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How many ordered pairs of positive integers satisfly the inequality 4x plus 5y20?
Gloriaxue
If the inequality is > (greater than) or >= (greater than or equal to), then there are an infinite number of solutions.
So let the inequality be < (less than) or <= (less than or equal to)
x = 1: 5y <= 16 so y = 1, 2 or 3
x = 2: 5y <= 12 so y = 1 or 2
x = 3: 5y <= 8 so y = 1
x >= 4: 5y <= 4 no solution.
So whether the inequality is < or <= there are 6 ordered pairs.
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∙ 13y ago
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