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Question:

If 2 ≤ |x – 1|×|y + 3| ≤ 5 and both x and y are negative integers, find the number of possible combinations of x and y.

Explanation:

Since 2 ≤ |x – 1| × |y + 3| ≤ 5, the value of |x – 1| × |y + 3| can be 2, 3, 4 or 5. Thus, we have the following cases.

Case 1: |x – 1| × |y + 3| = 2
⇒ |x – 1| = 2 and |y + 3| = 1
⇒ x = -1 and y = -4 or -2

Note: |x - 1| cannot be equal to 1 since x has to be a negative number.

∴ 2 possibilities.

Case 2: |x – 1| × |y + 3| = 3
⇒ |x – 1| = 3 and |y + 3| = 1
⇒ x = -2 and y = -4 or -2

∴ 2 possibilities.

Case 3: |x – 1| × |y + 3| = 4

⇒ |x – 1| = 4 and |y + 3| = 1
⇒ x = -3 and y = -4 or -2

⇒ |x – 1| = 2 and |y + 3| = 2
⇒ x = -1 and y = -1 or -5

∴ 4 possibilities.

Case 3: |x – 1| × |y + 3| = 5

⇒ |x – 1| = 5 and |y + 3| = 1
⇒ x = -4 and y = -4 or -2

∴ 2 possibilities.

∴ Total 2 + 2 + 4 + 2 = 10 possibilities.

Hence, option (e).