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

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

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

We move all terms to the left:

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

We multiply parentheses

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


We calculate terms in parentheses: -(2(y-3)+(y-3)+(y+1)), so:

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

We multiply parentheses

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

We get rid of parentheses

2y+y+y-3+1-6

We add all the numbers together, and all the variables

4y-8

Back to the equation:

-(4y-8)


We add all the numbers together, and all the variables

3y-(4y-8)-5=0

We get rid of parentheses

3y-4y+8-5=0

We add all the numbers together, and all the variables

-1y+3=0

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

-y=-3

y=-3/-1

y=+3