7(y+4)=6(3y-6)+42

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Solution for 7(y+4)=6(3y-6)+42 equation: 



7(y+4)=6(3y-6)+42

We move all terms to the left:

7(y+4)-(6(3y-6)+42)=0

We multiply parentheses

7y-(6(3y-6)+42)+28=0



We calculate terms in parentheses: -(6(3y-6)+42), so:

6(3y-6)+42

We multiply parentheses

18y-36+42

We add all the numbers together, and all the variables

18y+6

Back to the equation:

-(18y+6)



We get rid of parentheses

7y-18y-6+28=0

We add all the numbers together, and all the variables

-11y+22=0

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

-11y=-22

y=-22/-11

y=+2

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