Find three consecutive odd integers such that the sum of the smaller two is three times larger increased by seven

three consecutive odd integers: 2n-1 2n+1 2n+3

that the sum of the smaller two is three

times larger increased by seven: 2n-1 + 2n+1 = 3 (2n+3) + 7

4n = 6n + 9 + 7

4n = 6n + 16

4n - 6n = 16

-2n = 16

n = 16 / (-2)

n=-8

a) 2n - 1 = 2 (-8) - 1 = - 17

b) 2n+1 = 2 (-8) + 1 = - 15

c) 2n+1 = 2 (-8) + 3 = - 13

Ans. - 17; - 15; - 13