`2x + y ge 6, 3x + 4y le 12`

Explanation

To solve the system of linear inequalities given by \(2x + y \geq 6\) and \(3x + 4y \leq 12\), we will follow these steps:

### Step 1: Convert inequalities to equations
First, we convert the inequalities into equations to find the boundary lines.

1. For the first inequality \(2x + y = 6\).
2. For the second inequality \(3x + 4y = 12\).

### Step 2: Find intercepts for the first equation
To graph the line \(2x + y = 6\), we can find the x-intercept and y-intercept.

- **X-intercept**: Set \(y = 0\):
\[ 2x + 0 = 6 \implies x = 3 \quad \text{(Point: (3, 0))} \]

- **Y-intercept**: Set \(x = 0\):
\[ 2(0) + y = 6 \implies y = 6 \quad \text{(Point: (0, 6))} \]

### Step 3: Find intercepts for the second equation
Next, we find the intercepts for the line \(3x + 4y = 12\).

- **X-intercept**: Set \(y = 0\):
\[ 3x + 4(0) = 12 \implies x = 4 \quad \text{(Point: (4, 0))} \]

- **Y-intercept**: Set \(x = 0\):
\[ 3(0) + 4y = 12 \implies y = 3 \quad \text{(Point: (0, 3))} \]

### Step 4: Graph the lines
Now we can plot the lines on a graph:

1. The line \(2x + y = 6\) passes through points (3, 0) and (0, 6).
2. The line \(3x + 4y = 12\) passes through points (4, 0) and (0, 3).

### Step 5: Determine the region for each inequality
- For \(2x + y \geq 6\): The region above the line is included (shaded).
- For \(3x + 4y \leq 12\): The region below the line is included (shaded).

### Step 6: Test a point to find the feasible region
To find the feasible region, we can test a point that is not on either line. A common choice is the origin (0, 0).

- For \(2x + y \geq 6\):
\[ 2(0) + 0 \geq 6 \quad \text{(False)} \]

- For \(3x + 4y \leq 12\):
\[ 3(0) + 4(0) \leq 12 \quad \text{(True)} \]

Since the first inequality is false at the origin, the feasible region does not include the origin. Thus, we shade the area that does not include the origin.

### Step 7: Identify the feasible region
The feasible region is the area where the shaded regions of both inequalities overlap.

### Conclusion
The solution to the system of inequalities \(2x + y \geq 6\) and \(3x + 4y \leq 12\) is the area of intersection of the two shaded regions, which is bounded by the lines and does not include the origin.

---