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)

44

38

42

38, 42, 44

Answer 1

To solve the equation for the three consecutive even integers given by:

1. 1st integer = \(x\)
2. 2nd integer = \(x + 2\)
3. 3rd integer = \(x + 4\)

The equation is:

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

Now, let's simplify and solve for \(x\):

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

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

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

Now, divide both sides by 3:

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

Now that we have the value of \(x\), we can find the three consecutive even integers:

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

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

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

Hence, the correct answer is:

**42**