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

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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) About Me  (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.