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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

(y+1)-(2(y-1)+y+y)+7=0

We get rid of parentheses

y-(2(y-1)+y+y)+1+7=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

y-(4y-2)+8=0

We get rid of parentheses

y-4y+2+8=0

We add all the numbers together, and all the variables

-3y+10=0

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

-3y=-10

y=-10/-3

y=3+1/3