What are three consecutive integers whose sum is 96?

3 Answers
Mar 22, 2018

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

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.

Mar 22, 2018

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).