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

44
44

38, 42, 44
38, 42, 44

38
38

42

Answer 1

To solve the equation \( x + (x + 2) + (x + 4) = 120 \), we first simplify the left side:

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

Combining like terms, we get:

\[
3x + 6 = 120
\]

Next, we subtract 6 from both sides:

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

Now, we divide both sides by 3:

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

Now that we have the first integer \( x \), we can find the other two integers:

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 the three integers is:

\[
\boxed{42}
\]