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

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

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

We move all terms to the left:

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

We multiply parentheses

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


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

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

We multiply parentheses

2y+(y-1)-6

We get rid of parentheses

2y+y-1-6

We add all the numbers together, and all the variables

3y-7

Back to the equation:

-(3y-7)


We add all the numbers together, and all the variables

3y-(3y-7)-5=0

We get rid of parentheses

3y-3y+7-5=0

We add all the numbers together, and all the variables

2!=0

There is no solution for this equation