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