Find all pairs of consecutive even integers both of which are greater than 6 and their sum is less than 27.

Explanation

To solve the problem of finding all pairs of consecutive even integers that are both greater than 6 and whose sum is less than 27, we can follow these steps:

### Step 1: Define the Consecutive Even Integers
Let the first consecutive even integer be represented as \(2x\). The next consecutive even integer would then be \(2x + 2\).

### Step 2: Set Up the Inequalities
According to the problem, both integers must be greater than 6. Therefore, we can set up the following inequalities:
1. \(2x > 6\)
2. \(2x + 2 > 6\)

### Step 3: Solve the Inequalities
1. From the first inequality \(2x > 6\):
\[ x > 3 \]

2. From the second inequality \(2x + 2 > 6\):
\[ 2x > 4 \implies x > 2 \]

Since \(x > 3\) is the stricter condition, we will use this for our next steps.

### Step 4: Set Up the Sum Condition
Next, we know that the sum of the two integers must be less than 27:
\[ 2x + (2x + 2) < 27 \]
This simplifies to:
\[ 4x + 2 < 27 \]

### Step 5: Solve the Sum Inequality
Now, we can solve the inequality:
\[ 4x < 25 \implies x < \frac{25}{4} = 6.25 \]

### Step 6: Combine the Inequalities
Now we have two inequalities:
1. \(x > 3\)
2. \(x < 6.25\)

Thus, we can combine these to get:
\[ 3 < x < 6.25 \]

### Step 7: Identify Integer Values for \(x\)
Since \(x\) must be an integer, the possible integer values for \(x\) are:
- \(x = 4\)
- \(x = 5\)
- \(x = 6\)

### Step 8: Calculate the Consecutive Even Integers
Now we can find the corresponding pairs of consecutive even integers:
1. For \(x = 4\):
- First integer: \(2 \times 4 = 8\)
- Second integer: \(8 + 2 = 10\)
- Pair: (8, 10)

2. For \(x = 5\):
- First integer: \(2 \times 5 = 10\)
- Second integer: \(10 + 2 = 12\)
- Pair: (10, 12)

3. For \(x = 6\):
- First integer: \(2 \times 6 = 12\)
- Second integer: \(12 + 2 = 14\)
- Pair: (12, 14)

### Final Answer
The pairs of consecutive even integers that are both greater than 6 and whose sum is less than 27 are:
- (8, 10)
- (10, 12)
- (12, 14)