27-(4+8y-4(y+2))=-3(2y-5)-(5(y-1)-1y+7)

Solution for 27-(4+8y-4(y+2))=-3(2y-5)-(5(y-1)-1y+7) equation:

27-(4+8y-4(y+2))=-3(2y-5)-(5(y-1)-1y+7)

We move all terms to the left:

27-(4+8y-4(y+2))-(-3(2y-5)-(5(y-1)-1y+7))=0


We calculate terms in parentheses: -(4+8y-4(y+2)), so:

4+8y-4(y+2)

determiningTheFunctionDomain
8y-4(y+2)+4

We multiply parentheses

8y-4y-8+4

We add all the numbers together, and all the variables

4y-4

Back to the equation:

-(4y-4)



We calculate terms in parentheses: -(-3(2y-5)-(5(y-1)-1y+7)), so:

-3(2y-5)-(5(y-1)-1y+7)

We multiply parentheses

-6y-(5(y-1)-1y+7)+15


We calculate terms in parentheses: -(5(y-1)-1y+7), so:

5(y-1)-1y+7

We add all the numbers together, and all the variables

-1y+5(y-1)+7

We multiply parentheses

-1y+5y-5+7

We add all the numbers together, and all the variables

4y+2

Back to the equation:

-(4y+2)


We get rid of parentheses

-6y-4y-2+15

We add all the numbers together, and all the variables

-10y+13

Back to the equation:

-(-10y+13)


We get rid of parentheses

-4y+10y+4-13+27=0

We add all the numbers together, and all the variables

6y+18=0

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

6y=-18

y=-18/6

y=-3