Find five consecutive integers such that the sum of the first and 5 times the third is equal to 41 less than 3 times the sum of the second fourth and fifth

see below

Step-by-step explanation:

We'll cal the first integer x and then the rest of them will be x + 1, x + 2, x + 3 and x + 4. We can write x + 5 (x + 2) = 3 (x + 1 + x + 3 + x + 4) - 41.

x + 5x + 10 = 3 (3x + 8) - 41

6x + 10 = 9x + 24 - 41

6x + 10 = 9x - 17

3x = 27

x = 9

The numbers are 9, 10, 11, 12, 13.