3(y+4)=-3(4y-2)+7y

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Solution for 3(y+4)=-3(4y-2)+7y equation: 



3(y+4)=-3(4y-2)+7y

We move all terms to the left:

3(y+4)-(-3(4y-2)+7y)=0

We multiply parentheses

3y-(-3(4y-2)+7y)+12=0



We calculate terms in parentheses: -(-3(4y-2)+7y), so:

-3(4y-2)+7y

We add all the numbers together, and all the variables

7y-3(4y-2)

We multiply parentheses

7y-12y+6

We add all the numbers together, and all the variables

-5y+6

Back to the equation:

-(-5y+6)



We get rid of parentheses

3y+5y-6+12=0

We add all the numbers together, and all the variables

8y+6=0

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

8y=-6

y=-6/8

y=-3/4

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