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

 Jan 30, 2022
 #1
avatar+1209 
+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 pairs that solve this inequality.

 Jan 30, 2022
 #3
avatar+63 
+3

NEVERMIND, I just realized my mistake.

ShockFish  Jan 30, 2022
edited by ShockFish  Jan 30, 2022
 #2
avatar+63 
+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.

 

⚡⚡⚡

ShockFish

 Jan 30, 2022
edited by ShockFish  Jan 30, 2022

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