SOLUTION: Assuming the general expression for multiples to be 3N, 3(n+1), 3(n+2) etc. Find 4 consecutive multiples of 3 such that 5 times the sum of the 1st and 4th is 6 less than 13 time

Question 393594: Assuming the general expression for multiples to be 3N, 3(n+1), 3(n+2) etc.
Find 4 consecutive multiples of 3 such that 5 times the sum of the 1st and 4th is 6 less than 13 times the 3rd.
I have worked and reworked and can't come up with the right answer. Can you explain these steps to me?
Thanks so much!
Answer by robertb(5830) (Show Source): You can put this solution on YOUR website!
3n, 3(n+1), 3(n+2), and 3(n+3)
Then
5(3n + 3(n+3)) = 13(3(n+2)) - 6
==> 5(3n + 3n + 9) = 39(n+2) - 6
==> 5(6n + 9) = 39n + 78 - 6
==> 30n+ 45 = 39n + 72
==> -27 = 9n
==> n = -3
Then the numbers are -9, -6, -3, 0.
RELATED QUESTIONS
The general expression for consecutive multiples of seven is 7N, 7(N+1), 7(N+2), etc.... (answered by ankor@dixie-net.com)

help! The general expression for consecutive multiples of 6 is 6n, 6(n+1), 6(n+2) etc. (answered by checkley71)

The general expression for consecutive multiples of 6 is 6N, 6(N+1), 6(N+2), etc. Find... (answered by jorel1380)

The general expression for consecutive multiples of seven 7n, 7(n+1), 7(n+2)where n is... (answered by ankor@dixie-net.com)

The general expression for consecutive multiples of 5 is 5N, 5(N+1), 5(N+2), and so on,... (answered by stanbon)

There are 4 consecutive multiples of 3 such that the sum of the first 3 multiples is... (answered by Edwin McCravy)

multiples of 7 is 7N, 7(N+1), 7(N+2), and so on, where N is some unspecified integer.... (answered by ankor@dixie-net.com)

This is a problem from my math text. Find all sets of three consecutive multiples of 3 (answered by richwmiller)

find 3 consecutive multiples of 3 such that 6 times the first is 48 greater than 4 times... (answered by checkley77)