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

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

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

We move all terms to the left:

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

We multiply parentheses

3y-(2(y+1)+1)+12+6=0


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

2(y+1)+1

We multiply parentheses

2y+2+1

We add all the numbers together, and all the variables

2y+3

Back to the equation:

-(2y+3)


We add all the numbers together, and all the variables

3y-(2y+3)+18=0

We get rid of parentheses

3y-2y-3+18=0

We add all the numbers together, and all the variables

y+15=0

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

y=-15