Question

The sum of three consecutive integers is 114. what are the three integers?

Answer 1

Since your question asks for "three consecutive even integers" you can call them as follows:

x1 = first even integer

x2 = second even integer

x3 = third even integer

Since you know they're both consecutive and even, you know that:

x2 = x1 + 2

x3 = x2 + 2

This is because there's always a difference of 2 between consecutive even integers. Take a look at these examples:

6, 8, 10

102, 104, 106

514, 516, 518

No matter what the even integers are, as long as they're consecutive, they're spaced out by 2 each.

You know that the sum of your even integers adds up to 114, so:

x1 + x2 + x3 = 114

But since x3 = x2 + 2

x1 + x2 + (x2 + 2) = 114

And since x2 = x1 + 2

x1 + (x1 + 2) + ((x1 + 2) + 2) = 114

So your final equation will be:

x1 + x1 + 2 + x1 +2 + 2 = 114

3x1 + 6 = 114

3x1 = 108

x1 = 36

x2 = x1 + 2

x2 = 36 + 2 = 38

x3 = x2 + 2

x3 = 38 + 2 = 40

So your consecutive even integers are:

36, 38, and 40

If you add these all together you get:

36 + 38 + 40 = 114. So you know you did it correctly.

Hope this helps.

Answer 2

Answer: 37, 38, and 39

Step-by-step explanation: This problem states that the sum of 3 consecutive integers is 114 and it asks us to find the integers.

3 consecutive integers can be represented as followed.

X ⇒ first integer

X + 1 ⇒ second integer

X + 2 ⇒ third integer

Since the sum of our 3 consecutive integers is 114, we can set up an equation to represent this.

X + X + 1 + X + 2 = 114

We can simplify on the left side by combining the X's and the numbers.

3x + 3 = 114

-3 -3 ← subtract 3 from both sides of the equation

3x = 111

÷3 ÷3 ← divide both sides of the equation by 3

X = 37

X ⇒ first integer = 37

X + 1 ⇒ second integer = 38

X + 2 ⇒ third integer = 39

Therefore, our 3 consecutive integers are 37, 38, and 39.