Question

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

44
44

42
42

38, 42, 44

Answer 1

Let's solve the equation provided:

The equation is:

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

Combining like terms:

\[
x + x + 2 + x + 4 = 120
\]
\[
3x + 6 = 120
\]

Next, we will isolate \(x\):

\[
3x = 120 - 6
\]
\[
3x = 114
\]
\[
x = \frac{114}{3}
\]
\[
x = 38
\]

Now we can find the three consecutive even integers:

1st integer: \(x = 38\)

2nd integer: \(x + 2 = 38 + 2 = 40\)

3rd integer: \(x + 4 = 38 + 4 = 42\)

Now we have the integers: 38, 40, and 42. The largest of these three integers is **42**.

Thus, the answer is:

**42**