Question

The sum of three consecutive even integers (3 even integers in a row) is 120.
Use the following setup and given equation to find your solution:
1st integer: x
2nd integer: x + 2
3rd integer: x + 4
Equation: x + (x + 2) + (x + 4) = 120

Answer 1

Let's solve the problem using the setup provided.

We are given three consecutive even integers:
1. 1st integer: \( x \)
2. 2nd integer: \( x + 2 \)
3. 3rd integer: \( x + 4 \)

The equation based on the sum of these integers is:

\[
x + (x + 2) + (x + 4) = 120
\]

Now, we can simplify the left side of the equation:

\[
x + x + 2 + x + 4 = 120
\]

Combining like terms:

\[
3x + 6 = 120
\]

Next, we'll isolate \( x \) by first subtracting \( 6 \) from both sides:

\[
3x = 120 - 6
\]

\[
3x = 114
\]

Now, divide both sides by \( 3 \):

\[
x = \frac{114}{3}
\]

\[
x = 38
\]

Now we can find the three consecutive even integers:

1. 1st integer: \( x = 38 \)
2. 2nd integer: \( x + 2 = 38 + 2 = 40 \)
3. 3rd integer: \( x + 4 = 38 + 4 = 42 \)

Thus, the three consecutive even integers are \( 38, 40, \) and \( 42 \).

Let's verify the solution by calculating their sum:

\[
38 + 40 + 42 = 120
\]

Since the sum is correct, the solution is confirmed. Therefore, the three consecutive even integers are **38, 40,** and **42**.