Answer :
Let's graph the system of inequalities step-by-step and find the vertices of the feasible region.
#### Step 1: Identify and plot each inequality on the same coordinate plane.
1. First Inequality: [tex]\( 3y - x \leq 9 \)[/tex]
- Rewrite it in slope-intercept form: [tex]\( y \leq \frac{1}{3}x + 3 \)[/tex]
- Plot the line [tex]\( y = \frac{1}{3}x + 3 \)[/tex]. This line has a slope [tex]\( \frac{1}{3} \)[/tex] and a y-intercept at (0, 3).
- Since [tex]\( y \leq \frac{1}{3}x + 3 \)[/tex], shade the region below the line.
2. Second Inequality: [tex]\( y + 2x \leq 10 \)[/tex]
- Rewrite it in slope-intercept form: [tex]\( y \leq -2x + 10 \)[/tex]
- Plot the line [tex]\( y = -2x + 10 \)[/tex]. This line has a slope of [tex]\( -2 \)[/tex] and a y-intercept at (0, 10).
- Since [tex]\( y \leq -2x + 10 \)[/tex], shade the region below the line.
3. Third Inequality: [tex]\( y \geq 0 \)[/tex]
- This represents all points above the x-axis.
- Shade the region above the x-axis.
#### Step 2: Determine the vertices of the feasible region.
To find the vertices of the region where all the shaded areas overlap, we need to find the points of intersection of the boundary lines.
1. Intersection of [tex]\( 3y - x = 9 \)[/tex] and [tex]\( y + 2x = 10 \)[/tex]
- Solve the system:
[tex]\[ y = \frac{1}{3}x + 3 \][/tex]
[tex]\[ y = -2x + 10 \][/tex]
- Set the equations equal and solve for [tex]\( x \)[/tex]:
[tex]\[ \frac{1}{3}x + 3 = -2x + 10 \][/tex]
[tex]\[ \frac{1}{3}x + 2x = 10 - 3 \][/tex]
[tex]\[ \frac{7}{3}x = 7 \][/tex]
[tex]\[ x = 3 \][/tex]
- Substitute [tex]\( x = 3 \)[/tex] back into one of the equations:
[tex]\[ y = \frac{1}{3}(3) + 3 = 4 \][/tex]
- Intersection point: [tex]\( (3, 4) \)[/tex]
2. Intersection of [tex]\( 3y - x = 9 \)[/tex] and [tex]\( y = 0 \)[/tex]
- Substitute [tex]\( y = 0 \)[/tex] into [tex]\( 3y - x = 9 \)[/tex]:
[tex]\[ 3(0) - x = 9 \][/tex]
[tex]\[ -x = 9 \][/tex]
[tex]\[ x = -9 \][/tex]
- Intersection point: [tex]\( (-9, 0) \)[/tex]
3. Intersection of [tex]\( y + 2x = 10 \)[/tex] and [tex]\( y = 0 \)[/tex]
- Substitute [tex]\( y = 0 \)[/tex] into [tex]\( y + 2x = 10 \)[/tex]:
[tex]\[ 0 + 2x = 10 \][/tex]
[tex]\[ 2x = 10 \][/tex]
[tex]\[ x = 5 \][/tex]
- Intersection point: [tex]\( (5, 0) \)[/tex]
#### Step 3: Identify all distinct vertices.
Combining all the intersection points, we get the distinct vertices of the feasible region:
- [tex]\( (3, 4) \)[/tex]
- [tex]\( (-9, 0) \)[/tex]
- [tex]\( (5, 0) \)[/tex]
Thus, the coordinates of the vertices of the feasible region are:
[tex]\( (3, 4) \)[/tex], [tex]\( (-9, 0) \)[/tex], [tex]\( (5, 0) \)[/tex]
#### Step 1: Identify and plot each inequality on the same coordinate plane.
1. First Inequality: [tex]\( 3y - x \leq 9 \)[/tex]
- Rewrite it in slope-intercept form: [tex]\( y \leq \frac{1}{3}x + 3 \)[/tex]
- Plot the line [tex]\( y = \frac{1}{3}x + 3 \)[/tex]. This line has a slope [tex]\( \frac{1}{3} \)[/tex] and a y-intercept at (0, 3).
- Since [tex]\( y \leq \frac{1}{3}x + 3 \)[/tex], shade the region below the line.
2. Second Inequality: [tex]\( y + 2x \leq 10 \)[/tex]
- Rewrite it in slope-intercept form: [tex]\( y \leq -2x + 10 \)[/tex]
- Plot the line [tex]\( y = -2x + 10 \)[/tex]. This line has a slope of [tex]\( -2 \)[/tex] and a y-intercept at (0, 10).
- Since [tex]\( y \leq -2x + 10 \)[/tex], shade the region below the line.
3. Third Inequality: [tex]\( y \geq 0 \)[/tex]
- This represents all points above the x-axis.
- Shade the region above the x-axis.
#### Step 2: Determine the vertices of the feasible region.
To find the vertices of the region where all the shaded areas overlap, we need to find the points of intersection of the boundary lines.
1. Intersection of [tex]\( 3y - x = 9 \)[/tex] and [tex]\( y + 2x = 10 \)[/tex]
- Solve the system:
[tex]\[ y = \frac{1}{3}x + 3 \][/tex]
[tex]\[ y = -2x + 10 \][/tex]
- Set the equations equal and solve for [tex]\( x \)[/tex]:
[tex]\[ \frac{1}{3}x + 3 = -2x + 10 \][/tex]
[tex]\[ \frac{1}{3}x + 2x = 10 - 3 \][/tex]
[tex]\[ \frac{7}{3}x = 7 \][/tex]
[tex]\[ x = 3 \][/tex]
- Substitute [tex]\( x = 3 \)[/tex] back into one of the equations:
[tex]\[ y = \frac{1}{3}(3) + 3 = 4 \][/tex]
- Intersection point: [tex]\( (3, 4) \)[/tex]
2. Intersection of [tex]\( 3y - x = 9 \)[/tex] and [tex]\( y = 0 \)[/tex]
- Substitute [tex]\( y = 0 \)[/tex] into [tex]\( 3y - x = 9 \)[/tex]:
[tex]\[ 3(0) - x = 9 \][/tex]
[tex]\[ -x = 9 \][/tex]
[tex]\[ x = -9 \][/tex]
- Intersection point: [tex]\( (-9, 0) \)[/tex]
3. Intersection of [tex]\( y + 2x = 10 \)[/tex] and [tex]\( y = 0 \)[/tex]
- Substitute [tex]\( y = 0 \)[/tex] into [tex]\( y + 2x = 10 \)[/tex]:
[tex]\[ 0 + 2x = 10 \][/tex]
[tex]\[ 2x = 10 \][/tex]
[tex]\[ x = 5 \][/tex]
- Intersection point: [tex]\( (5, 0) \)[/tex]
#### Step 3: Identify all distinct vertices.
Combining all the intersection points, we get the distinct vertices of the feasible region:
- [tex]\( (3, 4) \)[/tex]
- [tex]\( (-9, 0) \)[/tex]
- [tex]\( (5, 0) \)[/tex]
Thus, the coordinates of the vertices of the feasible region are:
[tex]\( (3, 4) \)[/tex], [tex]\( (-9, 0) \)[/tex], [tex]\( (5, 0) \)[/tex]
