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

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

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

We move all terms to the left:

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

We multiply parentheses

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


We calculate terms in parentheses: -(4(y+1)-3), so:

4(y+1)-3

We multiply parentheses

4y+4-3

We add all the numbers together, and all the variables

4y+1

Back to the equation:

-(4y+1)


We add all the numbers together, and all the variables

8y-(4y+1)-6=0

We get rid of parentheses

8y-4y-1-6=0

We add all the numbers together, and all the variables

4y-7=0

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

4y=7

y=7/4

y=1+3/4