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)#.