Firstly, let's factorise each equation individually - to do this, find 2 numbers that when summed add to the value of the second term, and when multiplied give the value of the third term.
7 and 12 give us 4 and 3 (4+3=7, 4*3=12) -- 8 and 15 give us 5 and 3 (5+3=8, 5*3=15)
Now we can rewrite these equations as (y+4)(y+3) and (y+5)(y+3) respectively.
Putting this in a fraction: (y+4)(y+3)/(y+5)(y+3) -- We can clearly see that there is a y+3 on both sides of the fraction, and given there are no terms outside of the brackets being multiplied, we can directly cancel.
This gives us our final answer:
(y+4)/(y+5)
