6(5y-8)-20=11(2y-3)+3y

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Solution for 6(5y-8)-20=11(2y-3)+3y equation: 



6(5y-8)-20=11(2y-3)+3y

We move all terms to the left:

6(5y-8)-20-(11(2y-3)+3y)=0

We multiply parentheses

30y-(11(2y-3)+3y)-48-20=0



We calculate terms in parentheses: -(11(2y-3)+3y), so:

11(2y-3)+3y

We add all the numbers together, and all the variables

3y+11(2y-3)

We multiply parentheses

3y+22y-33

We add all the numbers together, and all the variables

25y-33

Back to the equation:

-(25y-33)



We add all the numbers together, and all the variables

30y-(25y-33)-68=0

We get rid of parentheses

30y-25y+33-68=0

We add all the numbers together, and all the variables

5y-35=0

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

5y=35

y=35/5

y=7

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