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

1 answer

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.