36-(1+7y-2(y+3))=-3(4y-7)-(9(y-1)-12y+13)

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Solution for 36-(1+7y-2(y+3))=-3(4y-7)-(9(y-1)-12y+13) equation: 



36-(1+7y-2(y+3))=-3(4y-7)-(9(y-1)-12y+13)

We move all terms to the left:

36-(1+7y-2(y+3))-(-3(4y-7)-(9(y-1)-12y+13))=0



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

1+7y-2(y+3)

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

We multiply parentheses

7y-2y-6+1

We add all the numbers together, and all the variables

5y-5

Back to the equation:

-(5y-5)





We calculate terms in parentheses: -(-3(4y-7)-(9(y-1)-12y+13)), so:

-3(4y-7)-(9(y-1)-12y+13)

We multiply parentheses

-12y-(9(y-1)-12y+13)+21



We calculate terms in parentheses: -(9(y-1)-12y+13), so:

9(y-1)-12y+13

We add all the numbers together, and all the variables

-12y+9(y-1)+13

We multiply parentheses

-12y+9y-9+13

We add all the numbers together, and all the variables

-3y+4

Back to the equation:

-(-3y+4)



We get rid of parentheses

-12y+3y-4+21

We add all the numbers together, and all the variables

-9y+17

Back to the equation:

-(-9y+17)



We get rid of parentheses

-5y+9y+5-17+36=0

We add all the numbers together, and all the variables

4y+24=0

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

4y=-24

y=-24/4

y=-6

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