2y-12=3(y-4)-y

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Solution for 2y-12=3(y-4)-y equation: 



2y-12=3(y-4)-y

We move all terms to the left:

2y-12-(3(y-4)-y)=0



We calculate terms in parentheses: -(3(y-4)-y), so:

3(y-4)-y

We add all the numbers together, and all the variables

-1y+3(y-4)

We multiply parentheses

-1y+3y-12

We add all the numbers together, and all the variables

2y-12

Back to the equation:

-(2y-12)



We get rid of parentheses

2y-2y+12-12=0

We add all the numbers together, and all the variables

=0

y=0/1

y=0

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