-(y-5)-3(2y-1)=3(y+1)-5

Solution for -(y-5)-3(2y-1)=3(y+1)-5 equation:

-(y-5)-3(2y-1)=3(y+1)-5

We move all terms to the left:

-(y-5)-3(2y-1)-(3(y+1)-5)=0

We multiply parentheses

-(y-5)-6y-(3(y+1)-5)+3=0

We get rid of parentheses

-y-6y-(3(y+1)-5)+5+3=0


We calculate terms in parentheses: -(3(y+1)-5), so:

3(y+1)-5

We multiply parentheses

3y+3-5

We add all the numbers together, and all the variables

3y-2

Back to the equation:

-(3y-2)


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-10y+10=0

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

-10y=-10

y=-10/-10

y=1