3+2(4y-1)=6(3+y)-19

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Solution for 3+2(4y-1)=6(3+y)-19 equation: 



3+2(4y-1)=6(3+y)-19

We move all terms to the left:

3+2(4y-1)-(6(3+y)-19)=0

We add all the numbers together, and all the variables

2(4y-1)-(6(y+3)-19)+3=0

We multiply parentheses

8y-(6(y+3)-19)-2+3=0



We calculate terms in parentheses: -(6(y+3)-19), so:

6(y+3)-19

We multiply parentheses

6y+18-19

We add all the numbers together, and all the variables

6y-1

Back to the equation:

-(6y-1)



We add all the numbers together, and all the variables

8y-(6y-1)+1=0

We get rid of parentheses

8y-6y+1+1=0

We add all the numbers together, and all the variables

2y+2=0

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

2y=-2

y=-2/2

y=-1

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