Question 333638: The general expression for consecutive multiples of seven is 7N, 7(N+1), 7(N+2), etc. Find three consecutive multiples of seven such that 3 times the first exceeds twice the third by 14.
Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! he general expression for consecutive multiples of seven is 7N, 7(N+1), 7(N+2), etc.
Find three consecutive multiples of seven such that 3 times the first exceeds
twice the third by 14.
:
3(7N) = 2(7(n+2) + 14
21N = 14(N+2) + 14
21N = 14N + 28 + 14
21N - 14n + 42
7N = 42, is the 1st multiple
then
42, 49, 56 are the 3 multiples
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