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To determine the solution region for the given system of inequalities, let's analyze each inequality separately.
### Inequality 1: [tex]\(2x + y > 2\)[/tex]
1. Rewrite the inequality as an equation to understand the boundary: [tex]\(2x + y = 2\)[/tex].
2. Find the intercepts:
- When [tex]\(x = 0\)[/tex]: [tex]\(2(0) + y = 2 \Rightarrow y = 2\)[/tex].
- When [tex]\(y = 0\)[/tex]: [tex]\(2x + 0 = 2 \Rightarrow x = 1\)[/tex].
3. Plot the line through points [tex]\((0, 2)\)[/tex] and [tex]\((1, 0)\)[/tex].
4. Determine the region where [tex]\(2x + y > 2\)[/tex]: substitute a test point not on the line, like [tex]\((0, 0)\)[/tex]:
- [tex]\(2(0) + 0 = 0 < 2\)[/tex], so the region above the line is the solution.
### Inequality 2: [tex]\(6x + 3y < 12\)[/tex]
1. Rewrite the inequality as an equation to understand the boundary: [tex]\(6x + 3y = 12\)[/tex].
2. Simplify the equation: [tex]\(2x + y = 4\)[/tex].
3. Find the intercepts:
- When [tex]\(x = 0\)[/tex]: [tex]\(2(0) + y = 4 \Rightarrow y = 4\)[/tex].
- When [tex]\(y = 0\)[/tex]: [tex]\(2x + 0 = 4 \Rightarrow x = 2\)[/tex].
4. Plot the line through points [tex]\((0, 4)\)[/tex] and [tex]\((2, 0)\)[/tex].
5. Determine the region where [tex]\(2x + y < 4\)[/tex]: substitute a test point not on the line, like [tex]\((0, 0)\)[/tex]:
- [tex]\(2(0) + 0 = 0 < 4\)[/tex], so the region below the line is the solution.
### Combine the Inequalities
To find the solution to the system, we need the region where both conditions are satisfied simultaneously:
1. The region above the line [tex]\(2x + y = 2\)[/tex].
2. The region below the line [tex]\(2x + y = 4\)[/tex].
Feasible Region:
- The feasible region lies between the two lines [tex]\(2x + y = 2\)[/tex] and [tex]\(2x + y = 4\)[/tex], including the areas above the first line but below the second line.
- Since both lines are parallel, the feasible region is a strip between these two lines.
Since we established the solution graphically, we can say:
True. This graph represents the solution region for the given system of linear inequalities. The shaded region (if plotted correctly) would indeed show the strip between the lines [tex]\(2x + y = 2\)[/tex] and [tex]\(2x + y = 4\)[/tex].
### Inequality 1: [tex]\(2x + y > 2\)[/tex]
1. Rewrite the inequality as an equation to understand the boundary: [tex]\(2x + y = 2\)[/tex].
2. Find the intercepts:
- When [tex]\(x = 0\)[/tex]: [tex]\(2(0) + y = 2 \Rightarrow y = 2\)[/tex].
- When [tex]\(y = 0\)[/tex]: [tex]\(2x + 0 = 2 \Rightarrow x = 1\)[/tex].
3. Plot the line through points [tex]\((0, 2)\)[/tex] and [tex]\((1, 0)\)[/tex].
4. Determine the region where [tex]\(2x + y > 2\)[/tex]: substitute a test point not on the line, like [tex]\((0, 0)\)[/tex]:
- [tex]\(2(0) + 0 = 0 < 2\)[/tex], so the region above the line is the solution.
### Inequality 2: [tex]\(6x + 3y < 12\)[/tex]
1. Rewrite the inequality as an equation to understand the boundary: [tex]\(6x + 3y = 12\)[/tex].
2. Simplify the equation: [tex]\(2x + y = 4\)[/tex].
3. Find the intercepts:
- When [tex]\(x = 0\)[/tex]: [tex]\(2(0) + y = 4 \Rightarrow y = 4\)[/tex].
- When [tex]\(y = 0\)[/tex]: [tex]\(2x + 0 = 4 \Rightarrow x = 2\)[/tex].
4. Plot the line through points [tex]\((0, 4)\)[/tex] and [tex]\((2, 0)\)[/tex].
5. Determine the region where [tex]\(2x + y < 4\)[/tex]: substitute a test point not on the line, like [tex]\((0, 0)\)[/tex]:
- [tex]\(2(0) + 0 = 0 < 4\)[/tex], so the region below the line is the solution.
### Combine the Inequalities
To find the solution to the system, we need the region where both conditions are satisfied simultaneously:
1. The region above the line [tex]\(2x + y = 2\)[/tex].
2. The region below the line [tex]\(2x + y = 4\)[/tex].
Feasible Region:
- The feasible region lies between the two lines [tex]\(2x + y = 2\)[/tex] and [tex]\(2x + y = 4\)[/tex], including the areas above the first line but below the second line.
- Since both lines are parallel, the feasible region is a strip between these two lines.
Since we established the solution graphically, we can say:
True. This graph represents the solution region for the given system of linear inequalities. The shaded region (if plotted correctly) would indeed show the strip between the lines [tex]\(2x + y = 2\)[/tex] and [tex]\(2x + y = 4\)[/tex].
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