2/3y-(1/(3+3y)=2

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Solution for 2/3y-(1/(3+3y)=2 equation: 



2/3y-(1/(3+3y)=2

We move all terms to the left:

2/3y-(1/(3+3y)-(2)=0



Domain of the equation: 3y!=0

y!=0/3

y!=0

y∈R





Domain of the equation: (3+3y)-2!=0

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

(3+3y)!=2

y∈R



We add all the numbers together, and all the variables

2/3y-(1/(3y+3)-2=0

We calculate fractions


(6y+4)/(9y^2+9y-2)+(-(1*3y)/(9y^2+9y-2)=0



We calculate terms in parentheses: +(-(1*3y)/(9y^2+9y-2), so:

-(1*3y)/(9y^2+9y-2

We add all the numbers together, and all the variables

-(+1*3y)/(9y^2+9y-2

We multiply all the terms by the denominator


-(+1*3y)

We get rid of parentheses

-1*3y

Wy multiply elements

-3y

Back to the equation:

+(-3y)



We get rid of parentheses

(6y+4)/(9y^2+9y-2)-3y=0

We multiply all the terms by the denominator


(6y+4)-3y*(9y^2+9y-2)=0

We multiply parentheses

-27y^3-27y^2+(6y+4)+6y=0

We get rid of parentheses

-27y^3-27y^2+6y+6y+4=0

We do not support eypression: y^3

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