Three consecutive even integers such that the sum of twice the smallest, one less than the middle, and four more than the largest is -15
Let the integers be:
n, n + 2, n + 4
2n + (n + 2) - 1 + (n + 4) + 4 = -15
4n - 1 + 10 = -15
4n + 9 = -15
4n = -15 - 9
4n = - 24
n = - 6 - the smallest number
-6 + 2 = - 4 - the middle number.
-6 + 4 = - 2 - the largest number.
