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.

Question

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.

i don't understand what you mean by that

What's some things I'll need for this class?