Question 936267
Find three consecutive odd integers such that the sum of the last two is 15 less than five times the first

let the integers be x , x+2, x+4

the sum of the last two= x+2+x+4 = 2x+6

5 times first = 5x

15 less than 5times first = 5x-15

equate

2x+6 = 5x-15

solve

3x=21
/3
x=7

the numbers are 7,9,11