3y+7(5y-4)=38y-28

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Solution for 3y+7(5y-4)=38y-28 equation: 



3y+7(5y-4)=38y-28

We move all terms to the left:

3y+7(5y-4)-(38y-28)=0

We multiply parentheses

3y+35y-(38y-28)-28=0

We get rid of parentheses

3y+35y-38y+28-28=0

We add all the numbers together, and all the variables

=0

y=0/1

y=0

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