Find 3 consecutive integers such that the sum of twice the smallest and 3 times the largest is 126

Find 3 consecutive integers such that the sum of twice the smallest and 3 times the largest is 126.

Let the first integer be n, the second integer be n + 1, and the third integer be n + 2. We have:

Sum of the smallest and 3 times the largest is 126:
n + 3(n + 2) = 126

Multiply through:
n + 3n + 6 = 126

Group like terms:
4n + 6 = 126

Type 4n + 6 = 126 into our calculator, we get n = 30. Which means the next two integers are 31 and 32.
{30, 31, 32}