You have not explained what the problem is. You have also not clarified whether the first and second terms are (3/2)y or 3/(2y)

It might help to realize that
3/(2y) - 3/(2y) = 0

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drwls

What happen to the 4 because the problem is 3 over 2y minus -3 over 2y+4= what. so what happen to the four

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Keaton

3/2y-3/2y+4

help me I must turn in tonight.

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Keaton

You should have written the "problem" with parentheses as
3/(2y) -3/(2y+4) . The way you wrote it, it is not clear what the denomonators are.

You still have not asked a question.

You cannot solve for y since it is not an equation. If you are supposed to rewite it with a common deominator, that would be
3(2y+4)-3(2y)/[2y(2y+4)]
= 12/[2y(2y+4)] = 6/[y(2y+4)]