3(y+1)+42=2(y-1)+y+y

Solution for 3(y+1)+42=2(y-1)+y+y equation:

3(y+1)+42=2(y-1)+y+y

We move all terms to the left:

3(y+1)+42-(2(y-1)+y+y)=0

We multiply parentheses

3y-(2(y-1)+y+y)+3+42=0


We calculate terms in parentheses: -(2(y-1)+y+y), so:

2(y-1)+y+y

We add all the numbers together, and all the variables

2y+2(y-1)

We multiply parentheses

2y+2y-2

We add all the numbers together, and all the variables

4y-2

Back to the equation:

-(4y-2)


We add all the numbers together, and all the variables

3y-(4y-2)+45=0

We get rid of parentheses

3y-4y+2+45=0

We add all the numbers together, and all the variables

-1y+47=0

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

-y=-47

y=-47/-1

y=+47