Find three consecutive integers such that the sum of the first and second is six more than the third

Question

also need the work shown

Answer 1

n = first

n + 1 = second

n + 2 = third

The wording of the problem translates into

n + (n + 1) = (n + 2) + 6

Can you solve for n, then get n+1 and n+2, and finish from here?