6(y+2)=-5(2y-4)+2y

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Solution for 6(y+2)=-5(2y-4)+2y equation: 



6(y+2)=-5(2y-4)+2y

We move all terms to the left:

6(y+2)-(-5(2y-4)+2y)=0

We multiply parentheses

6y-(-5(2y-4)+2y)+12=0



We calculate terms in parentheses: -(-5(2y-4)+2y), so:

-5(2y-4)+2y

We add all the numbers together, and all the variables

2y-5(2y-4)

We multiply parentheses

2y-10y+20

We add all the numbers together, and all the variables

-8y+20

Back to the equation:

-(-8y+20)



We get rid of parentheses

6y+8y-20+12=0

We add all the numbers together, and all the variables

14y-8=0

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

14y=8

y=8/14

y=4/7

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