Answer :
To determine which region contains the solution to the system of inequalities:
[tex]\[ \begin{array}{l} y \leq -\frac{1}{3}x + 3 \\ y \geq 3x + 2 \end{array} \][/tex]
### Step-by-Step Solution:
1. Graph the Inequalities:
- First Inequality: [tex]\( y \leq -\frac{1}{3}x + 3 \)[/tex]
- This is a line with a slope of [tex]\(-\frac{1}{3}\)[/tex] and a y-intercept of 3.
- It is a downward-sloping line.
- Since it's an inequality with [tex]\( \leq \)[/tex], shade the region below and on the line.
- Second Inequality: [tex]\( y \geq 3x + 2 \)[/tex]
- This is a line with a slope of [tex]\(3\)[/tex] and a y-intercept of 2.
- It is an upward-sloping line.
- Since it's an inequality with [tex]\( \geq \)[/tex], shade the region above and on the line.
2. Find the Intersection Point of the Lines:
- To find where the lines intersect, set the equations equal to each other:
[tex]\[ -\frac{1}{3}x + 3 = 3x + 2 \][/tex]
- Solve for [tex]\( x \)[/tex]:
[tex]\[ -\frac{1}{3}x + 3 = 3x + 2 \][/tex]
[tex]\[ 3 = 3x + 2 + \frac{1}{3} x \][/tex]
[tex]\[ 3 - 2 = 3x + \frac{1}{3} x \][/tex]
[tex]\[ 1 = 3x + \frac{1}{3} x \][/tex]
[tex]\[ 1 = \frac{10x}{3} \][/tex]
[tex]\[ x = 0.3 \quad \text{(or } x = \frac{3}{10} \text{)} \][/tex]
- Calculate the corresponding [tex]\( y \)[/tex]-value:
[tex]\[ y = 3(0.3) + 2 \][/tex]
[tex]\[ y = 0.9 + 2 \][/tex]
[tex]\[ y = 2.9 \][/tex]
- The intersection point is [tex]\((0.3, 2.9)\)[/tex].
3. Determine the Feasible Region:
- From the intersection point [tex]\((0.3, 2.9)\)[/tex], shade the regions that satisfy both inequalities.
- The inequality [tex]\( y \leq -\frac{1}{3}x + 3 \)[/tex] shades below the line.
- The inequality [tex]\( y \geq 3x + 2 \)[/tex] shades above the line.
- The feasible region (the region that satisfies both inequalities) is the area towards the upper left relative to the intersection point (0.3, 2.9).
By examining the description of the regions A, B, C, and D given in the question, it is clear that the region that contains the solution is:
C. Region A.
[tex]\[ \begin{array}{l} y \leq -\frac{1}{3}x + 3 \\ y \geq 3x + 2 \end{array} \][/tex]
### Step-by-Step Solution:
1. Graph the Inequalities:
- First Inequality: [tex]\( y \leq -\frac{1}{3}x + 3 \)[/tex]
- This is a line with a slope of [tex]\(-\frac{1}{3}\)[/tex] and a y-intercept of 3.
- It is a downward-sloping line.
- Since it's an inequality with [tex]\( \leq \)[/tex], shade the region below and on the line.
- Second Inequality: [tex]\( y \geq 3x + 2 \)[/tex]
- This is a line with a slope of [tex]\(3\)[/tex] and a y-intercept of 2.
- It is an upward-sloping line.
- Since it's an inequality with [tex]\( \geq \)[/tex], shade the region above and on the line.
2. Find the Intersection Point of the Lines:
- To find where the lines intersect, set the equations equal to each other:
[tex]\[ -\frac{1}{3}x + 3 = 3x + 2 \][/tex]
- Solve for [tex]\( x \)[/tex]:
[tex]\[ -\frac{1}{3}x + 3 = 3x + 2 \][/tex]
[tex]\[ 3 = 3x + 2 + \frac{1}{3} x \][/tex]
[tex]\[ 3 - 2 = 3x + \frac{1}{3} x \][/tex]
[tex]\[ 1 = 3x + \frac{1}{3} x \][/tex]
[tex]\[ 1 = \frac{10x}{3} \][/tex]
[tex]\[ x = 0.3 \quad \text{(or } x = \frac{3}{10} \text{)} \][/tex]
- Calculate the corresponding [tex]\( y \)[/tex]-value:
[tex]\[ y = 3(0.3) + 2 \][/tex]
[tex]\[ y = 0.9 + 2 \][/tex]
[tex]\[ y = 2.9 \][/tex]
- The intersection point is [tex]\((0.3, 2.9)\)[/tex].
3. Determine the Feasible Region:
- From the intersection point [tex]\((0.3, 2.9)\)[/tex], shade the regions that satisfy both inequalities.
- The inequality [tex]\( y \leq -\frac{1}{3}x + 3 \)[/tex] shades below the line.
- The inequality [tex]\( y \geq 3x + 2 \)[/tex] shades above the line.
- The feasible region (the region that satisfies both inequalities) is the area towards the upper left relative to the intersection point (0.3, 2.9).
By examining the description of the regions A, B, C, and D given in the question, it is clear that the region that contains the solution is:
C. Region A.
