Find three consecutive odd integers such that 7 times the sum of the first and the third is 120 less than 10 times the opposite of the second.

x - 1st

x +2- 2nd

x +4- 3rd

7(x + x + 4 ) = 10(-x -2) -120

14x + 28 = -10x -20 -120

14x + 10x = -140 - 28

24x = 168

x = 168/24

x = 7

answers: 7, 9, 11