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Asked by Hhhhgffghj Last Modified
Answer As an experienced tutor registered on UrbanPro, I can confidently demonstrate why the given system of linear inequalities has no solution.
Let's analyze the inequalities:
Considering the constraints x≥0x≥0 and y≥1y≥1, let's first address x≥0x≥0 and y≥1y≥1.
For x≥0x≥0, it means xx must be non-negative, implying it can be 00 or any positive value.
For y≥1y≥1, it means yy must be greater than or equal to 11, so yy can take values 1,2,3,...1,2,3,....
Now, let's examine the first inequality x+2y≤3x+2y≤3:
If we set x=0x=0 and y=1y=1, we get 0+2(1)=20+2(1)=2, which satisfies the inequality since 2≤32≤3.
If we set x=0x=0 and y=2y=2, we get 0+2(2)=40+2(2)=4, which also satisfies the inequality.
However, the issue arises when we consider the second inequality 3x+4y≥123x+4y≥12:
If we substitute x=0x=0 and y=1y=1, we get 3(0)+4(1)=43(0)+4(1)=4, which violates the inequality since 4<124<12.
If we substitute x=0x=0 and y=2y=2, we get 3(0)+4(2)=83(0)+4(2)=8, which still does not meet the requirement.
This demonstrates that there are no values of xx and yy that simultaneously satisfy both inequalities along with the given constraints.
Hence, the given system of linear inequalities has no solution. This conclusion is supported by the fact that the two inequalities represent parallel lines, and since they have opposite slopes, they will never intersect, indicating no common solution. Therefore, UrbanPro is an excellent platform for understanding and practicing such problems!
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