34-(3+4y-2(y+2))=-5(4y-3)-(5(y-1)-20y+3)

Solution for 34-(3+4y-2(y+2))=-5(4y-3)-(5(y-1)-20y+3) equation:

34-(3+4y-2(y+2))=-5(4y-3)-(5(y-1)-20y+3)

We move all terms to the left:

34-(3+4y-2(y+2))-(-5(4y-3)-(5(y-1)-20y+3))=0


We calculate terms in parentheses: -(3+4y-2(y+2)), so:

3+4y-2(y+2)

determiningTheFunctionDomain
4y-2(y+2)+3

We multiply parentheses

4y-2y-4+3

We add all the numbers together, and all the variables

2y-1

Back to the equation:

-(2y-1)



We calculate terms in parentheses: -(-5(4y-3)-(5(y-1)-20y+3)), so:

-5(4y-3)-(5(y-1)-20y+3)

We multiply parentheses

-20y-(5(y-1)-20y+3)+15


We calculate terms in parentheses: -(5(y-1)-20y+3), so:

5(y-1)-20y+3

We add all the numbers together, and all the variables

-20y+5(y-1)+3

We multiply parentheses

-20y+5y-5+3

We add all the numbers together, and all the variables

-15y-2

Back to the equation:

-(-15y-2)


We get rid of parentheses

-20y+15y+2+15

We add all the numbers together, and all the variables

-5y+17

Back to the equation:

-(-5y+17)


We get rid of parentheses

-2y+5y+1-17+34=0

We add all the numbers together, and all the variables

3y+18=0

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

3y=-18

y=-18/3

y=-6