Answer :
To graph the system of linear inequalities and identify the solution set, follow these steps:
### Step 1: Graph the boundary lines
First, we need to graph each inequality's boundary line by converting them to equalities.
#### Inequality 1: [tex]\( 3x + y \leq 2 \)[/tex]
Convert to an equality:
[tex]\[ 3x + y = 2 \][/tex]
To graph this line, find the intercepts:
- x-intercept (where [tex]\( y = 0 \)[/tex]):
[tex]\[ 3x + 0 = 2 \][/tex]
[tex]\[ x = \frac{2}{3} \][/tex]
- y-intercept (where [tex]\( x = 0 \)[/tex]):
[tex]\[ 3(0) + y = 2 \][/tex]
[tex]\[ y = 2 \][/tex]
Plot these points: [tex]\((\frac{2}{3}, 0)\)[/tex] and [tex]\((0, 2)\)[/tex]. Draw a straight line through them.
#### Inequality 2: [tex]\( -3x + y \leq -2 \)[/tex]
Convert to an equality:
[tex]\[ -3x + y = -2 \][/tex]
To graph this line, find the intercepts:
- x-intercept (where [tex]\( y = 0 \)[/tex]):
[tex]\[ -3x + 0 = -2 \][/tex]
[tex]\[ x = \frac{2}{3} \][/tex]
- y-intercept (where [tex]\( x = 0 \)[/tex]):
[tex]\[ -3(0) + y = -2 \][/tex]
[tex]\[ y = -2 \][/tex]
Plot these points: [tex]\((-\frac{2}{3}, 0)\)[/tex] and [tex]\((0, -2)\)[/tex]. Draw a straight line through them.
### Step 2: Shade the appropriate regions for each inequality
Use the test point method (typically the origin [tex]\((0, 0)\)[/tex]) to determine which side of each line to shade.
#### Inequality 1: [tex]\( 3x + y \leq 2 \)[/tex]
Test the point [tex]\((0, 0)\)[/tex]:
[tex]\[ 3(0) + 0 \leq 2 \][/tex]
[tex]\[ 0 \leq 2 \][/tex] (True)
Since the test point satisfies the inequality, shade the region below and including the line.
#### Inequality 2: [tex]\( -3x + y \leq -2 \)[/tex]
Test the point [tex]\((0, 0)\)[/tex]:
[tex]\[ -3(0) + 0 \leq -2 \][/tex]
[tex]\[ 0 \leq -2 \][/tex] (False)
Since the test point does not satisfy the inequality, shade the region above and including the line.
### Step 3: Find the intersection of the shaded regions
The solution set to the system of inequalities is the region where the shaded areas overlap. This overlapping region is the part of the plane that satisfies both inequalities simultaneously.
### Step 4: Final graphical representation
Your graph should show:
- The line [tex]\( 3x + y = 2 \)[/tex] with shading below it.
- The line [tex]\( -3x + y = -2 \)[/tex] with shading above it.
- The overlapping shaded region representing the solution set.
### Step 5: Verify vertices of the intersection region
Determine the points of intersection of the boundary lines:
Solve the system of equations:
[tex]\[ 3x + y = 2 \][/tex]
[tex]\[ -3x + y = -2 \][/tex]
Add the two equations:
[tex]\[ 3x + y - 3x + y = 2 - 2 \][/tex]
[tex]\[ 2y = 0 \][/tex]
[tex]\[ y = 0 \][/tex]
Substitute [tex]\( y = 0 \)[/tex] into one of the equations:
[tex]\[ 3x + 0 = 2 \][/tex]
[tex]\[ x = \frac{2}{3} \][/tex]
The intersection point is [tex]\(\left(\frac{2}{3}, 0\right)\)[/tex].
Therefore, graphing these inequalities accurately will give you the solution set illustrating the intersection region, which contains all the points that satisfy both inequalities.
### Step 1: Graph the boundary lines
First, we need to graph each inequality's boundary line by converting them to equalities.
#### Inequality 1: [tex]\( 3x + y \leq 2 \)[/tex]
Convert to an equality:
[tex]\[ 3x + y = 2 \][/tex]
To graph this line, find the intercepts:
- x-intercept (where [tex]\( y = 0 \)[/tex]):
[tex]\[ 3x + 0 = 2 \][/tex]
[tex]\[ x = \frac{2}{3} \][/tex]
- y-intercept (where [tex]\( x = 0 \)[/tex]):
[tex]\[ 3(0) + y = 2 \][/tex]
[tex]\[ y = 2 \][/tex]
Plot these points: [tex]\((\frac{2}{3}, 0)\)[/tex] and [tex]\((0, 2)\)[/tex]. Draw a straight line through them.
#### Inequality 2: [tex]\( -3x + y \leq -2 \)[/tex]
Convert to an equality:
[tex]\[ -3x + y = -2 \][/tex]
To graph this line, find the intercepts:
- x-intercept (where [tex]\( y = 0 \)[/tex]):
[tex]\[ -3x + 0 = -2 \][/tex]
[tex]\[ x = \frac{2}{3} \][/tex]
- y-intercept (where [tex]\( x = 0 \)[/tex]):
[tex]\[ -3(0) + y = -2 \][/tex]
[tex]\[ y = -2 \][/tex]
Plot these points: [tex]\((-\frac{2}{3}, 0)\)[/tex] and [tex]\((0, -2)\)[/tex]. Draw a straight line through them.
### Step 2: Shade the appropriate regions for each inequality
Use the test point method (typically the origin [tex]\((0, 0)\)[/tex]) to determine which side of each line to shade.
#### Inequality 1: [tex]\( 3x + y \leq 2 \)[/tex]
Test the point [tex]\((0, 0)\)[/tex]:
[tex]\[ 3(0) + 0 \leq 2 \][/tex]
[tex]\[ 0 \leq 2 \][/tex] (True)
Since the test point satisfies the inequality, shade the region below and including the line.
#### Inequality 2: [tex]\( -3x + y \leq -2 \)[/tex]
Test the point [tex]\((0, 0)\)[/tex]:
[tex]\[ -3(0) + 0 \leq -2 \][/tex]
[tex]\[ 0 \leq -2 \][/tex] (False)
Since the test point does not satisfy the inequality, shade the region above and including the line.
### Step 3: Find the intersection of the shaded regions
The solution set to the system of inequalities is the region where the shaded areas overlap. This overlapping region is the part of the plane that satisfies both inequalities simultaneously.
### Step 4: Final graphical representation
Your graph should show:
- The line [tex]\( 3x + y = 2 \)[/tex] with shading below it.
- The line [tex]\( -3x + y = -2 \)[/tex] with shading above it.
- The overlapping shaded region representing the solution set.
### Step 5: Verify vertices of the intersection region
Determine the points of intersection of the boundary lines:
Solve the system of equations:
[tex]\[ 3x + y = 2 \][/tex]
[tex]\[ -3x + y = -2 \][/tex]
Add the two equations:
[tex]\[ 3x + y - 3x + y = 2 - 2 \][/tex]
[tex]\[ 2y = 0 \][/tex]
[tex]\[ y = 0 \][/tex]
Substitute [tex]\( y = 0 \)[/tex] into one of the equations:
[tex]\[ 3x + 0 = 2 \][/tex]
[tex]\[ x = \frac{2}{3} \][/tex]
The intersection point is [tex]\(\left(\frac{2}{3}, 0\right)\)[/tex].
Therefore, graphing these inequalities accurately will give you the solution set illustrating the intersection region, which contains all the points that satisfy both inequalities.
