Finding consecutive integers

Question

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

Answer 1

3(f+s) = -15 + fourth

3(x +x+2) = x+6-15= x -9

6x +6 = x- 9

5x = -15

x = -3

-3, -1, 1, 3

Answer 2

The four consecutive integers can be x, x + 1, x + 2, and x + 3.

Three times the sum of the 1st and 2nd is 3(x + x + 1). 15 less than the 4th is x + 3 - 15

So the equation is 3(2x + 1) = x - 12

6x + 3 = x - 12 -> 5x = -15

So x = -3; x + 1 = -2; x + 2 = -1; x + 3 = 0 (Didn't say they had to be positive or non-zero!)

Check: 3(-3 + -2) = -15 and 0 - 15 = -15 (OK)