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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 consecutive even integers?

(1 point)
Responses

38
38

38, 42, 44
38, 42, 44

44
44

42

1 answer

Let's solve the equation \(x + (x + 2) + (x + 4) = 120\).

  1. Combine like terms: \[ x + x + 2 + x + 4 = 120 \] \[ 3x + 6 = 120 \]

  2. Subtract 6 from both sides: \[ 3x = 120 - 6 \] \[ 3x = 114 \]

  3. Divide by 3: \[ x = \frac{114}{3} \] \[ x = 38 \]

Now, we can find the three consecutive even integers:

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

The three consecutive even integers are 38, 40, and 42.

The largest of the three integers is 42.