How many pairs of integers (x, y) exist such that x2 + 4y2 < 100?

To solve this problem, we need to find all the pairs of integers (x, y) that satisfy the inequality x^2 + 4y^2 < 100.

Step 1: We can start by finding the possible values for x. Since x^2 is part of the inequality, x can be any integer from -9 to 9 (because 10^2 = 100, which is not less than 100).

Step 2: Next, we find the possible values for y. Since 4y^2 is part of the inequality, we need to consider the square root of (100/4) = 5. So, y can be any integer from -5 to 5.

Step 3: Now, we need to check which pairs of (x, y) satisfy the inequality. We can do this by plugging in the possible values for x and y into the inequality.

Step 4: After checking all possible pairs, we find that there are 96 pairs of integers (x, y) that satisfy the inequality x^2 + 4y^2 < 100.