Question

hi ppl need help with... Consectutive integers!

Answer 1

might need to give us a little more detail ....

Answer 2

the first one is I need to represent 3 integers I chose X,Y, and Z
so X+Y+Z+4=4Y

Answer 3

what does this website use that isn't working on my dDS COMPUTER?

Answer 4

dunno, but using google chrome helps alot

Answer 5

so i take it you need to find 3 consecutive integers that fit the bill for:

x + y + z + 4 = 4y

???

where does that expression come from?

Answer 6

my piece of paper

Answer 7

lol, k

well, i assume x y z are the numbers

consecutive integers are right next to each other: with 3 of them, i like to use:
(n-1), (n) , (n+1)

since y is in the middle, let x = n-1, y = n, and z = n+1

Answer 8

the -1 and +1 cancel giving us: 3n+4 = 4n

Answer 9

but why n+1 and n-1

Answer 10

thanks chrome fixed it

Answer 11

thats just my choice, what is important is that you have 3 numbers that are all in a cluster:

n, n+1, n+2

or

n-1 , n, n+1

or

n-2 , n-1 , n

or

n-k , n-(k+1) , n-(k+2)

Answer 12

<------ is what i am doing

Answer 13

ok then
WHAT IS THAT!

Answer 14

you just need to define some general way of expressing 3 numbers that are all in a row

Answer 15

if we start with some number, n

the next number is n+1

the number after that is n+1+1

Answer 16

did i mention my iq level is not all that high?

Answer 17

we end up after substituting into your equation ....

(n) + (n+1) + (n+1+1) + 4 = 4(n+1)

Answer 18

do u work for math-u-see cause that is word for word what it says in the boook

Answer 19

legit

Answer 20

lol, no i dont but it is a pretty basic concept

Answer 21

ok im going to make a new thread ok?

Answer 22

my original thought can be seen if we simply subtract 1 from n:

(n-1) + (n-1+1) + (n-1+1+1) + 4 = 4(n-1+1)

(n-1) + (n) + (n+1) + 4 = 4(n)

k, thats fine

Answer 23

ill give u a medal even tho i still dont understand

Answer 24

good luck