Certainly! Let's break down the solution to the system of inequalities step by step and understand what the graph of the solution would look like.
### Step 1: Analyze the Inequalities
We need to analyze the two inequalities separately:
1. \( 2x + y > 2 \)
2. \( 6x + 3y < 12 \)
### Step 2: Convert Inequalities to Equations
To visualize these inequalities, we first convert them to their corresponding equations:
1. \( 2x + y = 2 \)
2. \( 6x + 3y = 12 \)
### Step 3: Simplify the Second Equation
The second equation can be simplified. Divide every term by 3:
[tex]\[ \frac{6x}{3} + \frac{3y}{3} = \frac{12}{3} \][/tex]
[tex]\[ 2x + y = 4 \][/tex]
### Step 4: Find Intersection Points
To find the intersection point of the lines \( 2x + y = 2 \) and \( 2x + y = 4 \), we set the equations equal to each other:
[tex]\[ 2x + y = 2 \][/tex]
[tex]\[ 2x + y = 4 \][/tex]
Subtract the first equation from the second:
[tex]\[ (2x + y) - (2x + y) = 4 - 2 \][/tex]
[tex]\[ 0 = 2 \][/tex]
This result indicates that the lines are parallel and do not intersect, meaning there is no common solution (intersection point of the equations).
### Step 5: Graph the Lines
1. Graph of \( 2x + y = 2 \):
- Rewrite it as \( y = -2x + 2 \).
2. Graph of \( 2x + y = 4 \):
- Rewrite it as \( y = -2x + 4 \).
These two lines are parallel as they have the same slope (-2) but different y-intercepts.
### Step 6: Shade the Regions
1. Inequality \( 2x + y > 2 \):
- The solution region is above the line \( y = -2x + 2 \).
2. Inequality \( 6x + 3y < 12 \) or simplified \( 2x + y < 4 \):
- The solution region is below the line \( y = -2x + 4 \).
Since the two lines are parallel and there is no intersection:
- The solution to \( 2x + y > 2 \) would be the region above the line \( y = -2x + 2 \).
- The solution to \( 2x + y < 4 \) would be the region below the line \( y = -2x + 4 \).
### Step 7: Combine the Inequalities
The feasible region for the solution of the system of inequalities is where the shaded regions from both inequalities overlap. Due to the fact that these two lines are parallel and hence never intersect, this overlapping region does not exist.
### Conclusion
The solution to the system of inequalities [tex]\( 2x + y > 2 \)[/tex] and [tex]\( 6x + 3y < 12 \)[/tex] is an empty set. The graph would show the two parallel lines with no overlapping region indicating there is no (x, y) pair that satisfies both inequalities simultaneously.
