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 

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