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?

Answer 1

We start with the equation given for the sum of three consecutive even integers:

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

Now, we simplify the left side of the equation:

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

Combining like terms:

\[
3x + 6 = 120
\]

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

\[
3x = 120 - 6
\]

\[
3x = 114
\]

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

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

\[
x = 38
\]

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

1. First integer: \(x = 38\)
2. Second integer: \(x + 2 = 38 + 2 = 40\)
3. Third integer: \(x + 4 = 38 + 4 = 42\)

The largest of the three integers is:

\[
\boxed{42}
\]