SOLUTION: Eight times the sum of three consecutive even integers is 108 more than 21 times the smallest of three integers

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Question 406344: Eight times the sum of three consecutive even integers is 108 more than 21 times the smallest of three integers
Answer by ewatrrr(24785) About Me  (Show Source):
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Hi

Find three consecutive integers 

Let x, (x+1), (x+2) represent the three consecutive integers respectively

Question states***

  8[x+ (x+1)+ (x+2)] = 21x + 108

Solving for x

    8(3x + 3) = 21x + 108

      24x + 24 = 21x + 108

           3x = 84

             x = 28  The three consecutive integers are 28,29,30

          

CHECKING our Answer***

   8*87 = 696 = 588 + 108