Question

1. Find 3 consecutive integers whose sum is 33.

Answer 1

Answer: 10, 11, 12

Step-by-step explanation: Think of the integers like this:

1st integer: x

2nd integer: x+1

3rd integer: x+2

That is necessary because they are consecutive integers. Since the sum is 33, we need to create an equation.

x+x+1+x+2=33.

Simplify:

3x+3=33.

Opposite operations:

3x=-3+33.

To get the 3 close to the 33, we needed to make it negative, which is the opposite operation of the positive 3.

So,

3x=30.

Divide by 3:

x=10.

The first integer, x, equals 10.

To go with the guide that we already created,

1st integer: x=10

2nd integer: x+1=11

3rd integer:x+2=12.

Therefore, the three consecutive integers are 10, 11, and 12.

To check that, add them up. They all equal 33 and they are consecutive, which means this is the right answer!

Answer 2

10, 11, 12

Explanation:

For problems involving consecutive integers, I like to consider their average value. For three consecutive integers, their average will be the middle one. It will be ...

average = sum / (number of numbers) = 33/3 = 11

The middle number is 11, so the numbers are ...

10, 11, 12

_____

If you want to write an equation, you can let x represent the middle number. Then the other two are x-1 and x+1, and their sum is ...

(x-1) +(x) +(x+1) = 33 = 3x

x = 11 ⇒ x-1 = 10, x+1 = 12