2(3y-1)+y-4)=13

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Solution for 2(3y-1)+y-4)=13 equation: 



2(3y-1)+y-4)=13

We move all terms to the left:

2(3y-1)+y-4)-(13)=0

We add all the numbers together, and all the variables

y+2(3y-1)=0

We multiply parentheses

y+6y-2=0

We add all the numbers together, and all the variables

7y-2=0

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

7y=2

y=2/7

y=2/7

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