# algebra 2

Hi i posted this question:
the general expression for consecutive multiples of 6 is 6N, 6(N + 1), 6(N +2), etc. find three consecutive multiples of 6 such that 4 times the first exceeds twice the third by 12.

and than Bobpursley replied:
Let N be the first, so n+1 is next, etc.

4(6(n+1))-2(6(n+3))=12

so, find n, then 6(n+1) for the first, 6(n+2) for the second, and 6(n+3) for the third.

So now i don't understand what he means by that. help me out here

Did you solve for n?

If you did, then you need three consecutive numbers divisible by six. They will be:

6(n+1) is the first number.
6(n+2) will be the second number.
You figure out what the third number will be.

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