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To determine the region that contains the solution to the given system of inequalities, we'll proceed with graphing and analyzing each inequality step-by-step. Here is a detailed guide:
### Step 1: Graph the first inequality [tex]\( y \leq -\frac{1}{3}x + 3 \)[/tex]
1. Graph the boundary line [tex]\( y = -\frac{1}{3}x + 3 \)[/tex]:
- Determine the y-intercept: When [tex]\( x = 0 \)[/tex], [tex]\( y = 3 \)[/tex]. So the point (0, 3) is on the line.
- Determine the x-intercept: When [tex]\( y = 0 \)[/tex], solve [tex]\( 0 = -\frac{1}{3}x + 3 \)[/tex]. Multiplying both sides by 3, we get
[tex]\[ 0 = -x + 9 \implies x = 9 \][/tex]
So the point (9, 0) is on the line.
2. Draw the line: Connect the points (0, 3) and (9, 0). Since the inequality is [tex]\( y \leq -\frac{1}{3}x + 3 \)[/tex], draw the line solid to indicate that points on the line itself are included in the solution.
3. Shade the region: For the inequality [tex]\( y \leq -\frac{1}{3}x + 3 \)[/tex], we need to shade below the line.
### Step 2: Graph the second inequality [tex]\( y \geq 3x + 2 \)[/tex]
1. Graph the boundary line [tex]\( y = 3x + 2 \)[/tex]:
- Determine the y-intercept: When [tex]\( x = 0 \)[/tex], [tex]\( y = 2 \)[/tex]. So the point (0, 2) is on the line.
- Determine the x-intercept: When [tex]\( y = 0 \)[/tex], solve [tex]\( 0 = 3x + 2 \)[/tex]. So,
[tex]\[ 0 = 3x + 2 \implies 3x = -2 \implies x = -\frac{2}{3} \][/tex]
So the point [tex]\((- \frac{2}{3}, 0)\)[/tex] is on the line.
2. Draw the line: Connect the points (0, 2) and [tex]\((- \frac{2}{3}, 0)\)[/tex]. Since the inequality is [tex]\( y \geq 3x + 2 \)[/tex], draw the line solid to indicate that points on the line itself are included in the solution.
3. Shade the region: For the inequality [tex]\( y \geq 3x + 2 \)[/tex], we need to shade above the line.
### Step 3: Determine the solution region
1. Find the intersection: The solution to the system of inequalities is where the shaded regions overlap. First identify the coordinates at which the two lines intersect.
Set the two equations equal to each other:
[tex]\[ -\frac{1}{3}x + 3 = 3x + 2 \][/tex]
2. Solve for [tex]\( x \)[/tex]:
[tex]\[ -\frac{1}{3}x + 3 = 3x + 2 \][/tex]
Multiply every term by 3 to clear the fraction:
[tex]\[ -x + 9 = 9x + 6 \][/tex]
Combine like terms:
[tex]\[ 9 = 10x + 6 \implies 3 = 10x \implies x = \frac{3}{10} \][/tex]
3. Solve for [tex]\( y \)[/tex] using [tex]\( x = \frac{3}{10} \)[/tex]:
Substitute [tex]\( x \)[/tex] back into one of the original equations:
[tex]\[ y = 3\left(\frac{3}{10}\right) + 2 = \frac{9}{10} + 2 = \frac{29}{10} = 2.9 \][/tex]
Thus, the lines intersect at [tex]\(\left(\frac{3}{10}, 2.9 \right)\)[/tex].
4. Conclusion:
- From the graphed lines and the intersection, we see that the overlapping shaded region that satisfies both inequalities forms a region which we will call [tex]\( \text{Region A} \)[/tex].
The final graph should show this clearly and we can conclude:
Region A contains the solution to the system!
### Step 1: Graph the first inequality [tex]\( y \leq -\frac{1}{3}x + 3 \)[/tex]
1. Graph the boundary line [tex]\( y = -\frac{1}{3}x + 3 \)[/tex]:
- Determine the y-intercept: When [tex]\( x = 0 \)[/tex], [tex]\( y = 3 \)[/tex]. So the point (0, 3) is on the line.
- Determine the x-intercept: When [tex]\( y = 0 \)[/tex], solve [tex]\( 0 = -\frac{1}{3}x + 3 \)[/tex]. Multiplying both sides by 3, we get
[tex]\[ 0 = -x + 9 \implies x = 9 \][/tex]
So the point (9, 0) is on the line.
2. Draw the line: Connect the points (0, 3) and (9, 0). Since the inequality is [tex]\( y \leq -\frac{1}{3}x + 3 \)[/tex], draw the line solid to indicate that points on the line itself are included in the solution.
3. Shade the region: For the inequality [tex]\( y \leq -\frac{1}{3}x + 3 \)[/tex], we need to shade below the line.
### Step 2: Graph the second inequality [tex]\( y \geq 3x + 2 \)[/tex]
1. Graph the boundary line [tex]\( y = 3x + 2 \)[/tex]:
- Determine the y-intercept: When [tex]\( x = 0 \)[/tex], [tex]\( y = 2 \)[/tex]. So the point (0, 2) is on the line.
- Determine the x-intercept: When [tex]\( y = 0 \)[/tex], solve [tex]\( 0 = 3x + 2 \)[/tex]. So,
[tex]\[ 0 = 3x + 2 \implies 3x = -2 \implies x = -\frac{2}{3} \][/tex]
So the point [tex]\((- \frac{2}{3}, 0)\)[/tex] is on the line.
2. Draw the line: Connect the points (0, 2) and [tex]\((- \frac{2}{3}, 0)\)[/tex]. Since the inequality is [tex]\( y \geq 3x + 2 \)[/tex], draw the line solid to indicate that points on the line itself are included in the solution.
3. Shade the region: For the inequality [tex]\( y \geq 3x + 2 \)[/tex], we need to shade above the line.
### Step 3: Determine the solution region
1. Find the intersection: The solution to the system of inequalities is where the shaded regions overlap. First identify the coordinates at which the two lines intersect.
Set the two equations equal to each other:
[tex]\[ -\frac{1}{3}x + 3 = 3x + 2 \][/tex]
2. Solve for [tex]\( x \)[/tex]:
[tex]\[ -\frac{1}{3}x + 3 = 3x + 2 \][/tex]
Multiply every term by 3 to clear the fraction:
[tex]\[ -x + 9 = 9x + 6 \][/tex]
Combine like terms:
[tex]\[ 9 = 10x + 6 \implies 3 = 10x \implies x = \frac{3}{10} \][/tex]
3. Solve for [tex]\( y \)[/tex] using [tex]\( x = \frac{3}{10} \)[/tex]:
Substitute [tex]\( x \)[/tex] back into one of the original equations:
[tex]\[ y = 3\left(\frac{3}{10}\right) + 2 = \frac{9}{10} + 2 = \frac{29}{10} = 2.9 \][/tex]
Thus, the lines intersect at [tex]\(\left(\frac{3}{10}, 2.9 \right)\)[/tex].
4. Conclusion:
- From the graphed lines and the intersection, we see that the overlapping shaded region that satisfies both inequalities forms a region which we will call [tex]\( \text{Region A} \)[/tex].
The final graph should show this clearly and we can conclude:
Region A contains the solution to the system!
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