Question 1018435: Find three consecutive even integers such that the sum of twice the first, four times the second, and three times the third is 398
Answer by Cromlix(4381) (Show Source):
You can put this solution on YOUR website! Hi there,
3 consecutive even integers
n, n + 2 and n + 4.
2(n) + 4(n + 2) + 3(n + 4) = 398
2n + 4n + 8 + 3n + 12 = 398
Collect like terms
2n + 4n + 3n + 398 - 8 - 12
9n = 378
n = 42
42, 44, 46
Proof:
2(42) + 4(44) + 3(46) = 398
Hope this helps :-)
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