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

Explanation

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

### Step 1: Write down the inequalities
The inequalities given are:
1. \( x + 2y \leq 3 \)
2. \( 3x + 4y \geq 12 \)
3. \( x \geq 0 \)
4. \( y \geq 1 \)

### Step 2: Graph the first inequality \( x + 2y \leq 3 \)
First, we convert the inequality into an equation to find the boundary line:
- \( x + 2y = 3 \)

To find the intercepts:
- When \( x = 0 \):
\[ 0 + 2y = 3 \implies y = \frac{3}{2} \]
- When \( y = 0 \):
\[ x + 0 = 3 \implies x = 3 \]

Thus, the intercepts are \( (0, \frac{3}{2}) \) and \( (3, 0) \). Draw the line connecting these points and shade below it (since it is \( \leq \)).

### Step 3: Graph the second inequality \( 3x + 4y \geq 12 \)
Convert this inequality into an equation:
- \( 3x + 4y = 12 \)

To find the intercepts:
- When \( x = 0 \):
\[ 3(0) + 4y = 12 \implies y = 3 \]
- When \( y = 0 \):
\[ 3x + 0 = 12 \implies x = 4 \]

Thus, the intercepts are \( (0, 3) \) and \( (4, 0) \). Draw the line connecting these points and shade above it (since it is \( \geq \)).

### Step 4: Graph the constraints \( x \geq 0 \) and \( y \geq 1 \)
- The line \( x = 0 \) is the y-axis, and we shade to the right of this line.
- The line \( y = 1 \) is a horizontal line, and we shade above this line.

### Step 5: Identify the feasible region
Now, we need to find the area where all shaded regions overlap.

1. The area below the line \( x + 2y = 3 \).
2. The area above the line \( 3x + 4y = 12 \).
3. The area to the right of the y-axis (where \( x \geq 0 \)).
4. The area above the line \( y = 1 \).

### Step 6: Check for common solutions
Now we check if there is a common area that satisfies all inequalities.

- The line \( x + 2y = 3 \) intersects the line \( y = 1 \) at:
\[ x + 2(1) = 3 \implies x = 1 \]
So the point \( (1, 1) \) is on the boundary.

- The line \( 3x + 4y = 12 \) intersects the line \( y = 1 \) at:
\[ 3x + 4(1) = 12 \implies 3x = 8 \implies x = \frac{8}{3} \]
So the point \( \left(\frac{8}{3}, 1\right) \) is on the boundary.

### Step 7: Analyze the feasible region
After plotting all the lines and shading the appropriate regions, we find that there is no overlapping area that satisfies all the inequalities simultaneously.

### Conclusion
Thus, there is no feasible solution for the given system of inequalities.

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