3y+7(5y-4)+10=6(y-3)+32y

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Solution for 3y+7(5y-4)+10=6(y-3)+32y equation: 



3y+7(5y-4)+10=6(y-3)+32y

We move all terms to the left:

3y+7(5y-4)+10-(6(y-3)+32y)=0

We multiply parentheses

3y+35y-(6(y-3)+32y)-28+10=0



We calculate terms in parentheses: -(6(y-3)+32y), so:

6(y-3)+32y

We add all the numbers together, and all the variables

32y+6(y-3)

We multiply parentheses

32y+6y-18

We add all the numbers together, and all the variables

38y-18

Back to the equation:

-(38y-18)



We add all the numbers together, and all the variables

38y-(38y-18)-18=0

We get rid of parentheses

38y-38y+18-18=0

We add all the numbers together, and all the variables

=0

y=0/1

y=0

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