Question

What are three consecutive integers whose sum is 96?

Answer 1

I got `31,32 and33`

Explanation:

Call your integers:
`n`
`n+1`
`n+2`

you get:

`n+n+1+n+2=96`

rearrange:

`3n=93`
and so:
`n=93/3=31`

so our integers are:

`n=31`
`n+1=32`
`n+2=33`

Answer 2

You must symbolize the first integer with `x`.

Explanation:

Lets pretend the first number was `5`. What would you do to get to the immediate next integer? (Integers are whole numbers like `1, 2, 3` ) You would add `1`. So the next number is symbolized as "`x+1`".

How would you get from `5` to `7`? You would add `2` to the `x`. So the next number is written in symbols as "`x+2`."

Now add them all up like this: `x + x+1 + x+2 = 96`
Combine like terms: `3x +3 = 96`
Subtract the 3 from both sides `3x = 93`
Divide both sides by `3`: `x=32`
Answer: `x=32`.

BTW, "consecutive" means to come right after. In my pretend answer, `6` came right after `5`, and `7` came right after `6`.

Answer 3

31, 32, 33

Explanation:

If you represent the first integer with the letter `x`, then:
`x + (x+1) + (x+2) = 96`

This simplifies to:
`x + x + 1 + x + 2 = 96`
`x + x + 1 + x + 2 = 96`
`3x + 3 = 96`
`3x = 93`
`x = 31`

The first integer is 31. The next two consecutive integers are 32 `(x+1)` and 33 `(x+2)`.