Let's represent the three consecutive even integers as:

• The first integer: $x$
• The second integer: $x + 2$
• The third integer: $x + 4$

Step 1: Translate the problem into an equation

We know that:

• The sum of the first and third integers is $x + (x + 4) = 2x + 4$.
• Seven times the sum of the first and third integers is $7(2x + 4)$.
• Ten times the second integer is $10(x + 2)$.

According to the problem, 7 times the sum of the first and third integers is 48 less than 10 times the second integer. Therefore, the equation becomes:

$7(2x + 4) = 10(x + 2) - 48$

Step 2: Solve the equation

First, expand both sides:

$14x + 28 = 10x + 20 - 48$

Simplify the right-hand side:

$14x + 28 = 10x - 28$

Now, subtract $10x$ from both sides:

$4x + 28 = -28$

Next, subtract 28 from both sides:

$4x = -56$

Finally, divide by 4:

$x = -14$

Step 3: Find the integers

The first integer is $x = -14$. Therefore, the three consecutive even integers are:

• First integer: $-14$
• Second integer: $-12$
• Third integer: $-10$

Conclusion:

The three consecutive even integers are $-14$, $-12$, and $-10$.

Would you like further details on the steps, or do you have any questions?

Here are some related questions:

1. How would the equation change if they were consecutive odd integers?
2. What if the relationship involved a different number, say 5 times the sum instead of 7 times?
3. Can this problem be extended to four consecutive integers?
4. How does this equation-solving strategy apply to quadratic equations?
5. What would change if one of the integers were negative but the others positive?

Tip: Always carefully set up the equation based on the relationships described in word problems to avoid errors in translation.