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Question
Determine the region in the plane determined by the inequalities:
x ≤ 3y, x ≥ y
Solution
Given in equation are x ≤ 3y, x ≥ y
Suppose x = 3y
⇒`x/3` = y
| x | 0 | 3 | 6 | – 3 |
| y | 0 | 1 | 2 | – 1 |
If x = y
| x | 1 | 2 | – 1 | – 2 |
| y | 1 | 2 | – 1 | – 2 |
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