-20(y+1)+6(4y-3)=3(y-3)+12

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Solution for -20(y+1)+6(4y-3)=3(y-3)+12 equation: 



-20(y+1)+6(4y-3)=3(y-3)+12

We move all terms to the left:

-20(y+1)+6(4y-3)-(3(y-3)+12)=0

We multiply parentheses

-20y+24y-(3(y-3)+12)-20-18=0



We calculate terms in parentheses: -(3(y-3)+12), so:

3(y-3)+12

We multiply parentheses

3y-9+12

We add all the numbers together, and all the variables

3y+3

Back to the equation:

-(3y+3)



We add all the numbers together, and all the variables

4y-(3y+3)-38=0

We get rid of parentheses

4y-3y-3-38=0

We add all the numbers together, and all the variables

y-41=0

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

y=41

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