Solve the Following System of Inequalities Graphically
`3x + 2y le 6, x ge 1, y ge 2`

Explanation

To solve the system of inequalities graphically, we will follow these steps:

### Step 1: Rewrite the inequalities
We have the following inequalities:
1. \(3x + 2y \leq 6\)
2. \(x \geq 1\)
3. \(y \geq 2\)

### Step 2: Convert the first inequality to an equation
To graph the first inequality, we convert it to an equation:
\[3x + 2y = 6\]

### Step 3: Find intercepts for the line \(3x + 2y = 6\)
- **Find the y-intercept** (set \(x = 0\)):
\[ 3(0) + 2y = 6 \implies 2y = 6 \implies y = 3 \]
So, the y-intercept is \((0, 3)\).

- **Find the x-intercept** (set \(y = 0\)):
\[ 3x + 2(0) = 6 \implies 3x = 6 \implies x = 2 \]
So, the x-intercept is \((2, 0)\).

### Step 4: Plot the line \(3x + 2y = 6\)
- Plot the points \((0, 3)\) and \((2, 0)\) on the graph.
- Draw a straight line through these points. Since the inequality is \(\leq\), the line will be solid.

### Step 5: Determine the feasible region for \(3x + 2y \leq 6\)
- To find the feasible region, we can test a point not on the line, such as \((0, 0)\):
\[ 3(0) + 2(0) = 0 \leq 6 \quad \text{(True)} \]
Therefore, the region below the line (including the line) is the feasible region for this inequality.

### Step 6: Graph the second inequality \(x \geq 1\)
- This inequality represents a vertical line at \(x = 1\).
- Since the inequality is \(\geq\), the region to the right of this line is the feasible region.

### Step 7: Graph the third inequality \(y \geq 2\)
- This inequality represents a horizontal line at \(y = 2\).
- Since the inequality is \(\geq\), the region above this line is the feasible region.

### Step 8: Identify the feasible region
- Now, we need to find the region that satisfies all three inequalities:
- The area below the line \(3x + 2y = 6\)
- The area to the right of the line \(x = 1\)
- The area above the line \(y = 2\)

### Step 9: Check for common feasible region
- We can check for intersections:
- The line \(3x + 2y = 6\) intersects \(x = 1\):
\[ 3(1) + 2y = 6 \implies 3 + 2y = 6 \implies 2y = 3 \implies y = 1.5 \quad \text{(not feasible since } y \geq 2\text{)} \]
- The line \(3x + 2y = 6\) intersects \(y = 2\):
\[ 3x + 2(2) = 6 \implies 3x + 4 = 6 \implies 3x = 2 \implies x = \frac{2}{3} \quad \text{(not feasible since } x \geq 1\text{)} \]

### Conclusion
Since there are no points that satisfy all three inequalities simultaneously, the system of inequalities has no solution.

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