Find four consecutive odd integers such that three times the first and second is 15 less than the fourth.

Answer 1

x+ (x+2) + (x+4) + (x+6) are 4 consecutive odd integers

3(x + x+2) = x+6 -15

3(2x +2) = x -9

6x +6 = x -9

6x-x = -9-6

5x = -15

x = -15/5 = -3

the integers are -3, -1, 1, 3

3(-3 -1) = 3-15

3(-4) = -12