4(3y-5)-11=6(5y+1)+20

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Solution for 4(3y-5)-11=6(5y+1)+20 equation: 



4(3y-5)-11=6(5y+1)+20

We move all terms to the left:

4(3y-5)-11-(6(5y+1)+20)=0

We multiply parentheses

12y-(6(5y+1)+20)-20-11=0



We calculate terms in parentheses: -(6(5y+1)+20), so:

6(5y+1)+20

We multiply parentheses

30y+6+20

We add all the numbers together, and all the variables

30y+26

Back to the equation:

-(30y+26)



We add all the numbers together, and all the variables

12y-(30y+26)-31=0

We get rid of parentheses

12y-30y-26-31=0

We add all the numbers together, and all the variables

-18y-57=0

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

-18y=57

y=57/-18

y=-3+1/6

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