-10y-5(1+y)=3(2y-2)-20

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Solution for -10y-5(1+y)=3(2y-2)-20 equation: 



-10y-5(1+y)=3(2y-2)-20

We move all terms to the left:

-10y-5(1+y)-(3(2y-2)-20)=0

We add all the numbers together, and all the variables

-10y-5(y+1)-(3(2y-2)-20)=0

We multiply parentheses

-10y-5y-(3(2y-2)-20)-5=0



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

3(2y-2)-20

We multiply parentheses

6y-6-20

We add all the numbers together, and all the variables

6y-26

Back to the equation:

-(6y-26)



We add all the numbers together, and all the variables

-15y-(6y-26)-5=0

We get rid of parentheses

-15y-6y+26-5=0

We add all the numbers together, and all the variables

-21y+21=0

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

-21y=-21

y=-21/-21

y=1

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