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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

3y+4y-(2(y+1)+1)+8=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

7y-(2y+3)+8=0

We get rid of parentheses

7y-2y-3+8=0

We add all the numbers together, and all the variables

5y+5=0

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

5y=-5

y=-5/5

y=-1