13(3y-3)-y=-3(y-2)

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Solution for 13(3y-3)-y=-3(y-2) equation: 



13(3y-3)-y=-3(y-2)

We move all terms to the left:

13(3y-3)-y-(-3(y-2))=0

We add all the numbers together, and all the variables

-1y+13(3y-3)-(-3(y-2))=0

We multiply parentheses

-1y+39y-(-3(y-2))-39=0



We calculate terms in parentheses: -(-3(y-2)), so:

-3(y-2)

We multiply parentheses

-3y+6

Back to the equation:

-(-3y+6)



We add all the numbers together, and all the variables

38y-(-3y+6)-39=0

We get rid of parentheses

38y+3y-6-39=0

We add all the numbers together, and all the variables

41y-45=0

We move all terms containing y to the left, all other terms to the right

41y=45

y=45/41

y=1+4/41

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