Find all pairs of consecutive odd positive integers both of which are smaller than 10 such that their sum is more than 11.
The required pairs of consecutive odd positive integers are (5, 7) and (7, 9).
Explanation
This question is similar to Example 8 in the textbook, but with different conditions. Students need to set up inequalities based on the given conditions: both integers smaller than 10, and sum greater than 11. The key is to let x be the smaller odd integer, so the other is x+2. Then solve the system of inequalities and identify valid odd integers.
Solution Steps
Step 1: Let x be the smaller of the two consecutive odd positive integers. Then the other integer is x + 2.
Step 2: Since both integers are smaller than 10, we have x + 2 < 10, which gives x < 8.
Step 3: Since their sum is more than 11, we have x + (x + 2) > 11, which gives 2x + 2 > 11, or 2x > 9, or x > 4.5.
Step 4: Combining the inequalities: 4.5 < x < 8. Since x must be an odd positive integer, x can be 5 or 7.
Step 5: When x = 5, the pair is (5, 7). When x = 7, the pair is (7, 9). Both pairs satisfy all conditions: each integer is smaller than 10, and their sums (12 and 16 respectively) are greater than 11.