The sum of three consecutive even integers (3 even integers in a row) is 120.

Question

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:

\[ \boxed{42} \]