y-2(3y-5)=3(2+y)-20

Solution for y-2(3y-5)=3(2+y)-20 equation:

y-2(3y-5)=3(2+y)-20

We move all terms to the left:

y-2(3y-5)-(3(2+y)-20)=0

We add all the numbers together, and all the variables

y-2(3y-5)-(3(y+2)-20)=0

We multiply parentheses

y-6y-(3(y+2)-20)+10=0


We calculate terms in parentheses: -(3(y+2)-20), so:

3(y+2)-20

We multiply parentheses

3y+6-20

We add all the numbers together, and all the variables

3y-14

Back to the equation:

-(3y-14)


We add all the numbers together, and all the variables

-5y-(3y-14)+10=0

We get rid of parentheses

-5y-3y+14+10=0

We add all the numbers together, and all the variables

-8y+24=0

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

-8y=-24

y=-24/-8

y=+3