3y-30-(y+28)=3y-((2y-4)0

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Solution for 3y-30-(y+28)=3y-((2y-4)0 equation: 



3y-30-(y+28)=3y-((2y-4)0

We move all terms to the left:

3y-30-(y+28)-(3y-((2y-4)0)=0

We get rid of parentheses

3y-y-(3y-((2y-4)0)-30-28=0

We add all the numbers together, and all the variables

2y-(3y-((2y-4)0)-30-28=0

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