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


Solve the equation above. What is the largest of the three consecutive even integers?

(1 point)
Responses

38
38

38, 42, 44
38, 42, 44

44
44

42

Answer 1

Let's solve the equation \(x + (x + 2) + (x + 4) = 120\).

1. Combine like terms:
\[
x + x + 2 + x + 4 = 120
\]
\[
3x + 6 = 120
\]

2. Subtract 6 from both sides:
\[
3x = 120 - 6
\]
\[
3x = 114
\]

3. Divide by 3:
\[
x = \frac{114}{3}
\]
\[
x = 38
\]

Now, we can find the three consecutive even integers:
- 1st integer: \(x = 38\)
- 2nd integer: \(x + 2 = 38 + 2 = 40\)
- 3rd integer: \(x + 4 = 38 + 4 = 42\)

The three consecutive even integers are 38, 40, and 42.

The largest of the three integers is **42**.