-6(y-3y)+2=-3(3y-y)+4+17y

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Solution for -6(y-3y)+2=-3(3y-y)+4+17y equation: 



-6(y-3y)+2=-3(3y-y)+4+17y

We move all terms to the left:

-6(y-3y)+2-(-3(3y-y)+4+17y)=0

We add all the numbers together, and all the variables

-6(-2y)-(-3(+2y)+4+17y)+2=0

We multiply parentheses

12y-(-3(+2y)+4+17y)+2=0



We calculate terms in parentheses: -(-3(+2y)+4+17y), so:

-3(+2y)+4+17y

determiningTheFunctionDomain
-3(+2y)+17y+4

We add all the numbers together, and all the variables

17y-3(+2y)+4

We multiply parentheses

17y-6y+4

We add all the numbers together, and all the variables

11y+4

Back to the equation:

-(11y+4)



We get rid of parentheses

12y-11y-4+2=0

We add all the numbers together, and all the variables

y-2=0

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

y=2

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