Answer :
To determine whether there is a feasible region that satisfies both inequalities, we need to analyze each inequality step-by-step.
### Inequality 1: [tex]\( 2x + y > 2 \)[/tex]
1. Convert to equality for graphing: [tex]\( 2x + y = 2 \)[/tex]
- This is a straight line. To graph it, we'll find the intercepts:
- When [tex]\( x = 0 \)[/tex], [tex]\( y = 2 \)[/tex].
- When [tex]\( y = 0 \)[/tex], [tex]\( x = 1 \)[/tex].
2. Determine the region for the inequality: [tex]\( 2x + y > 2 \)[/tex]
- This represents the region above the line [tex]\( 2x + y = 2 \)[/tex].
### Inequality 2: [tex]\( 6x + 3y < 12 \)[/tex]
1. Convert to equality for graphing: [tex]\( 6x + 3y = 12 \)[/tex]
- Simplify this to [tex]\( 2x + y = 4 \)[/tex].
- This is another straight line. To graph it, we'll find the intercepts:
- When [tex]\( x = 0 \)[/tex], [tex]\( y = 4 \)[/tex].
- When [tex]\( y = 0 \)[/tex], [tex]\( x = 2 \)[/tex].
2. Determine the region for the inequality: [tex]\( 6x + 3y < 12 \)[/tex]
- This represents the region below the line [tex]\( 2x + y = 4 \)[/tex].
### Intersection and Feasibility
To find the feasible region, we need to locate where both conditions [tex]\( 2x + y > 2 \)[/tex] and [tex]\( 6x + 3y < 12 \)[/tex] are satisfied simultaneously. This is the region that lies:
- Above the line [tex]\( 2x + y = 2 \)[/tex], and
- Below the line [tex]\( 2x + y = 4 \)[/tex].
### Feasible Region
1. Draw the lines:
- Line 1 (2x + y = 2): A line that could intersect the y-axis at (0,2) and the x-axis at (1,0).
- Line 2 (2x + y = 4): A line that intersects the y-axis at (0,4) and the x-axis at (2,0).
2. Assess the regions:
- The region above [tex]\( 2x + y = 2 \)[/tex] does not include the line itself.
- The region below [tex]\( 2x + y = 4 \)[/tex] does not include the line itself.
3. Feasibility Check:
- If the feasible region exists, there would be an area between these two lines where both conditions are met.
### Conclusion
After analyzing the inequalities and their graphical representations, we see there indeed exists a region between the two lines where both inequalities are satisfied. Therefore, the correct answer is:
True
### Inequality 1: [tex]\( 2x + y > 2 \)[/tex]
1. Convert to equality for graphing: [tex]\( 2x + y = 2 \)[/tex]
- This is a straight line. To graph it, we'll find the intercepts:
- When [tex]\( x = 0 \)[/tex], [tex]\( y = 2 \)[/tex].
- When [tex]\( y = 0 \)[/tex], [tex]\( x = 1 \)[/tex].
2. Determine the region for the inequality: [tex]\( 2x + y > 2 \)[/tex]
- This represents the region above the line [tex]\( 2x + y = 2 \)[/tex].
### Inequality 2: [tex]\( 6x + 3y < 12 \)[/tex]
1. Convert to equality for graphing: [tex]\( 6x + 3y = 12 \)[/tex]
- Simplify this to [tex]\( 2x + y = 4 \)[/tex].
- This is another straight line. To graph it, we'll find the intercepts:
- When [tex]\( x = 0 \)[/tex], [tex]\( y = 4 \)[/tex].
- When [tex]\( y = 0 \)[/tex], [tex]\( x = 2 \)[/tex].
2. Determine the region for the inequality: [tex]\( 6x + 3y < 12 \)[/tex]
- This represents the region below the line [tex]\( 2x + y = 4 \)[/tex].
### Intersection and Feasibility
To find the feasible region, we need to locate where both conditions [tex]\( 2x + y > 2 \)[/tex] and [tex]\( 6x + 3y < 12 \)[/tex] are satisfied simultaneously. This is the region that lies:
- Above the line [tex]\( 2x + y = 2 \)[/tex], and
- Below the line [tex]\( 2x + y = 4 \)[/tex].
### Feasible Region
1. Draw the lines:
- Line 1 (2x + y = 2): A line that could intersect the y-axis at (0,2) and the x-axis at (1,0).
- Line 2 (2x + y = 4): A line that intersects the y-axis at (0,4) and the x-axis at (2,0).
2. Assess the regions:
- The region above [tex]\( 2x + y = 2 \)[/tex] does not include the line itself.
- The region below [tex]\( 2x + y = 4 \)[/tex] does not include the line itself.
3. Feasibility Check:
- If the feasible region exists, there would be an area between these two lines where both conditions are met.
### Conclusion
After analyzing the inequalities and their graphical representations, we see there indeed exists a region between the two lines where both inequalities are satisfied. Therefore, the correct answer is:
True