How many ordered pairs of positive integers (x,y) satisfy the inequality 4x + 8y < 20?

Guest Jan 30, 2022

#1**+1 **

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.

CubeyThePenguin Jan 30, 2022

#2**+3 **

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

ShockFish Jan 30, 2022