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

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

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

We move all terms to the left:

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


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

y∈R



Domain of the equation: y+1)!=0

y∈R


We get rid of parentheses

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

We calculate fractions


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


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

(3(y+3)*y)/(y+2)*y

We multiply all the terms by the denominator


(3(y+3)*y)


We calculate terms in parentheses: +(3(y+3)*y), so:

3(y+3)*y

We multiply parentheses

3y^2+9y

Back to the equation:

+(3y^2+9y)


We get rid of parentheses

3y^2+9y

Back to the equation:

+(3y^2+9y)


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We multiply all the terms by the denominator


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