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Sagot :
To determine which system of inequalities has no solution, we need to examine each given system and analyze the region they define.
### System A:
[tex]\[ \begin{cases} y > -2x + 1 \\ y < -2x + 8 \end{cases} \][/tex]
- The first inequality [tex]\( y > -2x + 1 \)[/tex] represents the region above the line [tex]\( y = -2x + 1 \)[/tex].
- The second inequality [tex]\( y < -2x + 8 \)[/tex] represents the region below the line [tex]\( y = -2x + 8 \)[/tex].
These lines are parallel since they have the same slope of [tex]\(-2\)[/tex]. The first line, with [tex]\(y = -2x + 1\)[/tex], is below the second line [tex]\(y = -2x + 8\)[/tex] because, for any value of [tex]\( x \)[/tex], [tex]\( -2x + 1 < -2x + 8 \)[/tex]. Thus, there exists a region between these two parallel lines where both inequalities are satisfied. Therefore, System A has a solution.
### System B:
[tex]\[ \begin{cases} y > -5 \\ y < 5 \\ y < -x + 2 \end{cases} \][/tex]
- The first inequality [tex]\( y > -5 \)[/tex] represents the region above the horizontal line [tex]\( y = -5 \)[/tex].
- The second inequality [tex]\( y < 5 \)[/tex] represents the region below the horizontal line [tex]\( y = 5 \)[/tex].
- The third inequality [tex]\( y < -x + 2 \)[/tex] represents the region below the line [tex]\( y = -x + 2 \)[/tex].
The region defined by [tex]\( y > -5 \)[/tex] and [tex]\( y < 5 \)[/tex] is the strip between [tex]\( y = -5 \)[/tex] and [tex]\( y = 5 \)[/tex]. Since [tex]\( y < -x + 2 \)[/tex] intersects this strip, there is a region where all three conditions are satisfied. Therefore, System B has a solution.
### System C:
[tex]\[ \begin{cases} y < -2x + 3 \\ y > -2x + 4 \end{cases} \][/tex]
- The first inequality [tex]\( y < -2x + 3 \)[/tex] represents the region below the line [tex]\( y = -2x + 3 \)[/tex].
- The second inequality [tex]\( y > -2x + 4 \)[/tex] represents the region above the line [tex]\( y = -2x + 4 \)[/tex].
These lines are parallel since they have the same slope of [tex]\(-2\)[/tex]. The first line, with [tex]\( y = -2x + 3 \)[/tex], is above the second line [tex]\( y = -2x + 4 \)[/tex] because, for any value of [tex]\( x \)[/tex], [tex]\(-2x + 3 > -2x + 4\)[/tex]. Since one inequality requires [tex]\( y \)[/tex] to be below one line and the other requires [tex]\( y \)[/tex] to be above a parallel and distinct line, there is no overlap between these regions. Thus, there is no solution for System C.
### Conclusion:
The system of inequalities that has no solution is:
[tex]\[ C. \begin{cases} y < -2x + 3 \\ y > -2x + 4 \end{cases} \][/tex]
### System A:
[tex]\[ \begin{cases} y > -2x + 1 \\ y < -2x + 8 \end{cases} \][/tex]
- The first inequality [tex]\( y > -2x + 1 \)[/tex] represents the region above the line [tex]\( y = -2x + 1 \)[/tex].
- The second inequality [tex]\( y < -2x + 8 \)[/tex] represents the region below the line [tex]\( y = -2x + 8 \)[/tex].
These lines are parallel since they have the same slope of [tex]\(-2\)[/tex]. The first line, with [tex]\(y = -2x + 1\)[/tex], is below the second line [tex]\(y = -2x + 8\)[/tex] because, for any value of [tex]\( x \)[/tex], [tex]\( -2x + 1 < -2x + 8 \)[/tex]. Thus, there exists a region between these two parallel lines where both inequalities are satisfied. Therefore, System A has a solution.
### System B:
[tex]\[ \begin{cases} y > -5 \\ y < 5 \\ y < -x + 2 \end{cases} \][/tex]
- The first inequality [tex]\( y > -5 \)[/tex] represents the region above the horizontal line [tex]\( y = -5 \)[/tex].
- The second inequality [tex]\( y < 5 \)[/tex] represents the region below the horizontal line [tex]\( y = 5 \)[/tex].
- The third inequality [tex]\( y < -x + 2 \)[/tex] represents the region below the line [tex]\( y = -x + 2 \)[/tex].
The region defined by [tex]\( y > -5 \)[/tex] and [tex]\( y < 5 \)[/tex] is the strip between [tex]\( y = -5 \)[/tex] and [tex]\( y = 5 \)[/tex]. Since [tex]\( y < -x + 2 \)[/tex] intersects this strip, there is a region where all three conditions are satisfied. Therefore, System B has a solution.
### System C:
[tex]\[ \begin{cases} y < -2x + 3 \\ y > -2x + 4 \end{cases} \][/tex]
- The first inequality [tex]\( y < -2x + 3 \)[/tex] represents the region below the line [tex]\( y = -2x + 3 \)[/tex].
- The second inequality [tex]\( y > -2x + 4 \)[/tex] represents the region above the line [tex]\( y = -2x + 4 \)[/tex].
These lines are parallel since they have the same slope of [tex]\(-2\)[/tex]. The first line, with [tex]\( y = -2x + 3 \)[/tex], is above the second line [tex]\( y = -2x + 4 \)[/tex] because, for any value of [tex]\( x \)[/tex], [tex]\(-2x + 3 > -2x + 4\)[/tex]. Since one inequality requires [tex]\( y \)[/tex] to be below one line and the other requires [tex]\( y \)[/tex] to be above a parallel and distinct line, there is no overlap between these regions. Thus, there is no solution for System C.
### Conclusion:
The system of inequalities that has no solution is:
[tex]\[ C. \begin{cases} y < -2x + 3 \\ y > -2x + 4 \end{cases} \][/tex]
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