How many ordered pairs of natural numbers satisfy the inequality 8x + ...
The inequality 8x - 2y < 40="" can="" be="" rewritten="" as="" 4x="" -="" y="" />< />
Let's consider the possible values for x and y.
When x = 1, the inequality becomes 4(1) - y < 20,="" which="" simplifies="" to="" -y="" />< 16.="" since="" y="" is="" a="" natural="" number,="" the="" only="" possible="" value="" is="" y="" />
When x = 2, the inequality becomes 4(2) - y < 20,="" which="" simplifies="" to="" 8="" -="" y="" />< 20.="" solving="" for="" y,="" we="" get="" y="" /> -12. Since y is a natural number, the possible values for y are 1, 2, 3, ..., 11, which is a total of 11 values.
Similarly, when x = 3, we have y > -8, giving us a total of 9 possible values for y.
Continuing this pattern, we find that when x = 1, there is 1 possible value for y. When x = 2, there are 11 possible values for y. When x = 3, there are 9 possible values for y. And so on.
Therefore, the total number of ordered pairs of natural numbers that satisfy the inequality is 1 + 11 + 9 + ... = 1 + 11 + 9 + 7 + 5 + 3 + 1 = 37.
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