SOLUTION: I need help with this problem. I am not sure if this problem falls under the category I put it in but I don't really know the category it is in: 1-2/3y=12/5-y/5

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Question 17596: I need help with this problem. I am not sure if this problem falls under the category I put it in but I don't really know the category it is in:
1-2/3y=12/5-y/5
Answer by whisper(27) About Me  (Show Source):
You can put this solution on YOUR website!
To solve this we will use two processes. We will need to separate constants to one side of the equation and also find common denominators. I am assuming that is (2/3)y
1 - (2/3)y = 12/5 - y/5
1 - 12/5 -(2/3)y = 12/5 - 12/5 - y/5
5/5 - 12/5 - (2/3)y = 0 - y/5
-7/5 - (2/3)y = -y/5
-7/5 - (2/3)y + (2/3)y = -y/5 + (2/3)y
-7/5 + 0 = -y/5 + (2/3)y
-7/5 = -y/5 + (2/3)y
At this point I am going to "clear" the denominator by multiplying through by 5.
5(-7/5) = 5[-y/5 + (2/3)y]
-7 = -y + (10/3)y
Find a common denominator for the y coefficients:
-7 = -(3/3)y + (10/3)y
-7 =-(7/3)y
Multiply both sides by the recipricol of -(7/3) which is (-3/7)
-7(-3/7)=-(7/3)[-(3/7)]
3 = 1y
3=y
Please don't hesitate to ask any further questions!