Pierre and Ernst found three consecutive even integers such a that 7 times the sum of the first and third was 48 less than 10 times the second. What were the integers

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.

Pierre and Ernst found three consecutive even integers such that 7 times the sum of the first and third was 48 less than 10 times the second. What were the integers?

Learn how to solve a problem involving three consecutive even integers where 7 times the sum of the first and third is 48 less than 10 times the second. This detailed step-by-step solution uses algebraic techniques suitable for students in Grades 6-8.

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