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Ryry45go
Answered

The general expressions for consecutive multiples of 3 are 3N, 3(N + 1), 3(N + 2), etc.,
where N is some unspecified integer. Find four consecutive multiples of 3 such that 5
times the sum of the first and fourth is 6 less than 13 times the third.

Answer :

LammettHashgo

These 4 consecutive multiples of 3 are of the form

3N, 3 (N + 1), 3 (N + 2), 3 (N + 3)

The given constraint is captured in the equation

5 (3N + 3 (N + 3)) = 13 (3 (N + 2)) - 6

Solve for N :

5 (3N + 3N + 9) = 13 (3N + 6) - 6

5 (6N + 9) = 13 (3N + 6) - 6

30N + 45 = 39N + 78 - 6

30N + 45 = 39N + 72

30N - 39N = 72 - 45

-9N = 27

N = 27/(-9)

N = -3

Then the 5 consecutive multiples of 3 are

-9, -6, -3, 0

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