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

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Solution for 6(y+3)=-4(3y-4)+2y equation: 



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

We move all terms to the left:

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

We multiply parentheses

6y-(-4(3y-4)+2y)+18=0



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

-4(3y-4)+2y

We add all the numbers together, and all the variables

2y-4(3y-4)

We multiply parentheses

2y-12y+16

We add all the numbers together, and all the variables

-10y+16

Back to the equation:

-(-10y+16)



We get rid of parentheses

6y+10y-16+18=0

We add all the numbers together, and all the variables

16y+2=0

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

16y=-2

y=-2/16

y=-1/8

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