Question 408661
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Hi
three consecutive odd integers
Let x, (x+2),(x+4) represent the three consecutive odd integers
Question states***
 x(x+2) = 5(x+4) + 8
Solving for x
 x^2 + 2x = 5x+ 20 + 8
 x^2-3x -28=0
factoring
 (x-7)(x+4)=0 Note:SUM of the inner product(-7x) and the outer product(4x) = -3x
 (x+4) = 0   x = -4  Extraneous solution 
 (x-7) = 0   x = 7   The three consecutive odd integers are 7,9,11

CHECKING our Answer***
  63 = 55 + 8