find two consecutive odd intergers that twice the larger is fifteen more than three times the smaller

11, 13

Step-by-step explanation:

Let n and n+2 be the consecutive odd integers

2 (n+2) = 3n + 15

2n + 4 = 3n + 15

n = 11

n + 2 = 13

11, 13

Step-by-step explanation:

an odd number can be represented by 2n+1

since they are consecutive, the larger odd number will be 2n + 1 + 2

now,

2 (2n+3) = 2n + 1 + 15

solving this eqn, we get n = 5

so the two numbers are (2n+1) = 11 and 11+2 = 13