solve the following system of inequation by graphical method : `5x +4y le 40, x ge 2, y ge 3`

Explanation

To solve the system of inequalities \(5x + 4y \leq 40\), \(x \geq 2\), and \(y \geq 3\) using the graphical method, follow these steps:

### Step 1: Convert the inequality into an equation
First, we will convert the inequality \(5x + 4y \leq 40\) into an equation to find the boundary line:
\[ 5x + 4y = 40 \]

### Step 2: Find intercepts
To graph the line, we need to find the x-intercept and y-intercept.

- **Finding the x-intercept**: Set \(y = 0\):
\[ 5x + 4(0) = 40 \implies 5x = 40 \implies x = 8 \]
So, the x-intercept is \((8, 0)\).

- **Finding the y-intercept**: Set \(x = 0\):
\[ 5(0) + 4y = 40 \implies 4y = 40 \implies y = 10 \]
So, the y-intercept is \((0, 10)\).

### Step 3: Graph the line
Plot the points \((8, 0)\) and \((0, 10)\) on the graph and draw a straight line through these points. Since the inequality is \(\leq\), the line will be solid.

### Step 4: Determine the shaded region for \(5x + 4y \leq 40\)
To determine which side of the line to shade, we can test a point not on the line, such as \((0, 0)\):
\[ 5(0) + 4(0) = 0 \leq 40 \quad \text{(True)} \]
Thus, we shade the region that includes the origin.

### Step 5: Graph the second inequality \(x \geq 2\)
This inequality represents a vertical line at \(x = 2\). Since it is \(\geq\), we will draw a solid line at \(x = 2\) and shade to the right of this line.

### Step 6: Graph the third inequality \(y \geq 3\)
This inequality represents a horizontal line at \(y = 3\). Since it is \(\geq\), we will draw a solid line at \(y = 3\) and shade above this line.

### Step 7: Identify the feasible region
Now, we look for the common shaded region that satisfies all three inequalities. This region will be bounded by the lines \(5x + 4y = 40\), \(x = 2\), and \(y = 3\).

### Step 8: Determine the vertices of the feasible region
The vertices of the feasible region can be found by solving the equations of the lines:
1. Intersection of \(5x + 4y = 40\) and \(x = 2\):
\[ 5(2) + 4y = 40 \implies 10 + 4y = 40 \implies 4y = 30 \implies y = 7.5 \]
So, one vertex is \((2, 7.5)\).

2. Intersection of \(5x + 4y = 40\) and \(y = 3\):
\[ 5x + 4(3) = 40 \implies 5x + 12 = 40 \implies 5x = 28 \implies x = 5.6 \]
So, another vertex is \((5.6, 3)\).

3. The point where \(x = 2\) and \(y = 3\) is also a vertex: \((2, 3)\).

### Final Solution
The feasible region is a triangle formed by the points \((2, 3)\), \((2, 7.5)\), and \((5.6, 3)\). This region represents the solution to the system of inequalities.