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| 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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