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 integers?

(1 point)
Responses

38
38

38, 42, 44
38, 42, 44

42
42

44

Answer 1

To solve the equation \( x + (x + 2) + (x + 4) = 120 \), we first combine like terms:

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

This simplifies to:

\[
3x + 6 = 120
\]

Next, we isolate \( 3x \) by subtracting 6 from both sides:

\[
3x = 120 - 6
\]
\[
3x = 114
\]

Now, we divide by 3 to solve for \( x \):

\[
x = \frac{114}{3}
\]
\[
x = 38
\]

Now that we have the first integer, we can find the other two:

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

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

The largest of these integers is **42**.

So the correct response is:

**42**