How many ordered pairs of positive integers (x,y) satisfy the inequality 4x + 8y < 20?
This question has been answered before: https://web2.0calc.com/questions/help-plz_55355
Otherwise, let's solve it here.
y = 1: x = 1, 2
y = 2: no solutions for x (you are solving 4x + 16 < 20, but x has to be less than 1)
So, there are only 2 pairs that solve this inequality.
We can simplify the inequality first:
Dividing by 4,
x + 2y < 5
Remember the problem says positive integers. Therefore, we can count the limited number of possibilties:
(1, 1)
(2, 1)
(1, 2) doesn't work because 5 is not smaller than 5.
That's it, because if any of the numbers were larger, x + 2y would be bigger than 5.
Thus the number of ordered pairs of positive integers that satisfy the inequality is 2.
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ShockFish