Find all pairs of consecutive even positive integers both of which are greater than 5 such that their sum is less than 23.

To find pairs of consecutive even positive integers greater than 5, whose sum is less than 23, we can start by defining the integers. Let the first integer be represented as x. The next consecutive even integer would then be x + 2.

Setting Up the Equation

The sum of these two integers can be expressed as:

x + (x + 2) < 23

This simplifies to:

2x + 2 < 23

Solving the Inequality

Next, we can solve for x:

• Subtract 2 from both sides: 2x < 21
• Divide by 2: x < 10.5

Identifying Valid Integers

Since x must be an even integer greater than 5, the possible values for x are 6, 8, and 10.

Calculating the Pairs

Now, let's find the pairs:

• If x = 6: The pair is (6, 8) and their sum is 14.
• If x = 8: The pair is (8, 10) and their sum is 18.
• If x = 10: The pair is (10, 12) and their sum is 22.

Final Results

All pairs of consecutive even positive integers greater than 5, with their sums less than 23, are:

• (6, 8)
• (8, 10)
• (10, 12)