now, we need to find the common area that satisfies all inequalities.

1. The first inequality \( x + 2y \leq 3 \) gives a region below the line.
2. The second inequality \( 3x + 4y \geq 12 \) gives a region above the line.
3. The constraints \( x \geq 0 \) and \( y \geq 1 \) restrict the solution to the first quadrant and above the line \( y = 1 \).

### Step 5: Check for common solutions

1. **Check intersection points**:
- The lines \( x + 2y = 3 \) and \( 3x + 4y = 12 \) can be solved simultaneously to find points of intersection.
- However, checking the origin \( (0,0) \) in both inequalities shows that \( (0,0) \) does not satisfy \( 3x + 4y \geq 12 \).
- Similarly, checking other points will show that there is no overlap in the shaded regions.

### Conclusion

Since there is no common area that satisfies all inequalities, we conclude that there is **no feasible solution** for the given linear inequalities.

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