3consecutive odd integers that the sum of the smallest and 4 times the largest is 61

  9, 11, 13

Step-by-step explanation:

Let n represent the smallest of the integers. Then n+4 is the largest of them. The given total is ...

  n + 4(n+4) = 61

  5n +16 = 61

  5n = 45 . . . . . . subtract 16

  n = 9 . . . . . . . . divide by 5

The smallest of the integers is 9, so the other two are 11 and 13.

you need to give us more information since we can't answer

-8/2y-8=5/y+4-7y+8/y^2-16

we move all terms to the left:

-8/2y-8-(5/y+4-7y+8/y^2-16)=0

we can not solve this equation