Find three consecutive numbers such that the sum of one-fourth the first and one-fifth the second is five less than one-seventh the third.

-10, - 11 and - 12

-12, - 13 and - 14

-14, - 15 and - 16

N, n+1, n+2 are the numbers.

n/4 + (n+1) / 5 = (n+2) / 7-5 make a common denominator of 140

(35n+28n+28) / 140 = (20n+40-700) / 140 multiply both sides by 140

35n+28n+28=20n+40-700

63n+28=20n-660

43n+28=-660

43n=-688

n=-16

since we started with n, n+1, and n+2 the numbers are:

-16, - 15, - 14