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Question
Determine the region in the plane determined by the inequalities:
y ≥ 2x, −2x + 3y ≤ 6
Solution
Suppose y = 2x
| x | 1 | − 1 | 2 | − 2 |
| y | 2 | − 2 | 4 | − 4 |
– 2x + 3y = 6
– 2x = 6 – 3y
x = `(6 - 3y)/(-2)`
x = `- 6/2 - (3y)/(-2)`
x = `- 3 + (3y)/2`
3 + x = `(3y)/2`
| x | 0 | – 3 |
| y | 2 | 0 |
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