High school algebra 1 honors class

Question

Find three consecutive integers such that the sum of the first, twice the second, and three times the third is -76.

Answer 1

let n be the first number then (n+1) and (n+2) are its successors.

Then

n + 2(n+1) + 3(n+2) = -76

n + 2n + 3n + 2 + 6 = -76

6n + 8 = - 76

6n = - 84

n = -14

Thus -14, -13 and -12 are the solutions.