Question

the sum of three integral numbers is 117 find the numbers.

Answer 1

3x + (x+1) + (x +2) = 117?.... right

Answer 2

another good one to guess and check.
i assume there is something else like "consecutive'

Answer 3

1 + 1 +115 works

Answer 4

oh yeah sorry.. :/

Answer 5

so lets try 25 + 26 + 27

Answer 6

too small

Answer 7

yeah thats wrong

Answer 8

35 + 36 + 37?

Answer 9

still too small, but not by much

Answer 10

38+39+40?

Answer 11

Lol i feel like im cheating by doing that... :P

Answer 12

bingo.
now for the algebra. if you put one number as n,
the the other two are n +1 and n+2
then put
\[n+n+1+n+2=117\]
\[3n+3=117\] etc

Answer 13

nice to know the answer in advance. then we can make sure the algebra is right
\[3n+3=117\]
\[3n=114\]
\[n=114\div 3\] whatever that is

Answer 14

alright i have another for you... im good at the algebra part its just the unecessary words confuse me.. :/

Answer 15

do i get to guess?

Answer 16

haha sure.. :P

Answer 17

One number is 800 less then 3 times the other and there sum is 1,200 find the numbers

Answer 18

grrr

Answer 19

ok i guess i can guess lets try 500 and 1500-800

Answer 20

wow got it on the first try!!

Answer 21

Nice job! :P

Answer 22

now algebra...
one number is n, other is 3n-800

Answer 23

i would give you another medal if i could.. :P

Answer 24

now i am sure you can do the algebra for this one
\[n + 3n-800=1200\] etc

Answer 25

actually the point of guessing is to make sure that the algebra makes sense

Answer 26

okie dokie.. :D

Answer 27

i guess 500
where does the next number come from? i multiply by 3 and subtract 800

Answer 28

so if i guess n, the next number is 3n - 800
and that is what i have
\[n+3n-800=1200\]

Answer 29

ohh see i took 800 from 1200 and etc

Answer 30

so you solve via
\[4n-800=1200\]
\[4n=1200+800=2000\]
\[n=2000\div 4=500\]

Answer 31

okay now you confuddled me....