(2y-(3y-4)+5-6)+10y=(12y-12)+36

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Solution for (2y-(3y-4)+5-6)+10y=(12y-12)+36 equation: 



(2y-(3y-4)+5-6)+10y=(12y-12)+36

We move all terms to the left:

(2y-(3y-4)+5-6)+10y-((12y-12)+36)=0

We add all the numbers together, and all the variables

10y+(2y-(3y-4)+5-6)-((12y-12)+36)=0



We calculate terms in parentheses: +(2y-(3y-4)+5-6), so:

2y-(3y-4)+5-6

We add all the numbers together, and all the variables

2y-(3y-4)-1

We get rid of parentheses

2y-3y+4-1

We add all the numbers together, and all the variables

-1y+3

Back to the equation:

+(-1y+3)





We calculate terms in parentheses: -((12y-12)+36), so:

(12y-12)+36

We get rid of parentheses

12y-12+36

We add all the numbers together, and all the variables

12y+24

Back to the equation:

-(12y+24)



We get rid of parentheses

10y-1y-12y+3-24=0

We add all the numbers together, and all the variables

-3y-21=0

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

-3y=21

y=21/-3

y=-7

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