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Question 390544: multiples of 7 is 7N, 7(N+1), 7(N+2), and so on, where N is some unspecified integer. Find three consecutive multiples of 7 such that the product of -3 and the sum of the first and third is 21 less than 5 times the opposite of the second
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! multiples of 7 is 7N, 7(N+1), 7(N+2), and so on, where N is some unspecified integer.
Find three consecutive multiples of 7 such that the product of -3 and the sum
of the first and third is 21 less than 5 times the opposite of the second
:
Let the three integers be: 7n, 7(n+1), 7(n+2)
:
-3(7n + 7(n+2)) = 5(-7(n+1) - 21
:
-3(7n + 7n + 14)) = 5(-7(n+1) - 21
:
-3(14n + 14)) = 5(-7(n+1) - 21
:
-42m - 42 = 5(-7n - 7) - 21
:
-42n - 42 = -35n - 35 - 21
:
-42n - 42 = -35n - 56
Get rid of all those negatives, multiply by -1
42n + 42 = 35n + 56
:
42n - 35n = 56 - 42
:
7n = 14
:
n = 2
:
The three integers 14, 21, 28
:
:
See if that checks out in the statement;
"the product of -3 and the sum of the first and third is 21 less than 5 times the opposite of the second"
-3(14+28) = 5(-21) - 21
-3(42) = -105 - 21
-126 = - 126; confirms our solution of n=2
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