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 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) 384244 44 38 42

Answer 1

Let's solve the equation step by step.

Given:

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

The equation representing the sum of these three consecutive even integers is:

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

Now, simplify the left side:

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

Combine like terms:

\[
3x + 6 = 120
\]

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

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

Now, divide by 3 to find \( x \):

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

Now, we can determine the three consecutive even integers:

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

The largest of the three integers is \( 42 \).

Therefore, the answer is **42**.