math
posted by Keaton
3/2y3/2y+4
could you help me please

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

Keaton
3/2y3/2y+4
help me I must turn in tonight. 
drwls
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)]
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