Data Sufficiency

I need your help on this one because I got it right, however I believe the 2nd premise was luck because I did it differently and it may be wrong.

I rewrote "is the ratio of x to y greater than 3" as x/y > 3? Yes or no question

1. 2x/3y > 2 so here multiply both sides by 3 to cancel out the 3y which results in 3(2x/3y) > 3(2) = 2x/y > 6, then I multiplied both sides by ½ (or divided by 2) to cancel the 2 which results in ½(2x/y > ½(6) = x/y > 3. This is yes so sufficient.
2. x = 3y + 2, here I divided both sides by y to remove the y from the right side which results in (x = 3y + 2)/y = x/y = 3 + 2 or 5. Yes as well. After seeing the explanation, I'm not entirely sure that I can remove the y from the right side unless the 2 also had y as the denominator. So something like x = 3y + 2y or 5y.
Please let me know if this is confusing, I can right it out on scratch paper. Thanks for your help.

The correction you made is right (and your initial process for statement #2, not)... if I divide (3y + 2) by y, I have to distribe that y to both parts (think of this as the equvalent of multiplying (3y+2) by (1/y)...you would distribute that 1/y to both the 3y and the 2). Another way to think of why we can't get (3y + 2)/y to equal 5 is that if that were true, it would have to check back in multiplication. In other words, if (3y+2)/y = 5, then it must also be the case that 5*y = 3y +2. But it doesn't, since 5*y just equals 5y. So yep, remember to distribute. (3y+2)/y should come out to 3 + 2/y (this now equals x/y). At that point, we need to capitalize on the given in the question that x and y are positive, because if y were allowed to be negative, 3 + 2/y could still be less than 3... but since y is positive, 2/y will be positive, so when we add it to 3 we'll definitely get something greater than 3.