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Sagot :
To solve this system of inequalities, we need to analyze and graph the two lines represented by:
1. [tex]\( y = -5x + 2 \)[/tex]
2. [tex]\( y = 3x - 1.5 \)[/tex]
### Step-by-Step Solution
1. Find the intersection point of the two lines:
To find where the two lines intersect, we set the equations equal to each other:
[tex]\[ -5x + 2 = 3x - 1.5 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ -5x - 3x + 2 = -1.5 \\ -8x + 2 = -1.5 \\ -8x = -1.5 - 2 \\ -8x = -3.5 \\ x = \frac{-3.5}{-8} \\ x = 0.4375 \][/tex]
Now, solve for [tex]\( y \)[/tex] using [tex]\( x = 0.4375 \)[/tex] in either of the original equations. Using [tex]\( y = -5x + 2 \)[/tex]:
[tex]\[ y = -5(0.4375) + 2 \\ y = -2.1875 + 2 \\ y = -0.1875 \][/tex]
The intersection point is [tex]\( (0.4375, -0.1875) \)[/tex].
2. Graph the lines:
- The line [tex]\( y = -5x + 2 \)[/tex]: This line has a slope of -5 and a y-intercept at [tex]\( (0, 2) \)[/tex].
- The line [tex]\( y = 3x - 1.5 \)[/tex]: This line has a slope of 3 and a y-intercept at [tex]\( (0, -1.5) \)[/tex].
3. Shading the regions:
- For the inequality [tex]\( y \geq -5x + 2 \)[/tex], shade the region above or on the line [tex]\( y = -5x + 2 \)[/tex].
- For the inequality [tex]\( y > 3x - 1.5 \)[/tex], shade the region strictly above the line [tex]\( y = 3x - 1.5 \)[/tex] (note that this does not include the line itself).
4. Determine the feasible region:
- The feasible region is the area that satisfies both inequalities simultaneously. This is the region above the line [tex]\( y = -5x + 2 \)[/tex] AND strictly above [tex]\( y = 3x - 1.5 \)[/tex].
### Final Graph
The graph should include:
- The line [tex]\( y = -5x + 2 \)[/tex] in solid, indicating the boundary for the [tex]\(\geq\)[/tex] inequality, with shading above the line including the line itself.
- The line [tex]\( y = 3x - 1.5 \)[/tex] in dashed, indicating the boundary for the [tex]\(>\)[/tex] inequality, with shading strictly above the line excluding the line itself.
- The intersection point at [tex]\( (0.4375, -0.1875) \)[/tex].
Look for the graph that illustrates the above description.
1. [tex]\( y = -5x + 2 \)[/tex]
2. [tex]\( y = 3x - 1.5 \)[/tex]
### Step-by-Step Solution
1. Find the intersection point of the two lines:
To find where the two lines intersect, we set the equations equal to each other:
[tex]\[ -5x + 2 = 3x - 1.5 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ -5x - 3x + 2 = -1.5 \\ -8x + 2 = -1.5 \\ -8x = -1.5 - 2 \\ -8x = -3.5 \\ x = \frac{-3.5}{-8} \\ x = 0.4375 \][/tex]
Now, solve for [tex]\( y \)[/tex] using [tex]\( x = 0.4375 \)[/tex] in either of the original equations. Using [tex]\( y = -5x + 2 \)[/tex]:
[tex]\[ y = -5(0.4375) + 2 \\ y = -2.1875 + 2 \\ y = -0.1875 \][/tex]
The intersection point is [tex]\( (0.4375, -0.1875) \)[/tex].
2. Graph the lines:
- The line [tex]\( y = -5x + 2 \)[/tex]: This line has a slope of -5 and a y-intercept at [tex]\( (0, 2) \)[/tex].
- The line [tex]\( y = 3x - 1.5 \)[/tex]: This line has a slope of 3 and a y-intercept at [tex]\( (0, -1.5) \)[/tex].
3. Shading the regions:
- For the inequality [tex]\( y \geq -5x + 2 \)[/tex], shade the region above or on the line [tex]\( y = -5x + 2 \)[/tex].
- For the inequality [tex]\( y > 3x - 1.5 \)[/tex], shade the region strictly above the line [tex]\( y = 3x - 1.5 \)[/tex] (note that this does not include the line itself).
4. Determine the feasible region:
- The feasible region is the area that satisfies both inequalities simultaneously. This is the region above the line [tex]\( y = -5x + 2 \)[/tex] AND strictly above [tex]\( y = 3x - 1.5 \)[/tex].
### Final Graph
The graph should include:
- The line [tex]\( y = -5x + 2 \)[/tex] in solid, indicating the boundary for the [tex]\(\geq\)[/tex] inequality, with shading above the line including the line itself.
- The line [tex]\( y = 3x - 1.5 \)[/tex] in dashed, indicating the boundary for the [tex]\(>\)[/tex] inequality, with shading strictly above the line excluding the line itself.
- The intersection point at [tex]\( (0.4375, -0.1875) \)[/tex].
Look for the graph that illustrates the above description.
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