Find the pair of consecutive even positive integers which are greater than 5 and are such that their sum is less than 20.

Answer

To find the pair of consecutive even positive integers greater than 5 such that their sum is less than 20, we can follow these steps:

### Step 1: Define the integers
Let the first even positive integer be \( x \). Since we are looking for consecutive even integers, the next even integer will be \( x + 2 \).

### Step 2: Set up the inequality for the sum
According to the problem, the sum of these two integers must be less than 20. Therefore, we can write the inequality:
\[ x + (x + 2) < 20 \]

### Step 3: Simplify the inequality
Combine like terms:
\[ 2x + 2 < 20 \]

### Step 4: Isolate \( x \)
Subtract 2 from both sides:
\[ 2x < 18 \]
Now, divide both sides by 2:
\[ x < 9 \]

### Step 5: Consider the lower bound
Since we are looking for positive even integers greater than 5, we have:
\[ x > 5 \]

### Step 6: Determine possible values for \( x \)
Now we have the inequalities:
\[ 5 < x < 9 \]
The even integers that satisfy this condition are 6 and 8.

### Step 7: Find the pairs
1. If \( x = 6 \):
- The consecutive even integers are \( 6 \) and \( 8 \).
- Their sum is \( 6 + 8 = 14 \), which is less than 20.

2. If \( x = 8 \):
- The consecutive even integers are \( 8 \) and \( 10 \).
- Their sum is \( 8 + 10 = 18 \), which is also less than 20.

### Step 8: Check for any further pairs
If we try \( x = 10 \):
- The consecutive even integers would be \( 10 \) and \( 12 \).
- Their sum is \( 10 + 12 = 22 \), which exceeds 20.

### Conclusion
The pairs of consecutive even positive integers greater than 5 and whose sum is less than 20 are:
- \( (6, 8) \)
- \( (8, 10) \)

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